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Solving Equations Solving open sentences using inverse operations.

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Presentation on theme: "Solving Equations Solving open sentences using inverse operations."— Presentation transcript:

1 Solving Equations Solving open sentences using inverse operations.

2 What will happen if you add or subtract an equal amount of weight on both sides of the scales? Solving equations is like balancing scales, we must always keep the sides equal.

3 Solving equations is just a matter of undoing operations that are being done to the variable. In a simple equation, this may mean that we only have to undo one operation, as in the following example. Solve the following equation for x x + 3 = 8 x + 3 = 8 the variable is x x + 3 – 3 = 8 – 3 we are adding 3 to the variable, so to get rid of the added 3, we do the opposite--- subtract 3. x = 5 remember to do this to both sides of the equation.

4 In an equation which has more than one operation, we have to undo the operations in the correct order. We start with the operation the farthest away from the variable. Solve the following equation: 5x – 2 =13 5x – 2 = 13 The variable is x 5x – 2 + 2 = 13 + 2 We are multiplying it by 5, and subtracting 2 First, undo the subtracting by adding 2. 5x = 15 Then, undo the multiplication by dividing by 5. 5 5 x = 3

5 Suppose there are variables on both sides of the equation. The trick now, is to get the variables on the same side by adding or subtracting them. Solve for x in the equation 4x + 5 = x – 4 We have two terms with the variable, 4x and x. 4x + 5 = x - 4 We’ll move the variable 4x – x = x – x - 4 with the smaller 3x + 5 = -4 coefficient, x. To do this we have to look at the sign in front of the variable we’re moving. Since the is no sign we know it is +. To move this Variable we do the opposite, so we’’ll subtract x from both sides.

6 Now we proceed as before: 3x + 5 = -4 3x + 5 – 5 = -4 – 5 Subtract 5 from both sides. 3x = -9 3 3 Divide both sides by 3. x = -3

7 With any math there are new vocabulary words and rules we must follow. Let’s look at some of the new terms and rules before we move on.

8 Solving Equations by Adding or Subtracting Equation – a mathematical sentence that shows two expressions are equal. Solve – to find the answer or solution. Solution – the value that makes an equation true. Inverse operations – operations that “undo” each other; addition and subtraction, multiplication and division. Isolate the variable – to get the variable on one side of an equation or inequality by itself in order to solve. Open sentence – an equation that contains at least one variable.

9 Addition Property of Equality – states you can add the same amount to both sides of an equation and the equation remains true. 2 + 3 = 5 2 + 3 + 4 = 5 + 4 9 = 9 ? true Subtraction Property of Equality – states you can subtract the same amount from both sides of an equation and the equation remains true. 4 + 7 = 11 4 + 7 – 3 = 11 – 3 8 = 8 ? true

10 Addition and subtraction are inverse operations, which means they “undo” each other. To solve an equation, use inverse operations to isolate the variable, or to get the variable on one side of the equal sign by itself. x + 4 = 9 subtract 4 from both sides x + 4 – 4 = 9 – 4 Subtraction property of equality x + 0 = 5 Identity Property of Zero: x + 0 = 5 check: x + 4 = 9 5 + 4 = 9 substitute 5 for x 9 = 9 ? true

11 w – 3 = 9 Add 3 to both sides w – 3 + 3 = 9 + 3 Addition Property of Equality w + 0 = 12 Identity Property of Zero: w + 0 = w check: 12 – 3 = 9 Substitute 12 for w 9 = 9 ? True It is very important to write all the steps and check your answer each time you solve an equation.

12 Solving Equations by Multiplication or Division Multiplication Property of Equality – states you can multiply the same amount on both sides of an equation and the equation remains true. 4 · 3 = 12 2 · 4 · 3 = 12 · 2 24 = 24 Division Property of Equality – states you can divide the same amount on both sides of an equation and the equation remains true. 4 · 3 = 12 2 12 = 6 2

13 Multiplication and Division are inverse operations, which means they “undo” each other. To solve an equation, use inverse operations to isolate the variable, or get the variable on one side of the equal sign by itself. 7x = 35 Divide both sides by 7. 7x = 35 Division Property of Equality 7 7 1x = 5 1 · x = x X = 5 Check: 7x = 35 7 (5) = 35 substitute 5 for x 35 = 35 ? true

14 n ÷ 5 = 7 Multiply both sides by 5 n ÷ 5 · 5 = 7 · 5 Multiplication Property of Equality n = 35 check: n ÷ 5 = 7 35 ÷ 5 = 7 Substitute 35 for n 7 = 7 ? True It is very important to write all the steps and check your solution each time you solve an equation.

15 Sometimes it is necessary to solve equations by using 2 or more inverse operations. For instance, the equation 6x – 2 = 10. Always start with the operation that is the farthest away from the variable. 6x – 2 = 10 Add 2 to both sides first. 6x – 2 + 2 = 10 + 2 Addition Property of Equality 6x = 12 Divide both sides by 6 6 6 Division Property of Equality x = 2 Check: 6x – 2 = 10 6(2) – 2 = 10 Substitute 2 for x 12 – 2 = 10 10 = 10 ? true

16 Solving equations Get you pencil and calculator ready and try these problems. 1)m + 15 = 25 2)50 = h – 3 3)4d = 144 4)x/3 = 18 5)S + 2 = 13 6)4x + 3 =19 7)y/2 – 5 = 1 8)26 = 3f + 10f 9)4(2x -1) + 3x = 11 10)144 = 12h

17 Evaluating and solving simple expressions and equations, using order of operations, and using variables to solve real- world problems is the first step to becoming “good” at math. These skills lay the foundation for studies of algebra, geometry, and statistics.

18 Using Formulas Formulas are equations used to show relationships between quantities.

19 Using Formulas (equations) A formula or equation shows the relationship among certain quantities. The formula below can be used to find the miles per gallon achieved by a car. number of miles ÷ # of equals miles per driven gallons gas gallon m ÷ g = mpg You drove 294 miles before stopping to get gas. Your gas tank holds 12 gallons of gas. What gas mileage does your car get? 294 ÷ 12 = 24.5 mpg

20 The formula was distance traveled by a moving object is d = rt, