1.Simple algebra

# 1.Simple algebra ## Contact with students and Diagnostic Assessment ### Introduction Welcome to your first mathematics lesson for Form Three! This session is a little different from a regular lesson where you learn new topics. Today, we'll take some time to connect, understand your previous experiences with mathematics, and conduct a "Diagnostic Assessment" on simple algebra. A diagnostic assessment is like a check-up. It's not about giving you a grade, but about identifying what you already know well and where you might need a little extra support or review from your Form One and Form Two algebra topics. This will help your teacher understand how best to support your learning journey in Form Three mathematics. ### Key Concepts: Review of Fundamental Algebraic Concepts Before we begin the diagnostic assessment, let's quickly recap some essential algebraic concepts you've encountered previously. #### 1. What is Algebra? Algebra is a branch of mathematics that uses letters (called variables) and symbols to represent numbers and quantities in equations and formulas. It allows us to solve problems by finding unknown values. #### 2. Algebraic Terms * **Variable:** A letter that represents an unknown number or a quantity that can change. Common variables are \(x\), \(y\), \(a\), \(b\). * *Example:* In \(3x + 5\), \(x\) is the variable. * **Constant:** A fixed numerical value that does not change. * *Example:* In \(3x + 5\), \(5\) is the constant. * **Coefficient:** The number multiplied by a variable in an algebraic term. * *Example:* In \(3x + 5\), \(3\) is the coefficient of \(x\). * **Term:** A single number, a single variable, or a product of numbers and variables. Terms are separated by addition or subtraction signs. * *Example:* In \(4x^2 - 2y + 7\), \(4x^2\), \(-2y\), and \(7\) are terms. * **Like Terms:** Terms that have the exact same variables raised to the same powers. * *Example:* \(5x\) and \(2x\) are like terms. \(3y^2\) and \(-y^2\) are like terms. \(4x\) and \(4x^2\) are NOT like terms. #### 3. Simplifying Algebraic Expressions Simplifying an expression means combining like terms. * **Rule:** You can only add or subtract like terms. * *Example:* \(3x + 5y - x + 2y = (3x - x) + (5y + 2y) = 2x + 7y\) #### 4. Expanding Algebraic Expressions Expanding an expression involves multiplying a term outside parentheses by each term inside the parentheses (using the distributive law). * **Distributive Law:** \[ a(b + c) = ab + ac \] * *Example:* \(2(x + 3) = 2x + 6\) * *Example:* \(-3(y - 4) = -3y + 12\) (Remember: minus times minus is plus) #### 5. Solving Linear Equations Solving an equation means finding the value of the unknown variable that makes the equation true. * **Principle:** Whatever operation you perform on one side of the equation, you must perform the same operation on the other side to maintain balance. * **Steps (general):** 1. Simplify both sides of the equation (if necessary). 2. Collect all terms with the variable on one side and constants on the other side. 3. Isolate the variable by performing the inverse operation. * *Example:* Solve \(2x - 5 = 11\) * Add 5 to both sides: \(2x = 16\) * Divide by 2: \(x = 8\) #### 6. Substitution Substitution involves replacing variables in an algebraic expression or formula with given numerical values and then calculating the result. * *Example:* If \(x = 3\) and \(y = -2\), find the value of \(4x - y\). * \(4(3) - (-2) = 12 + 2 = 14\) ### Vocabulary * **Algebra:** A branch of mathematics dealing with symbols and the rules for manipulating these symbols. * **Variable:** A letter (e.g., \(x\), \(y\)) representing an unknown numerical value. * **Constant:** A fixed numerical value (e.g., \(5\), \(-7\)). * **Coefficient:** The numerical factor of a term containing a variable (e.g., \(3\) in \(3x\)). * **Term:** A single number, variable, or product of numbers and variables (e.g., \(5x^2\), \(-2y\), \(8\)). * **Expression:** A combination of terms connected by mathematical operations (e.g., \(3x + 5y - 7\)). It does NOT contain an equals sign. * **Equation:** A mathematical statement that two expressions are equal, containing an equals sign (e.g., \(2x + 3 = 7\)). * **Like Terms:** Terms that have the same variables raised to the same powers (e.g., \(4x\) and \(-2x\)). * **Simplify:** To combine like terms in an expression to make it shorter and easier to understand. * **Expand:** To multiply out parts of an algebraic expression, often removing parentheses using the distributive law. * **Factorize:** The reverse of expanding; to write an expression as a product of its factors. * **Solve:** To find the value(s) of the variable(s) that make an equation true. * **Substitute:** To replace a variable with a numerical value. ### Examples Here are some examples demonstrating the concepts we've just reviewed: ##### Example 1: Simplifying Expressions Simplify: \(5a + 7b - 2a - 3b\) * Group like terms: \((5a - 2a) + (7b - 3b)\) * Combine like terms: \(3a + 4b\) ##### Example 2: Expanding Expressions Expand: \(-4(2x - 5y)\) * Multiply \(-4\) by each term inside: \((-4)(2x) + (-4)(-5y)\) * Result: \(-8x + 20y\) ##### Example 3: Solving Linear Equations Solve for \(m\): \(3m + 8 = 20\) * Subtract 8 from both sides: \(3m = 20 - 8\) * \(3m = 12\) * Divide both sides by 3: \(m = \frac{12}{3}\) * Result: \(m = 4\) ##### Example 4: Substitution Given \(p = 5\) and \(q = -3\), evaluate \(2p - 3q\). * Substitute the values: \(2(5) - 3(-3)\) * Calculate: \(10 - (-9)\) * \(10 + 9\) * Result: \(19\) ### Exercises (Diagnostic Assessment) Please answer the following questions to the best of your ability. This is to help your teacher understand your current understanding of simple algebra. Show your working clearly. 1. **Simplifying Expressions:** a. Simplify: \(8x + 3y - 5x + y\) b. Simplify: \(4a^2 - 2a + 7 + a^2 + 3a - 1\) c. Simplify: \(3(2m + n) - 2(m - n)\) 2. **Expanding Expressions:** a. Expand: \(5(3p - 4)\) b. Expand: \(-2x(x + 7)\) c. Expand: \((y + 3)(y - 2)\) 3. **Solving Linear Equations:** a. Solve for \(k\): \(k + 9 = 23\) b. Solve for \(x\): \(4x - 7 = 13\) c. Solve for \(y\): \(2(y + 3) = 18\) d. Solve for \(m\): \(3m - 5 = m + 7\) 4. **Substitution:** a. If \(a = 4\) and \(b = -1\), find the value of \(2a + 5b\). b. If \(x = -2\) and \(y = 3\), find the value of \(x^2 - 3y\). 5. **Forming Algebraic Expressions:** a. A number \(n\) is increased by 5, then the result is multiplied by 3. Write an expression. b. The cost of a pen is \(x\) shillings and a pencil is \(y\) shillings. Write an expression for the total cost of 4 pens and 2 pencils. ### Assignment Continue practicing these concepts at home. This will help you reinforce your understanding. 1. **Simplify the following expressions:** a. \(10p - 7q + 3p - q\) b. \(5(x + 2y) - 3(x - y)\) 2. **Expand the following expressions:** a. \(7(2a - b)\) b. \(-x(5x + 3)\) 3. **Solve for the unknown variable in each equation:** a. \(m - 12 = 8\) b. \(6k + 1 = 25\) c. \(4(x - 2) = 20\) d. \(5y + 2 = 2y + 11\) 4. **Evaluate the expressions given the values:** a. If \(s = 6\) and \(t = -2\), find \(3s + 4t\). b. If \(a = -3\) and \(b = -1\), find \(a^2 - b^2\). ### Summary Today's session was an important first step in Form Three mathematics. We reviewed fundamental concepts of simple algebra, including understanding variables, constants, simplifying and expanding expressions, solving linear equations, and substitution. The diagnostic assessment you completed will help your teacher identify your strengths and areas where you might need more revision. Remember, this assessment is a tool to help you learn better, not a final judgment of your abilities. We will use this information to plan our upcoming lessons and ensure everyone builds a strong foundation in algebra. Keep practicing, and don't hesitate to ask questions! ## Building and solving equations from real life problems ### Introduction Welcome to this lesson on building and solving equations from real-life problems! In our daily lives, we often encounter situations that require us to find an unknown quantity. Algebra provides us with a powerful tool to solve these problems by translating them into mathematical equations. This lesson will teach you how to set up these equations from word problems and then solve them using algebraic methods. ### Key Concepts #### What is an Equation? An equation is a mathematical statement that shows two expressions are equal. It always contains an equals sign (\( = \)). For example, \( 2x + 3 = 7 \) is an equation because it states that the expression \( 2x + 3 \) has the same value as the number \( 7 \). #### Variables and Constants * **Variables:** These are symbols, usually letters (like \( x, y, a, b \)), that represent unknown numerical values. The value of a variable can change or vary depending on the problem. * **Constants:** These are fixed numerical values that do not change. For example, in the equation \( 2x + 3 = 7 \), the numbers \( 2, 3 \), and \( 7 \) are constants. #### Translating Word Problems into Algebraic Expressions Before forming an equation, you need to be able to translate parts of a word problem into algebraic expressions. Here are some common keywords and what they mean: * **Addition (+):** sum of, increased by, more than, total, added to * Example: "five more than a number" translates to \( x + 5 \) * **Subtraction (-):** difference of, decreased by, less than, subtracted from, reduced by * Example: "a number decreased by three" translates to \( x - 3 \) * **Multiplication (\( \times \) or implied):** product of, times, multiplied by, twice, thrice * Example: "twice a number" translates to \( 2x \) * **Division (\( \div \) or fraction bar):** quotient of, divided by, half of, third of * Example: "a number divided by four" translates to \( \frac{x}{4} \) * **Equals (=):** is, was, will be, results in, gives * Example: "a number increased by two is ten" translates to \( x + 2 = 10 \) #### Forming Equations from Real-Life Scenarios To successfully build an equation from a word problem, follow these steps: 1. **Understand the problem:** Read the problem carefully. Identify what information is given (knowns) and what you need to find (unknowns). 2. **Assign a variable:** Choose a letter (e.g., \( x \)) to represent the unknown quantity that you need to find. 3. **Translate the words into an equation:** Use the keywords and your assigned variable to write a mathematical equation that represents the relationships described in the problem. 4. **Solve the equation:** Use appropriate algebraic techniques to find the value of the variable. 5. **Check your answer:** Substitute the value you found back into the original word problem (or the equation) to ensure it makes sense and satisfies all conditions. This step is crucial to catch any errors. #### Steps to Solve Linear Equations Once you have formed an equation, you need to solve it. For linear equations (where the highest power of the variable is 1), follow these general steps: 1. **Simplify both sides (if necessary):** This involves combining like terms or using the distributive property if there are parentheses. 2. **Isolate the variable term:** Use inverse operations (addition or subtraction) to move all terms containing the variable to one side of the equation and all constant terms to the other side. 3. **Isolate the variable:** Use inverse operations (multiplication or division) to get the variable by itself with a coefficient of 1. * Remember: Whatever operation you perform on one side of the equation, you must perform the exact same operation on the other side to maintain equality. #### Checking Solutions To check if your solution is correct, substitute the value you found for the variable back into the original equation. If both sides of the equation are equal after substitution, then your solution is correct. ### Vocabulary * **Algebraic Expression:** A combination of variables, constants, and mathematical operation symbols (e.g., \( 5x + 7 \)). It does not contain an equals sign. * **Equation:** A mathematical statement that asserts the equality of two expressions (e.g., \( 5x + 7 = 22 \)). * **Variable:** A symbol (usually a letter) representing an unknown numerical value. * **Constant:** A fixed numerical value that does not change. * **Coefficient:** The numerical factor that multiplies a variable in an algebraic term (e.g., in \( 3y \), 3 is the coefficient). * **Term:** A single number, a single variable, or a product/quotient of numbers and variables within an expression or equation. * **Solution (of an equation):** The value(s) of the variable(s) that make the equation a true statement. ### Examples #### Example 1: Basic Translation and Solution **Problem:** When 7 is added to a certain number, the result is 15. Find the number. **Solution:** 1. **Understand:** We need to find an unknown number. We know that if we add 7 to it, we get 15. 2. **Assign a variable:** Let the number be \( x \). 3. **Translate to an equation:** "7 is added to a number" is \( x + 7 \). "the result is 15" means \( = 15 \). So, the equation is: \[ x + 7 = 15 \] 4. **Solve the equation:** Subtract 7 from both sides: \[ x + 7 - 7 = 15 - 7 \] \[ x = 8 \] 5. **Check your answer:** Substitute \( x = 8 \) back into the original problem: \( 8 + 7 = 15 \). This is true. So the number is 8. #### Example 2: Perimeter Problem **Problem:** The perimeter of a rectangular garden is 30 meters. If the length of the garden is 5 meters more than its width, find the dimensions (length and width) of the garden. **Solution:** 1. **Understand:** We know the perimeter and a relationship between length and width. We need to find the specific length and width. 2. **Assign a variable:** Let the width of the garden be \( w \) meters. Since the length is 5 meters more than the width, the length will be \( l = w + 5 \) meters. 3. **Translate to an equation:** The formula for the perimeter of a rectangle is \( P = 2(l + w) \). We are given \( P = 30 \). Substitute \( l = w + 5 \) into the perimeter formula: \[ 30 = 2((w + 5) + w) \] 4. **Solve the equation:** Simplify inside the parenthesis: \[ 30 = 2(2w + 5) \] Distribute the 2: \[ 30 = 4w + 10 \] Subtract 10 from both sides: \[ 30 - 10 = 4w + 10 - 10 \] \[ 20 = 4w \] Divide both sides by 4: \[ \frac{20}{4} = \frac{4w}{4} \] \[ w = 5 \] So, the width is 5 meters. Now find the length: \( l = w + 5 = 5 + 5 = 10 \) meters. 5. **Check your answer:** If width is 5m and length is 10m, then \( l = w + 5 \) (\( 10 = 5 + 5 \) is true). The perimeter is \( 2(10 + 5) = 2(15) = 30 \) meters. This matches the given information. The dimensions are 10m by 5m. #### Example 3: Age Problem **Problem:** A father is three times as old as his son. In 12 years, the sum of their ages will be 84. How old are they now? **Solution:** 1. **Understand:** We need to find their current ages. We have a relationship between their current ages and their ages in 12 years. 2. **Assign a variable:** Let the son's current age be \( s \) years. Since the father is three times as old, the father's current age is \( 3s \) years. In 12 years: Son's age: \( s + 12 \) Father's age: \( 3s + 12 \) 3. **Translate to an equation:** The sum of their ages in 12 years will be 84. \[ (s + 12) + (3s + 12) = 84 \] 4. **Solve the equation:** Combine like terms: \[ s + 3s + 12 + 12 = 84 \] \[ 4s + 24 = 84 \] Subtract 24 from both sides: \[ 4s + 24 - 24 = 84 - 24 \] \[ 4s = 60 \] Divide both sides by 4: \[ \frac{4s}{4} = \frac{60}{4} \] \[ s = 15 \] So, the son's current age is 15 years. The father's current age is \( 3s = 3 \times 15 = 45 \) years. 5. **Check your answer:** Son is 15, Father is 45. (45 is 3 times 15, true). In 12 years: Son will be \( 15 + 12 = 27 \). Father will be \( 45 + 12 = 57 \). Sum of their ages in 12 years: \( 27 + 57 = 84 \). This is true. So their current ages are 15 and 45. ### Exercises Solve the following problems by building and solving equations: 1. When 10 is subtracted from three times a certain number, the result is 29. Find the number. 2. The sum of three consecutive integers is 84. Find the three integers. 3. A shop sells apples for 30 shillings each and oranges for 40 shillings each. If a customer buys 5 more apples than oranges and spends a total of 550 shillings, how many apples and how many oranges did they buy? ### Assignment Continue practicing at home with these problems: 1. The perimeter of an isosceles triangle is 40 cm. The base of the triangle is 4 cm longer than each of the equal sides. Find the length of each side of the triangle. 2. Sarah is twice as old as her brother, Tom. Five years ago, the sum of their ages was 25. How old are Sarah and Tom now? 3. A rectangular field has a length that is 6 meters less than twice its width. If the perimeter of the field is 120 meters, find its length and width. ### Summary In this lesson, you have learned the crucial skill of translating real-life problems into algebraic equations. We covered: * Identifying variables and constants. * Using keywords to form algebraic expressions. * Following a systematic approach to build equations from word problems. * Applying inverse operations to solve linear equations. * The importance of checking your solutions to ensure they are correct and make sense in the context of the problem. Mastering these steps will enable you to tackle a wide range of problems using the power of algebra! ## Solving quadratic equations by factorization ### Introduction Welcome to this lesson on solving quadratic equations! In Form Two, you learned about linear equations, which involve variables raised to the power of one. Today, we're stepping up to quadratic equations, where the highest power of the variable is two. We will explore a powerful method called **factorization** to find the solutions to these equations. Understanding quadratic equations is crucial as they appear in many real-world situations, from projectile motion to economics. ### Key Concepts #### What is a Quadratic Equation? A quadratic equation is a polynomial equation of the second degree. This means the highest power of the unknown variable (usually \( x \)) in the equation is 2. #### Standard Form of a Quadratic Equation The standard form of a quadratic equation is: \[ ax^2 + bx + c = 0 \] Where: * \( a, b, \) and \( c \) are constants (numbers). * \( a \) cannot be equal to zero (\( a \neq 0 \)), otherwise, it would be a linear equation. * \( x \) is the unknown variable. **Examples:** * \( x^2 - 5x + 6 = 0 \) (Here, \( a=1, b=-5, c=6 \)) * \( 2y^2 + 7y = 0 \) (Here, \( a=2, b=7, c=0 \)) * \( z^2 - 9 = 0 \) (Here, \( a=1, b=0, c=-9 \)) #### What is Factorization? Factorization is the process of breaking down an algebraic expression into a product of simpler expressions (its factors). For example, the number 12 can be factorized into \( 3 \times 4 \). Similarly, a quadratic expression like \( x^2 + 5x + 6 \) can be factorized into \( (x+2)(x+3) \). #### The Zero Product Property (Null Factor Law) This is a fundamental property used in solving equations by factorization. It states that: If the product of two or more factors is zero, then at least one of the factors must be zero. In mathematical terms, if \( A \times B = 0 \), then either \( A = 0 \) or \( B = 0 \) (or both). #### Steps to Solve Quadratic Equations by Factorization Follow these steps to solve any quadratic equation using factorization: 1. **Write in Standard Form:** Rearrange the equation so that all terms are on one side, and the other side is zero. Ensure it's in the form \( ax^2 + bx + c = 0 \). 2. **Factorize the Quadratic Expression:** Factorize the expression \( ax^2 + bx + c \) into two linear factors. You might use techniques like finding common factors, grouping, or trinomial factorization (finding two numbers that multiply to \( ac \) and add to \( b \)). 3. **Apply the Zero Product Property:** Set each of the factorized expressions equal to zero. 4. **Solve the Linear Equations:** Solve each of the resulting linear equations for the variable \( x \). These values are the solutions (or roots) of the quadratic equation. 5. **Check Your Solutions (Optional but Recommended):** Substitute each solution back into the original quadratic equation to verify that it makes the equation true. ### Vocabulary * **Quadratic Equation:** An equation where the highest power of the variable is 2, typically in the form \( ax^2 + bx + c = 0 \). * **Factorization:** The process of breaking down an algebraic expression into a product of simpler expressions (its factors). * **Roots / Solutions:** The values of the variable that satisfy the equation (make it true). A quadratic equation typically has two roots. * **Zero Product Property (Null Factor Law):** If \( A \times B = 0 \), then \( A=0 \) or \( B=0 \). * **Standard Form:** The customary way to write an equation or expression, for quadratic equations it is \( ax^2 + bx + c = 0 \). * **Factor:** An expression that divides another expression evenly. ### Examples Let's work through some examples step-by-step: #### Example 1: Simple Trinomial Solve: \( x^2 - 5x + 6 = 0 \) **Solution:** 1. **Standard Form:** The equation is already in standard form. 2. **Factorize:** We need two numbers that multiply to \( 6 \) and add to \( -5 \). These numbers are \( -2 \) and \( -3 \). So, \( (x-2)(x-3) = 0 \) 3. **Zero Product Property:** Set each factor to zero. \( x-2 = 0 \) or \( x-3 = 0 \) 4. **Solve:** \( x = 2 \) or \( x = 3 \) Thus, the solutions are \( x = 2 \) and \( x = 3 \). #### Example 2: Common Factor Solve: \( 3x^2 + 9x = 0 \) **Solution:** 1. **Standard Form:** The equation is already in standard form (\( c=0 \)). 2. **Factorize:** Find the common factor, which is \( 3x \). \( 3x(x+3) = 0 \) 3. **Zero Product Property:** Set each factor to zero. \( 3x = 0 \) or \( x+3 = 0 \) 4. **Solve:** \( x = 0 \) or \( x = -3 \) Thus, the solutions are \( x = 0 \) and \( x = -3 \). #### Example 3: Trinomial with \( a \neq 1 \) Solve: \( 2x^2 + 7x + 3 = 0 \) **Solution:** 1. **Standard Form:** The equation is in standard form. 2. **Factorize:** For \( ax^2 + bx + c \), we look for two numbers that multiply to \( ac \) (which is \( 2 \times 3 = 6 \)) and add to \( b \) (which is \( 7 \)). These numbers are \( 1 \) and \( 6 \). Rewrite the middle term: \( 2x^2 + x + 6x + 3 = 0 \) Factor by grouping: \( x(2x+1) + 3(2x+1) = 0 \) \( (2x+1)(x+3) = 0 \) 3. **Zero Product Property:** Set each factor to zero. \( 2x+1 = 0 \) or \( x+3 = 0 \) 4. **Solve:** \( 2x = -1 \implies x = -rac{1}{2} \) or \( x = -3 \) Thus, the solutions are \( x = -rac{1}{2} \) and \( x = -3 \). #### Example 4: Difference of Two Squares Solve: \( x^2 - 16 = 0 \) **Solution:** 1. **Standard Form:** The equation is in standard form (\( b=0 \)). 2. **Factorize:** This is a difference of two squares, \( a^2 - b^2 = (a-b)(a+b) \). So, \( x^2 - 4^2 = (x-4)(x+4) = 0 \) 3. **Zero Product Property:** Set each factor to zero. \( x-4 = 0 \) or \( x+4 = 0 \) 4. **Solve:** \( x = 4 \) or \( x = -4 \) Thus, the solutions are \( x = 4 \) and \( x = -4 \). ### Exercises Solve the following quadratic equations by factorization. Show all your steps. 1. \( x^2 + 7x + 10 = 0 \) 2. \( y^2 - 3y - 18 = 0 \) 3. \( 4p^2 - 20p = 0 \) 4. \( 3m^2 + 11m + 6 = 0 \) 5. \( z^2 - 49 = 0 \) ### Assignment Continue practicing at home by solving these quadratic equations: 1. \( x^2 - 8x + 15 = 0 \) 2. \( 2x^2 + 5x - 3 = 0 \) 3. \( 5t^2 - 25t = 0 \) 4. \( (x+1)(x-5) = 16 \) (Hint: Expand and rearrange to standard form first) 5. \( 6x^2 - x - 2 = 0 \) ### Summary In this lesson, you have learned how to solve quadratic equations using the factorization method. Key takeaways include: * A quadratic equation is of the form \( ax^2 + bx + c = 0 \), where \( a \neq 0 \). * **Factorization** involves breaking the quadratic expression into a product of linear factors. * The **Zero Product Property** is crucial: if a product of factors is zero, then at least one factor must be zero. * The process involves converting the equation to standard form, factorizing, setting each factor to zero, and solving the resulting linear equations to find the roots. This method provides an efficient way to find the two solutions (roots) that satisfy a quadratic equation. Keep practicing to master this essential skill! ## Solving quadratic equations by of quadratic formula ### Introduction Welcome to this lesson on solving quadratic equations! In Form Two, you learned how to solve linear equations, which involve variables raised to the power of one. Now, we're stepping up to quadratic equations, which involve a variable raised to the power of two. While some quadratic equations can be solved by factoring, not all of them can. This is where the powerful **quadratic formula** comes in handy. It's a universal tool that can solve *any* quadratic equation, regardless of how complex it seems. ### Key Concepts #### What is a Quadratic Equation? A quadratic equation is an equation of the second degree, meaning it contains at least one term where the variable is squared. The **standard form** of a quadratic equation is: \[ ax^2 + bx + c = 0 \] Where: * \( x \) represents an unknown variable. * \( a \), \( b \), and \( c \) are known numbers, with \( a \neq 0 \). * \( a \) is the coefficient of the \( x^2 \) term. * \( b \) is the coefficient of the \( x \) term. * \( c \) is the constant term. **Examples of Quadratic Equations:** * \( 2x^2 + 3x - 5 = 0 \) (Here, \( a=2, b=3, c=-5 \)) * \( x^2 - 4 = 0 \) (Here, \( a=1, b=0, c=-4 \)) * \( -3x^2 + 7x = 0 \) (Here, \( a=-3, b=7, c=0 \)) #### Why Use the Quadratic Formula? While factoring is a useful method for solving some quadratic equations, it doesn't work for all of them, especially when the roots (solutions) are irrational numbers or when the equation is difficult to factor. The quadratic formula provides a direct and guaranteed way to find the solutions for any quadratic equation in standard form. #### The Quadratic Formula The solutions (also called roots) for a quadratic equation in the form \( ax^2 + bx + c = 0 \) are given by the quadratic formula: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] This formula will give you two possible values for \( x \), corresponding to the \( + \) and \( - \) signs before the square root. #### Steps to Use the Quadratic Formula 1. **Ensure the equation is in standard form:** \( ax^2 + bx + c = 0 \). If it's not, rearrange it first. 2. **Identify the values of \( a \), \( b \), and \( c \):** Be careful with the signs. 3. **Substitute the values** of \( a \), \( b \), and \( c \) into the quadratic formula. 4. **Simplify the expression** under the square root (the discriminant). 5. **Calculate the square root** if possible. 6. **Calculate the two possible values of \( x \)** (one using \( + \) and one using \( - \)). #### The Discriminant The expression inside the square root in the quadratic formula, \( b^2 - 4ac \), is called the **discriminant**. It tells us about the nature of the roots without actually solving the equation: * If \( b^2 - 4ac > 0 \): There are two distinct real roots. * If \( b^2 - 4ac = 0 \): There is exactly one real root (a repeated root). * If \( b^2 - 4ac < 0 \): There are no real roots (the roots are complex numbers, which you will learn about in higher grades). ### Vocabulary * **Quadratic Equation:** An equation where the highest power of the variable is 2, typically in the form \( ax^2 + bx + c = 0 \). * **Standard Form (of a quadratic equation):** The arrangement \( ax^2 + bx + c = 0 \). * **Coefficient:** A numerical or constant quantity placed before and multiplying the variable in an algebraic term (e.g., \( a \) and \( b \) in \( ax^2 + bx + c \)). * **Constant Term:** A term in an algebraic expression that has a value that is fixed or cannot change (e.g., \( c \) in \( ax^2 + bx + c \)). * **Discriminant:** The expression \( b^2 - 4ac \) from the quadratic formula, which determines the nature of the roots. * **Roots (or Solutions):** The values of the variable that satisfy the equation (i.e., make the equation true). For a quadratic equation, there can be up to two real roots. ### Examples **Example 1:** Solve \( x^2 + 5x + 6 = 0 \) using the quadratic formula. 1. **Identify \( a, b, c \):** \( a = 1 \) \( b = 5 \) \( c = 6 \) 2. **Substitute into the formula:** \[ x = \frac{-5 \pm \sqrt{5^2 - 4(1)(6)}}{2(1)} \] 3. **Simplify:** \[ x = \frac{-5 \pm \sqrt{25 - 24}}{2} \] \[ x = \frac{-5 \pm \sqrt{1}}{2} \] \[ x = \frac{-5 \pm 1}{2} \] 4. **Calculate the two roots:** \( x_1 = \frac{-5 + 1}{2} = \frac{-4}{2} = -2 \) \( x_2 = \frac{-5 - 1}{2} = \frac{-6}{2} = -3 \) The solutions are \( x = -2 \) and \( x = -3 \). **Example 2:** Solve \( 2x^2 - 3x - 7 = 0 \) using the quadratic formula. 1. **Identify \( a, b, c \):** \( a = 2 \) \( b = -3 \) \( c = -7 \) 2. **Substitute into the formula:** \[ x = \frac{-(-3) \pm \sqrt{(-3)^2 - 4(2)(-7)}}{2(2)} \] 3. **Simplify:** \[ x = \frac{3 \pm \sqrt{9 - (-56)}}{4} \] \[ x = \frac{3 \pm \sqrt{9 + 56}}{4} \] \[ x = \frac{3 \pm \sqrt{65}}{4} \] 4. **Calculate the two roots (leave in radical form or approximate to decimal places if specified):** \( x_1 = \frac{3 + \sqrt{65}}{4} \) \( x_2 = \frac{3 - \sqrt{65}}{4} \) (Approximately, \( \sqrt{65} \approx 8.06 \), so \( x_1 \approx \frac{3+8.06}{4} = \frac{11.06}{4} \approx 2.765 \) and \( x_2 \approx \frac{3-8.06}{4} = \frac{-5.06}{4} \approx -1.265 \)) **Example 3:** Solve \( x^2 - 6x + 9 = 0 \) using the quadratic formula. 1. **Identify \( a, b, c \):** \( a = 1 \) \( b = -6 \) \( c = 9 \) 2. **Substitute into the formula:** \[ x = \frac{-(-6) \pm \sqrt{(-6)^2 - 4(1)(9)}}{2(1)} \] 3. **Simplify:** \[ x = \frac{6 \pm \sqrt{36 - 36}}{2} \] \[ x = \frac{6 \pm \sqrt{0}}{2} \] \[ x = \frac{6 \pm 0}{2} \] 4. **Calculate the root:** \( x = \frac{6}{2} = 3 \) Here, there is one real root (a repeated root). ### Exercises Solve the following quadratic equations using the quadratic formula. Show all your steps. 1. \( x^2 + 7x + 10 = 0 \) 2. \( 3x^2 - 5x + 2 = 0 \) 3. \( x^2 - 2x - 1 = 0 \) (Leave your answer in simplest radical form) 4. \( 4x^2 + 12x + 9 = 0 \) 5. \( 2x^2 + x - 3 = 0 \) ### Assignment Continue practicing at home! Solve the following problems using the quadratic formula. Round your answers to two decimal places where necessary. 1. \( x^2 - 11x + 28 = 0 \) 2. \( 5x^2 + 8x - 4 = 0 \) 3. \( 2x^2 + 4x - 5 = 0 \) 4. \( -x^2 + 6x - 7 = 0 \) (Hint: You can multiply the entire equation by -1 to make 'a' positive) 5. \( 3x^2 - x = 8 \) (Hint: First, rearrange into standard form) ### Summary In this lesson, we learned about the quadratic formula, a powerful tool for solving any quadratic equation in the form \( ax^2 + bx + c = 0 \). We identified the coefficients \( a \), \( b \), and \( c \) and substituted them into the formula: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] We also explored the **discriminant**, \( b^2 - 4ac \), which helps us determine the number and type of roots (solutions) an equation will have. Mastering this formula is crucial for solving quadratic equations that cannot be easily factored, providing you with a reliable method for a wide range of problems. ## Algebraic expressions (Simplification) ### Introduction Welcome to this lesson on algebraic expressions! In mathematics, algebra is a branch where letters (called variables) are used to represent unknown numbers. This allows us to write general rules and solve problems where values can change. In this lesson, we will focus on simplifying algebraic expressions, which means making them easier and shorter to work with without changing their value. ### Key Concepts #### What is an Algebraic Expression? An algebraic expression is a combination of variables, numbers, and arithmetic operations (addition, subtraction, multiplication, division). Unlike an equation, an expression does not have an equals sign. * **Example:** \( 3x + 5y - 7 \) is an algebraic expression. #### Components of an Algebraic Expression Every algebraic expression is made up of different parts: * **Terms:** These are parts of an expression separated by addition or subtraction signs. Each term can be a constant, a variable, or a product of a constant and one or more variables. * In \( 3x + 5y - 7 \), the terms are \( 3x \), \( 5y \), and \( -7 \). * **Variables:** These are the letters that represent unknown values. Their values can change. * In \( 3x + 5y - 7 \), \( x \) and \( y \) are variables. * **Constants:** These are terms that have a fixed numerical value. They do not contain any variables. * In \( 3x + 5y - 7 \), \( -7 \) is the constant. * **Coefficients:** This is the numerical factor of a term that contains a variable. It is the number multiplied by the variable(s). * In \( 3x + 5y - 7 \), \( 3 \) is the coefficient of \( x \), and \( 5 \) is the coefficient of \( y \). #### Like and Unlike Terms Understanding like and unlike terms is crucial for simplification. * **Like Terms:** These are terms that have the exact same variables raised to the exact same powers. Only their numerical coefficients can be different. * **Examples of Like Terms:** * \( 5x \) and \( -2x \) * \( 3y^2 \) and \( 7y^2 \) * \( -ab \) and \( 6ab \) * **Unlike Terms:** These are terms that have different variables or the same variables raised to different powers. * **Examples of Unlike Terms:** * \( 4x \) and \( 4y \) (different variables) * \( 2x \) and \( 2x^2 \) (same variable, different powers) * \( 5ab \) and \( 5bc \) (different combinations of variables) #### Simplification: Combining Like Terms To simplify an algebraic expression, we combine its like terms. We can only add or subtract like terms. When combining like terms, we add or subtract their coefficients while keeping the variable part the same. * **Rule:** \( ax + bx = (a+b)x \) and \( ax - bx = (a-b)x \) * **Steps for Simplification:** 1. **Identify Like Terms:** Group terms that have the same variables and powers together. 2. **Combine Coefficients:** Add or subtract the numerical coefficients of the like terms. 3. **Rewrite the Expression:** Write the simplified terms, making sure unlike terms are not combined. * **Using Parentheses (Brackets):** * If a plus sign precedes the parentheses, remove the parentheses and keep the signs of the terms inside the same. * Example: \( a + (b - c) = a + b - c \) * If a minus sign precedes the parentheses, remove the parentheses and change the sign of each term inside. * Example: \( a - (b - c) = a - b + c \) #### Order of Operations (BODMAS/PEMDAS) When an expression involves multiple operations, remember the order of operations: * **B**rackets (Parentheses) * **O**rders (Exponents/Powers/Roots) * **D**ivision and **M**ultiplication (from left to right) * **A**ddition and **S**ubtraction (from left to right) ### Vocabulary * **Algebraic Expression:** A mathematical phrase containing numbers, variables, and operation symbols. * **Variable:** A letter (e.g., \( x, y, a \)) representing an unknown number. * **Constant:** A term in an expression that has a fixed numerical value (a number without a variable). * **Coefficient:** The numerical factor of a term that contains a variable (the number multiplied by the variable). * **Term:** Parts of an algebraic expression separated by addition or subtraction signs. * **Like Terms:** Terms that have the same variables raised to the same powers. * **Unlike Terms:** Terms that have different variables or different powers of the same variables. * **Simplification:** The process of reducing an algebraic expression to its simplest form by combining like terms. ### Examples #### Example 1: Basic Simplification Simplify: \( 7a + 3b - 2a + 5b \) **Solution:** 1. Identify like terms: * Like terms with \( a \): \( 7a \) and \( -2a \) * Like terms with \( b \): \( 3b \) and \( 5b \) 2. Group like terms: * \( (7a - 2a) + (3b + 5b) \) 3. Combine coefficients: * \( (7 - 2)a + (3 + 5)b \) * \( 5a + 8b \) So, \( 7a + 3b - 2a + 5b = 5a + 8b \). #### Example 2: Simplification with Powers Simplify: \( 4x^2 - 3x + 2x^2 + 5x - 7 \) **Solution:** 1. Identify like terms: * Like terms with \( x^2 \): \( 4x^2 \) and \( 2x^2 \) * Like terms with \( x \): \( -3x \) and \( 5x \) * Constant term: \( -7 \) 2. Group like terms: * \( (4x^2 + 2x^2) + (-3x + 5x) - 7 \) 3. Combine coefficients: * \( (4 + 2)x^2 + (-3 + 5)x - 7 \) * \( 6x^2 + 2x - 7 \) So, \( 4x^2 - 3x + 2x^2 + 5x - 7 = 6x^2 + 2x - 7 \). #### Example 3: Simplification with Parentheses Simplify: \( (3p + 2q) - (p - 4q) \) **Solution:** 1. Remove parentheses. Remember to change signs for terms inside the second parenthesis because of the minus sign outside it. * \( 3p + 2q - p + 4q \) 2. Identify and group like terms: * \( (3p - p) + (2q + 4q) \) 3. Combine coefficients: * \( (3 - 1)p + (2 + 4)q \) * \( 2p + 6q \) So, \( (3p + 2q) - (p - 4q) = 2p + 6q \). #### Example 4: More Complex Simplification Simplify: \( 5(2x - 3y) + 2(x + 4y) \) **Solution:** 1. Distribute the numbers outside the parentheses to each term inside: * \( 5 imes 2x - 5 imes 3y + 2 imes x + 2 imes 4y \) * \( 10x - 15y + 2x + 8y \) 2. Identify and group like terms: * \( (10x + 2x) + (-15y + 8y) \) 3. Combine coefficients: * \( (10 + 2)x + (-15 + 8)y \) * \( 12x - 7y \) So, \( 5(2x - 3y) + 2(x + 4y) = 12x - 7y \). ### Exercises Simplify the following algebraic expressions: 1. \( 8x + 5y - 3x + 2y \) 2. \( 6a - 9b + a - 4b \) 3. \( 10p^2 + 7p - 3p^2 - 2p + 5 \) 4. \( (4m - 2n) + (3m + 5n) \) 5. \( (11c + 6d) - (5c - 3d) \) 6. \( 3(4r + s) + 2(r - 5s) \) 7. \( 7xy + 4x - 2xy + 3x - y \) 8. \( (x^2 + 5x - 3) - (2x^2 - x + 4) \) ### Assignment Simplify these expressions and show all your steps: 1. \( 12a + 7b - 5c - 8a + 2b + 3c \) 2. \( (9x^2 - 4x + 6) + (2x^2 + 5x - 1) \) 3. \( 5(3y - 2) - 4(y + 7) \) 4. \( -3(2p + 5q) + 6(p - q) \) 5. \( 2(a^2 + 3ab) - 3(ab - b^2) - 4a^2 \) ### Summary In this lesson, we learned how to simplify algebraic expressions. The key steps involve identifying **like terms** (terms with the same variables raised to the same powers) and then **combining their coefficients** through addition or subtraction. We also reviewed how to handle parentheses by distributing the terms or changing signs when a minus sign precedes them. Remember that simplification makes expressions easier to understand and work with, and it's a fundamental skill in algebra. Keep practicing to master it! ## Expansions of algebraic expressions. # Expansions of algebraic expressions.\n\n## Introduction\nWelcome to this lesson on expansions of algebraic expressions! In mathematics, you often encounter expressions within brackets. To simplify or solve equations involving these expressions, we need to 'expand' them. Expansion is the process of multiplying out the terms in brackets. This lesson will teach you how to expand single and double brackets, including some special cases that can save you time.\n\n## Key Concepts\n### What is Algebraic Expansion?\n* Algebraic expansion is the process of removing brackets from an algebraic expression by multiplying each term inside the bracket by the term(s) outside the bracket.\n* The goal is to transform an expression with brackets into an equivalent expression without brackets.\n\n### Expanding a Single Bracket\n* To expand an expression with a single bracket, multiply the term outside the bracket by *every* term inside the bracket.\n* The distributive property of multiplication over addition is key here: \( a(b+c) = ab + ac \).\n* Remember to pay attention to signs (positive and negative).\n\n#### Example 1 (Single Bracket):\nExpand \( 3(x+2) \).\n\( 3(x+2) = 3 \times x + 3 \times 2 = 3x + 6 \)\n\n#### Example 2 (Single Bracket with Subtraction):\nExpand \( -2(y-5) \).\n\( -2(y-5) = -2 \times y + (-2) \times (-5) = -2y + 10 \)\n\n### Expanding Two Brackets\n* When expanding two brackets, you need to multiply *each* term in the first bracket by *each* term in the second bracket.\n* A common method is FOIL (First, Outer, Inner, Last) for binomials.\n * **F**irst: Multiply the first terms in each bracket.\n * **O**uter: Multiply the outer terms.\n * **I**nner: Multiply the inner terms.\n * **L**ast: Multiply the last terms in each bracket.\n* Then, combine any like terms.\n\n#### Example 3 (Two Brackets):\nExpand \( (x+3)(x+2) \).\n* **F**irst: \( x \times x = x^2 \)\n* **O**uter: \( x \times 2 = 2x \)\n* **I**nner: \( 3 \times x = 3x \)\n* **L**ast: \( 3 \times 2 = 6 \)\n* Combine: \( x^2 + 2x + 3x + 6 = x^2 + 5x + 6 \)\nSo, \( (x+3)(x+2) = x^2+5x+6 \).\n\n### Special Products\nThese are common expansions that are good to remember as they appear frequently.\n\n#### 1. Square of a Sum\n* The square of a sum means multiplying a binomial by itself.\n* Formula: \[ (a+b)^2 = a^2 + 2ab + b^2 \]\n\n#### Example 4 (Square of a Sum):\nExpand \( (y+4)^2 \).\nUsing the formula: \( a=y, b=4 \)\n\( (y+4)^2 = y^2 + 2(y)(4) + 4^2 = y^2 + 8y + 16 \)\n\n#### 2. Square of a Difference\n* Similar to the square of a sum, but with a subtraction.\n* Formula: \[ (a-b)^2 = a^2 - 2ab + b^2 \]\n\n#### Example 5 (Square of a Difference):\nExpand \( (2x-3)^2 \).\nUsing the formula: \( a=2x, b=3 \)\n\( (2x-3)^2 = (2x)^2 - 2(2x)(3) + 3^2 = 4x^2 - 12x + 9 \)\n\n#### 3. Difference of Two Squares\n* This occurs when you multiply two identical binomials, one with a plus sign and one with a minus sign.\n* Formula: \[ (a+b)(a-b) = a^2 - b^2 \]\n\n#### Example 6 (Difference of Two Squares):\nExpand \( (p+5)(p-5) \).\nUsing the formula: \( a=p, b=5 \)\n\( (p+5)(p-5) = p^2 - 5^2 = p^2 - 25 \)\n\n## Vocabulary\n* **Algebraic Expression:** A mathematical phrase that can contain numbers, variables, and operation signs.\n* **Expansion:** The process of removing brackets from an algebraic expression by multiplication.\n* **Term:** A single number, variable, or number and variable multiplied together in an expression.\n* **Like Terms:** Terms that have the same variables raised to the same powers. They can be combined.\n* **Binomial:** An algebraic expression with two terms.\n* **Distributive Property:** \( a(b+c) = ab+ac \).\n\n## Examples\n1. Expand \( 5(2m-n) \)\n \( 5(2m-n) = 5 \times 2m + 5 \times (-n) = 10m - 5n \)\n2. Expand \( (y-7)(y+1) \)\n \( (y-7)(y+1) = y^2 + y - 7y - 7 = y^2 - 6y - 7 \)\n3. Expand \( (3x+y)^2 \)\n \( (3x+y)^2 = (3x)^2 + 2(3x)(y) + y^2 = 9x^2 + 6xy + y^2 \)\n4. Expand \( (4p-3)(4p+3) \)\n \( (4p-3)(4p+3) = (4p)^2 - 3^2 = 16p^2 - 9 \)\n\n## Exercises\nExpand the following expressions:\n1. \( 4(3a+7) \)\n2. \( -6(2b-c) \)\n3. \( x(x^2+2x-1) \)\n4. \( (m+4)(m+5) \)\n5. \( (2y-1)(y+3) \)\n6. \( (a+9)^2 \)\n7. \( (3x-5)^2 \)\n8. \( (k+10)(k-10) \)\n9. \( (5p-2q)(5p+2q) \)\n10. \( (x-2)(x^2+2x+4) \) (Hint: This is a product of a binomial and a trinomial, apply distributive property thoroughly).\n\n## Assignment\n1. Expand and simplify: \( 7(3x-2) - 4(x+5) \)\n2. Expand and simplify: \( (2m+3)(m-4) - (m+1)^2 \)\n3. Expand: \( (3x-4y)^2 \)\n4. Expand: \( (x^2+y^2)(x^2-y^2) \)\n5. A rectangular garden has a length of \( (x+5) \) meters and a width of \( (x-2) \) meters. Find the area of the garden in terms of \( x \). (Area = length \( \times \) width)\n\n## Summary\n* Algebraic expansion is about removing brackets by multiplication.\n* For single brackets, multiply the term outside by every term inside: \( a(b+c) = ab+ac \).\n* For two brackets (binomials), multiply each term in the first bracket by each term in the second (FOIL method).\n* Remember the special products:\n * \[ (a+b)^2 = a^2 + 2ab + b^2 \]\n * \[ (a-b)^2 = a^2 - 2ab + b^2 \]\n * \[ (a+b)(a-b) = a^2 - b^2 \]\n* Always combine like terms after expansion to simplify the expression. ## Factorisation of algebraic expressions. ### Introduction Welcome to this lesson on factorisation of algebraic expressions! In previous lessons, you learned how to expand algebraic expressions by multiplying factors together. Factorisation is the reverse process: it's about breaking down an algebraic expression into a product of its factors. Think of it like taking a number, say 12, and writing it as a product of its prime factors: \( 2 \times 2 \times 3 \). In algebra, we do the same thing with expressions. Mastering factorisation is crucial as it helps us simplify expressions, solve equations, and work with fractions in algebra. ### Key Concepts #### What is Factorisation? Factorisation is the process of expressing a polynomial as a product of two or more polynomials. When an expression is factorised, it is written as a multiplication of its factors. For example: * Expanding: \( 3(x+2) = 3x + 6 \) * Factorising: \( 3x + 6 = 3(x+2) \) #### Methods of Factorisation ##### 1. Factorising by Taking Out Common Factors This is the most basic form of factorisation. We look for a common factor (a number, a variable, or both) that divides all terms in the expression. This common factor is then written outside a bracket, and the remaining terms form the expression inside the bracket. **Steps:** 1. Identify the highest common factor (HCF) of the numerical coefficients. 2. Identify the common variables and their lowest powers. 3. Write the HCF (numbers and variables) outside the bracket. 4. Divide each term in the original expression by the HCF and write the results inside the bracket. ##### 2. Factorising by Grouping This method is typically used for expressions with four terms. We group the terms into pairs and then factorise each pair by taking out common factors. If done correctly, a common binomial factor will emerge, which can then be factored out. **Steps:** 1. Group the four terms into two pairs. 2. Factorise each pair by taking out their respective common factors. 3. If the binomial expressions inside the brackets are identical, then that binomial is a common factor. 4. Factor out the common binomial expression. ##### 3. Factorising the Difference of Two Squares This method applies to expressions that are a subtraction of two perfect squares. The general form is \( a^2 - b^2 \). **Formula:** \[ a^2 - b^2 = (a-b)(a+b) \] **Steps:** 1. Identify if the expression has two terms and if both are perfect squares. 2. Determine the square root of the first term (this is 'a'). 3. Determine the square root of the second term (this is 'b'). 4. Apply the formula: \( (a-b)(a+b) \). ##### 4. Factorising Quadratic Expressions (\( ax^2 + bx + c \)) Quadratic expressions are trinomials (expressions with three terms) where the highest power of the variable is 2. The general form is \( ax^2 + bx + c \), where \( a \), \( b \), and \( c \) are constants and \( a \neq 0 \). ###### a. When \( a = 1 \) (i.e., \( x^2 + bx + c \)) **Steps:** 1. Look for two numbers that: * Multiply to give \( c \) (the constant term). * Add up to give \( b \) (the coefficient of the \( x \) term). 2. Let these two numbers be \( p \) and \( q \). The factorised form will be \( (x+p)(x+q) \). ###### b. When \( a \neq 1 \) (i.e., \( ax^2 + bx + c \)) **Steps (by splitting the middle term):** 1. Find the product \( ac \). 2. Look for two numbers that: * Multiply to give \( ac \). * Add up to give \( b \). 3. Rewrite the middle term (\( bx \)) as the sum of two terms using the two numbers found in step 2. 4. Factorise the resulting four-term expression by grouping. ### Vocabulary * **Factor:** A number or algebraic expression that divides another number or expression evenly without a remainder. When factorised, factors are multiplied together to get the original expression. * **Expression:** A collection of numbers, variables, and operation symbols (like \( + \), \( - \), \( \times \), \( \div \)). * **Algebraic Expression:** An expression that contains at least one variable. * **Term:** A single number or variable, or numbers and variables multiplied together. Terms are separated by \( + \) or \( - \) signs. * **Coefficient:** The numerical part of a term that multiplies a variable (e.g., 5 is the coefficient in \( 5x \)). * **Quadratic Expression:** An algebraic expression of the form \( ax^2 + bx + c \), where \( a \neq 0 \). * **Binomial:** An algebraic expression consisting of two terms (e.g., \( x+3 \)). * **Trinomial:** An algebraic expression consisting of three terms (e.g., \( x^2 + 5x + 6 \)). ### Examples #### 1. Common Factors **Example 1:** Factorise \( 12xy - 18x \) * HCF of 12 and 18 is 6. * Common variable is \( x \). * Common factor is \( 6x \). \[ 12xy - 18x = 6x(2y) - 6x(3) = 6x(2y - 3) \] #### 2. Grouping **Example 2:** Factorise \( 3ab - 6a + 5b - 10 \) \[\begin{aligned} 3ab - 6a + 5b - 10 &= (3ab - 6a) + (5b - 10) \\ &= 3a(b - 2) + 5(b - 2) \\ &= (b - 2)(3a + 5) \end{aligned}\] #### 3. Difference of Two Squares **Example 3:** Factorise \( 49y^2 - 16 \) * \( 49y^2 = (7y)^2 \) * \( 16 = (4)^2 \) \[ 49y^2 - 16 = (7y)^2 - (4)^2 = (7y - 4)(7y + 4) \] #### 4. Quadratic Expressions **Example 4 (a=1):** Factorise \( m^2 - 7m + 10 \) * We need two numbers that multiply to 10 and add to -7. * These numbers are -2 and -5 (since \( -2 \times -5 = 10 \) and \( -2 + (-5) = -7 \)). \[ m^2 - 7m + 10 = (m - 2)(m - 5) \] **Example 5 (a \( \neq \) 1):** Factorise \( 3p^2 - 10p + 8 \) * Product \( ac = 3 \times 8 = 24 \). * We need two numbers that multiply to 24 and add to -10. * These numbers are -4 and -6 (since \( -4 \times -6 = 24 \) and \( -4 + (-6) = -10 \)). * Split the middle term: \[\begin{aligned} 3p^2 - 10p + 8 &= 3p^2 - 4p - 6p + 8 \\ &= (3p^2 - 4p) - (6p - 8) \\ &= p(3p - 4) - 2(3p - 4) \\ &= (3p - 4)(p - 2) \end{aligned}\] ### Exercises Factorise the following expressions completely: 1. \( 5x + 15 \) 2. \( 2y^2 - 8y \) 3. \( ab + 4a + 3b + 12 \) 4. \( x^2 - 100 \) 5. \( 9x^2 - 25y^2 \) 6. \( x^2 + 8x + 15 \) 7. \( y^2 - 10y + 24 \) 8. \( 2x^2 + 5x + 2 \) 9. \( 6x^2 - 13x + 5 \) ### Assignment Continue practicing at home! Factorise the following expressions completely: 1. \( 7mn - 21m \) 2. \( 4p^2q + 8pq^2 - 12p^2q^2 \) 3. \( xy - 3x + 2y - 6 \) 4. \( (x+y)^2 - 1 \) 5. \( 16 - (a-b)^2 \) 6. \( x^2 + 11x + 28 \) 7. \( m^2 - 13m + 36 \) 8. \( 3x^2 + 14x - 5 \) 9. \( 4y^2 - 9y + 2 \) 10. \( 5a^2 + 13a + 6 \) ### Summary Today, we explored the important concept of factorisation in algebra, which is the reverse process of expansion. We learned several key methods to factorise algebraic expressions: * **Taking Out Common Factors:** Identifying and factoring out the highest common factor from all terms. * **Grouping:** Used for four-term expressions, pairing terms, and finding a common binomial factor. * **Difference of Two Squares:** Applying the formula \( a^2 - b^2 = (a-b)(a+b) \) for expressions involving the subtraction of two perfect squares. * **Quadratic Expressions:** Factorising trinomials of the form \( ax^2 + bx + c \) by finding two numbers that satisfy specific product and sum conditions, and then either directly forming binomial factors (when \( a=1 \)) or splitting the middle term and using grouping (when \( a \neq 1 \)). Remember that practice is key to mastering factorisation. Keep working through examples and exercises to build your confidence and skill! ## Factorising quadratic trinomials ### Introduction Welcome to this lesson on factorising quadratic trinomials! Factorising is a fundamental skill in algebra that allows us to break down complex expressions into simpler, multiplied components. It's like finding the building blocks of an algebraic expression. In this lesson, we will focus specifically on **quadratic trinomials**, which are expressions with three terms where the highest power of the variable is two. Mastering this will greatly help you in solving quadratic equations and simplifying algebraic fractions. ### Key Concepts #### What is a Quadratic Trinomial? A **quadratic trinomial** is an algebraic expression that can be written in the general form: \[ ax^2 + bx + c \] Where: * \( x \) is the variable. * \( a \), \( b \), and \( c \) are constant numbers (coefficients). * \( a \) cannot be zero, otherwise it wouldn't be quadratic (the \( x^2 \) term would disappear). * It has three terms: a term with \( x^2 \), a term with \( x \), and a constant term. #### Factorising Trinomials where \( a = 1 \) When the coefficient of \( x^2 \) is 1 (i.e., \( a=1 \)), the trinomial takes the form \( x^2 + bx + c \). To factorise such a trinomial, we need to find two numbers, let's call them \( p \) and \( q \), such that: 1. Their product is equal to \( c \) (the constant term): \( p \times q = c \) 2. Their sum is equal to \( b \) (the coefficient of \( x \)): \( p + q = b \) Once you find these two numbers, the factorised form of the trinomial will be: \[ (x + p)(x + q) \] #### Factorising Trinomials where \( a \neq 1 \) (The AC Method/Splitting the Middle Term) When the coefficient of \( x^2 \) is not 1 (i.e., \( a eq 1 \)), we use a method often called the 'AC Method' or 'Splitting the Middle Term'. Follow these steps: 1. **Find the product \( ac \):** Multiply the coefficient of \( x^2 \) (\( a \)) by the constant term (\( c \)). 2. **Find two numbers:** Identify two numbers, let's call them \( p \) and \( q \), such that: * Their product is equal to \( ac \): \( p \times q = ac \) * Their sum is equal to \( b \) (the coefficient of \( x \)): \( p + q = b \) 3. **Split the middle term:** Rewrite the original trinomial, replacing the middle term \( bx \) with \( px + qx \). * The expression becomes: \( ax^2 + px + qx + c \) 4. **Factor by grouping:** Group the first two terms and the last two terms, then factor out the greatest common factor (GCF) from each pair. * \( (ax^2 + px) + (qx + c) \) * This should result in a common binomial factor. 5. **Write the factors:** Factor out the common binomial factor to get the final factorised form. ### Vocabulary * **Factorising:** The process of breaking down an expression into a product of simpler expressions (factors). * **Quadratic Trinomial:** An algebraic expression with three terms, where the highest power of the variable is 2, generally in the form \( ax^2 + bx + c \). * **Coefficient:** A numerical or constant quantity placed before and multiplying the variable in an algebraic expression (e.g., \( a \) and \( b \) in \( ax^2 + bx + c \)). * **Constant Term:** A term in an algebraic expression that has a fixed value and does not contain any variables (e.g., \( c \) in \( ax^2 + bx + c \)). * **Binomial:** An algebraic expression consisting of two terms (e.g., \( x+3 \)). ### Examples #### Example 1: Factorising \( x^2 + 7x + 10 \) Here, \( a=1 \), \( b=7 \), and \( c=10 \). We need two numbers that multiply to 10 and add to 7. The numbers are 2 and 5. So, \( p=2 \) and \( q=5 \). \[ x^2 + 7x + 10 = (x + 2)(x + 5) \] #### Example 2: Factorising \( x^2 - 3x - 18 \) Here, \( a=1 \), \( b=-3 \), and \( c=-18 \). We need two numbers that multiply to -18 and add to -3. The numbers are 3 and -6. So, \( p=3 \) and \( q=-6 \). \[ x^2 - 3x - 18 = (x + 3)(x - 6) \] #### Example 3: Factorising \( 2x^2 + 11x + 5 \) Here, \( a=2 \), \( b=11 \), and \( c=5 \). 1. **Find \( ac \):** \( a \times c = 2 \times 5 = 10 \). 2. **Find two numbers:** We need two numbers that multiply to 10 and add to 11. The numbers are 1 and 10. So, \( p=1 \) and \( q=10 \). 3. **Split the middle term:** Rewrite \( 11x \) as \( 1x + 10x \). \( 2x^2 + x + 10x + 5 \) 4. **Factor by grouping:** \( (2x^2 + x) + (10x + 5) \) Factor out GCF from each group: \( x(2x + 1) + 5(2x + 1) \) 5. **Write the factors:** Notice the common binomial factor \( (2x+1) \). \( (x + 5)(2x + 1) \] #### Example 4: Factorising \( 3x^2 - 10x + 8 \) Here, \( a=3 \), \( b=-10 \), and \( c=8 \). 1. **Find \( ac \):** \( a \times c = 3 \times 8 = 24 \). 2. **Find two numbers:** We need two numbers that multiply to 24 and add to -10. The numbers are -4 and -6. So, \( p=-4 \) and \( q=-6 \). 3. **Split the middle term:** Rewrite \( -10x \) as \( -4x - 6x \). \( 3x^2 - 4x - 6x + 8 \) 4. **Factor by grouping:** \( (3x^2 - 4x) + (-6x + 8) \) Factor out GCF from each group: \( x(3x - 4) - 2(3x - 4) \) (Note: factor out -2 to get the common binomial) 5. **Write the factors:** \( (x - 2)(3x - 4) \] ### Exercises Factorise the following quadratic trinomials: 1. \( x^2 + 9x + 18 \) 2. \( y^2 - 5y + 6 \) 3. \( m^2 - m - 12 \) 4. \( x^2 + 4x - 21 \) 5. \( 2x^2 + 7x + 3 \) 6. \( 3y^2 - 14y + 8 \) 7. \( 5a^2 + 13a + 6 \) 8. \( 4p^2 - 4p - 3 \) ### Assignment Continue practicing factorising quadratic trinomials. Factorise the following expressions: 1. \( x^2 + 10x + 24 \) 2. \( b^2 - 11b + 28 \) 3. \( k^2 + 2k - 15 \) 4. \( x^2 - 7x - 30 \) 5. \( 6x^2 + 19x + 10 \) 6. \( 4y^2 - 17y + 4 \) 7. \( 2m^2 - m - 15 \) 8. \( 8c^2 + 2c - 1 \) ### Summary In this lesson, you learned how to factorise quadratic trinomials. We covered two main cases: 1. **When \( a=1 \) (form \( x^2 + bx + c \)):** Find two numbers that multiply to \( c \) and add to \( b \). The factors are \( (x+p)(x+q) \). 2. **When \( a \neq 1 \) (form \( ax^2 + bx + c \)):** Use the AC Method. Find two numbers that multiply to \( ac \) and add to \( b \). Split the middle term using these numbers, and then factor by grouping. This involves rewriting \( bx \) as \( px + qx \) and then factoring out common terms from pairs of terms. Factorising is a crucial skill that will be used in many other areas of mathematics. Keep practicing to master these techniques! ## Factorising difference of two squares ### Introduction Welcome to this lesson on factorising the difference of two squares! In algebra, factorising means breaking down an expression into a product of simpler expressions (its factors). You've already learned how to factorise common terms or quadratic expressions. Today, we'll focus on a special type of factorisation that is very useful and appears frequently in mathematics: the difference of two squares. This method allows us to quickly factorise expressions where one perfect square is subtracted from another perfect square. Mastering this will make solving algebraic problems much easier. ### Key Concepts #### What is Factorising? Factorising is the reverse process of expanding brackets. For example, if we expand \( (x+2)(x-2) \), we get \( x^2 - 4 \). Factorising \( x^2 - 4 \) would mean turning it back into \( (x+2)(x-2) \). #### What is a "Difference"? The term "difference" in mathematics means the result of a subtraction. So, when we talk about the "difference of two squares," we mean one square number or algebraic term being subtracted from another. #### What are "Two Squares"? "Two squares" refers to two terms that are perfect squares. A perfect square is a number or an algebraic expression that can be written as the square of another number or expression. For example: * Numbers: \( 4 \) is a perfect square because \( 4 = 2^2 \). \( 9 = 3^2 \). \( 25 = 5^2 \). * Algebraic terms: \( x^2 \) is a perfect square. \( 4y^2 = (2y)^2 \) is a perfect square. \( 9p^4 = (3p^2)^2 \) is a perfect square. #### The Formula for Difference of Two Squares This is the most important concept of this lesson. The formula states that: \[ a^2 - b^2 = (a - b)(a + b) \] Where: * \( a^2 \) is the first perfect square term. * \( b^2 \) is the second perfect square term. * \( a \) is the square root of the first term. * \( b \) is the square root of the second term. #### How to Apply the Formula To use the formula \( a^2 - b^2 = (a - b)(a + b) \), you need to follow these steps: 1. **Identify the two terms:** Make sure the expression consists of two terms separated by a subtraction sign. 2. **Check if they are perfect squares:** Determine if both terms are perfect squares. If they are, find their square roots. * For \( a^2 \), its square root is \( a \). * For \( b^2 \), its square root is \( b \). 3. **Apply the formula:** Once you have \( a \) and \( b \), substitute them into the formula \( (a - b)(a + b) \). #### Important Notes * The formula *only* works for a **difference** (subtraction). It does not work for the sum of two squares (e.g., \( a^2 + b^2 \) cannot be factorised using this method with real numbers). * Always look for a common factor first before applying this method. Sometimes, an expression might not look like a difference of two squares initially, but it becomes one after factorising out a common factor. ### Vocabulary * **Factorise:** To express a number or an algebraic expression as a product of its factors. * **Difference:** The result of a subtraction. * **Perfect Square:** A number or algebraic term that is the square of another number or expression (e.g., 9 is a perfect square because \( 3^2 = 9 \); \( x^2 \) is a perfect square because it is \( x imes x \)). * **Term:** A single number or variable, or numbers and variables multiplied together, separated by \( + \) or \( - \) signs in an expression. ### Examples Let's work through some examples to see how the formula is applied. #### Example 1: Factorise \( x^2 - 25 \) 1. **Identify terms:** The terms are \( x^2 \) and \( 25 \). 2. **Check for perfect squares:** * \( x^2 \) is a perfect square, and its square root is \( x \) (so \( a = x \)). * \( 25 \) is a perfect square, and its square root is \( 5 \) (so \( b = 5 \)). 3. **Apply formula:** Substitute \( a=x \) and \( b=5 \) into \( (a-b)(a+b) \). \[ x^2 - 25 = (x - 5)(x + 5) \] #### Example 2: Factorise \( 4y^2 - 81 \) 1. **Identify terms:** The terms are \( 4y^2 \) and \( 81 \). 2. **Check for perfect squares:** * \( 4y^2 \) is a perfect square, and its square root is \( 2y \) (so \( a = 2y \)). * \( 81 \) is a perfect square, and its square root is \( 9 \) (so \( b = 9 \)). 3. **Apply formula:** Substitute \( a=2y \) and \( b=9 \) into \( (a-b)(a+b) \). \[ 4y^2 - 81 = (2y - 9)(2y + 9) \] #### Example 3: Factorise \( 100 - p^2 \) 1. **Identify terms:** The terms are \( 100 \) and \( p^2 \). 2. **Check for perfect squares:** * \( 100 \) is a perfect square, and its square root is \( 10 \) (so \( a = 10 \)). * \( p^2 \) is a perfect square, and its square root is \( p \) (so \( b = p \)). 3. **Apply formula:** Substitute \( a=10 \) and \( b=p \) into \( (a-b)(a+b) \). \[ 100 - p^2 = (10 - p)(10 + p) \] #### Example 4: Factorise \( 3x^2 - 27 \) 1. **Identify terms:** The terms are \( 3x^2 \) and \( 27 \). 2. **Check for perfect squares:** Neither \( 3x^2 \) nor \( 27 \) are perfect squares. However, notice that both terms share a common factor, \( 3 \). 3. **Factor out common factor first:** \( 3x^2 - 27 = 3(x^2 - 9) \) 4. **Now, factorise the bracketed term:** \( x^2 - 9 \) is a difference of two squares! * \( x^2 \) is \( x^2 \), so \( a = x \). * \( 9 \) is \( 3^2 \), so \( b = 3 \). 5. **Apply formula to the bracketed term:** \( x^2 - 9 = (x - 3)(x + 3) \) 6. **Combine with the common factor:** \[ 3x^2 - 27 = 3(x - 3)(x + 3) \] ### Exercises Factorise the following expressions: 1. \( y^2 - 49 \) 2. \( 9m^2 - 1 \) 3. \( 36 - n^2 \) 4. \( 16a^2 - 25b^2 \) 5. \( 2p^2 - 18 \) 6. \( 4x^2 - 100 \) 7. \( x^4 - y^4 \) (Hint: this might require factorising twice!) ### Assignment Continue practicing at home. Factorise the following expressions fully: 1. \( c^2 - 64 \) 2. \( 49x^2 - 121 \) 3. \( 1 - 81k^2 \) 4. \( 144y^2 - x^2 \) 5. \( 50 - 2r^2 \) 6. \( 3a^2 - 75b^2 \) 7. \( (x+1)^2 - 9 \) 8. \( x^8 - 1 \) ### Summary Today, we learned about a powerful technique for factorising algebraic expressions called the **difference of two squares**. The key points to remember are: * The formula is \( a^2 - b^2 = (a - b)(a + b) \). * It applies only when one perfect square term is subtracted from another perfect square term. * Always identify the square roots of the two terms, \( a \) and \( b \), before applying the formula. * Remember to look for common factors first, as this can simplify the expression and reveal a difference of two squares. ## Solving simple linear equations ### Introduction Welcome to this lesson on solving simple linear equations! In mathematics, equations are like puzzles where you need to find the value of an unknown piece. Linear equations are a fundamental part of algebra, and learning how to solve them will help you tackle more complex problems in the future. This lesson will guide you through the step-by-step process of finding these unknown values. ### Key Concepts #### What is an Equation? An equation is a mathematical statement that shows two expressions are equal. It always contains an **equality sign** \( = \), meaning what is on the left side has the same value as what is on the right side. * Example: \( x + 5 = 10 \) #### What is a Linear Equation? A linear equation is an equation where the highest power of the variable is 1. This means you will not see terms like \( x^2 \), \( y^3 \), or \( \/\sqrt{x} \) in a linear equation. * General form of a linear equation in one variable: \( ax + b = c \), where \( a \), \( b \), and \( c \) are constants (numbers) and \( x \) is the variable. #### What Does "Solving" an Equation Mean? To solve an equation means to find the value of the unknown variable that makes the equation true. This value is called the **solution** of the equation. When you substitute the solution back into the original equation, both sides should be equal. #### Inverse Operations To solve equations, we use inverse (opposite) operations to isolate the variable. These operations "undo" each other. * Addition (\( + \)) and Subtraction (\( - \)) are inverse operations. * Multiplication (\( \times \)) and Division (\( \div \)) are inverse operations. #### The Golden Rule of Equations Whatever operation you perform on one side of the equation, you **must** perform the exact same operation on the other side. This ensures the equation remains balanced and true. #### Steps to Solve Simple Linear Equations ##### 1. One-Step Equations These equations require only one inverse operation to solve. * **Type 1: Addition/Subtraction** * If a number is added to the variable, subtract that number from both sides. * If a number is subtracted from the variable, add that number to both sides. * **Type 2: Multiplication/Division** * If the variable is multiplied by a number, divide both sides by that number. * If the variable is divided by a number, multiply both sides by that number. ##### 2. Two-Step Equations These equations require two inverse operations to solve. Generally, you should: 1. Undo any addition or subtraction first. 2. Undo any multiplication or division second. ##### 3. Equations with Variables on Both Sides When the variable appears on both sides of the equality sign: 1. Collect all terms with the variable on one side (usually the side where the coefficient will remain positive). 2. Collect all constant terms on the other side. 3. Solve the resulting two-step (or one-step) equation. ##### 4. Equations with Brackets If an equation contains brackets: 1. Use the **distributive property** to expand and remove the brackets first. * Example: \( a(b + c) = ab + ac \) 2. Simplify both sides of the equation. 3. Proceed to solve as a two-step equation or an equation with variables on both sides. ### Vocabulary * **Equation:** A mathematical statement showing that two expressions are equal. * **Linear Equation:** An equation where the highest power of the variable is 1. * **Variable:** An unknown value represented by a letter (e.g., \( x, y, z \)). * **Constant:** A fixed numerical value (e.g., \( 5, -2, rac{1}{2} \)). * **Coefficient:** The number multiplied by a variable (e.g., in \( 3x \), \( 3 \) is the coefficient). * **Solution:** The value of the variable that makes the equation true. * **Inverse Operations:** Operations that undo each other (e.g., addition and subtraction). * **Term:** A single number or variable, or numbers and variables multiplied together (e.g., \( 3x \), \( 5 \), \( -y \)). * **Distributive Property:** \( a(b + c) = ab + ac \). It tells us to multiply a term outside the bracket by each term inside the bracket. ### Examples #### Example 1: One-Step Equation (Addition/Subtraction) Solve \( x - 7 = 15 \). * **Solution:** To isolate \( x \), add \( 7 \) to both sides. \[ x - 7 + 7 = 15 + 7 \] \[ x = 22 \] #### Example 2: One-Step Equation (Multiplication/Division) Solve \( 5y = 30 \). * **Solution:** To isolate \( y \), divide both sides by \( 5 \). \[ \frac{5y}{5} = \frac{30}{5} \] \[ y = 6 \] #### Example 3: Two-Step Equation Solve \( 3x + 4 = 19 \). * **Solution:** 1. Subtract \( 4 \) from both sides: \[ 3x + 4 - 4 = 19 - 4 \] \[ 3x = 15 \] 2. Divide both sides by \( 3 \): \[ \frac{3x}{3} = \frac{15}{3} \] \[ x = 5 \] #### Example 4: Equation with Variable in Denominator Solve \( \frac{m}{2} - 3 = 5 \). * **Solution:** 1. Add \( 3 \) to both sides: \[ \frac{m}{2} - 3 + 3 = 5 + 3 \] \[ \frac{m}{2} = 8 \] 2. Multiply both sides by \( 2 \): \[ \frac{m}{2} \times 2 = 8 \times 2 \] \[ m = 16 \] #### Example 5: Equations with Variables on Both Sides Solve \( 5x - 8 = 2x + 7 \). * **Solution:** 1. Subtract \( 2x \) from both sides (to collect variables on the left): \[ 5x - 2x - 8 = 2x - 2x + 7 \] \[ 3x - 8 = 7 \] 2. Add \( 8 \) to both sides (to collect constants on the right): \[ 3x - 8 + 8 = 7 + 8 \] \[ 3x = 15 \] 3. Divide both sides by \( 3 \): \[ \frac{3x}{3} = \frac{15}{3} \] \[ x = 5 \] #### Example 6: Equations with Brackets Solve \( 4(z - 2) = 12 \). * **Solution:** 1. Apply the distributive property to remove brackets: \[ 4 \times z - 4 \times 2 = 12 \] \[ 4z - 8 = 12 \] 2. Add \( 8 \) to both sides: \[ 4z - 8 + 8 = 12 + 8 \] \[ 4z = 20 \] 3. Divide both sides by \( 4 \): \[ \frac{4z}{4} = \frac{20}{4} \] \[ z = 5 \] ### Exercises Solve the following linear equations: 1. \( x + 9 = 21 \) 2. \( y - 12 = 5 \) 3. \( 7m = 42 \) 4. \( \frac{p}{5} = 10 \) 5. \( 2a + 7 = 15 \) 6. \( \frac{b}{3} - 4 = 2 \) 7. \( 6k - 5 = 3k + 10 \) 8. \( 3(w + 4) = 21 \) ### Assignment Solve the following linear equations. Show all your working steps. 1. \( c + 15 = 40 \) 2. \( 18 - d = 7 \) 3. \( -4e = 24 \) 4. \( \frac{f}{-2} = 9 \) 5. \( 5g - 11 = 29 \) 6. \( \frac{h}{4} + 6 = 1 \) 7. \( 7(j - 3) = 14 \) 8. \( 4p + 1 = p + 13 \) 9. \( 2(q + 5) = 3q - 1 \) 10. \( 2(r - 1) + 3r = 18 \) ### Summary In this lesson, you have learned the essential steps to solve simple linear equations. Remember these key points: * An equation states that two expressions are equal, marked by the \( = \) sign. * Linear equations have variables raised only to the power of 1. * Solving an equation means finding the value of the unknown variable that makes the equation true. * Always use **inverse operations** (addition/subtraction, multiplication/division) to isolate the variable. * Whatever you do to one side of the equation, you **must** do to the other side to keep it balanced. * Follow the order: clear brackets (distributive property), then combine like terms, then isolate the variable using inverse operations (undo addition/subtraction, then undo multiplication/division). Keep practicing these steps, and solving linear equations will become second nature to you! ## Solving simultaneous linear equations by elimination method ### Introduction Welcome to this lesson on solving simultaneous linear equations! In mathematics, sometimes we encounter problems where we have two or more unknown values that are related by several equations. When we need to find the values that satisfy all these equations at the same time, we call them **simultaneous equations**. In Form Three, you will learn different methods to solve these equations. Today, we will focus on the **elimination method**, which is a powerful way to find those unknown values. ### Key Concepts #### What are Simultaneous Linear Equations? Simultaneous linear equations are a set of two or more linear equations (equations where the highest power of any variable is 1) that involve the same variables. The goal is to find a set of values for the variables that satisfy *all* the equations concurrently. For example, consider these two equations: \[ x + y = 5 \] \[ x - y = 1 \] Here, we are looking for a single value for \( x \) and a single value for \( y \) that make both equations true simultaneously. For instance, \( x=3 \) and \( y=2 \) would satisfy both equations: \( 3+2=5 \) and \( 3-2=1 \). #### What is the Elimination Method? The elimination method is a technique used to solve simultaneous linear equations. The main idea is to eliminate (get rid of) one of the variables by adding or subtracting the two equations. This leaves you with a single equation that has only one variable, which you can then solve easily. #### Steps for Solving by Elimination To solve simultaneous linear equations using the elimination method, follow these steps: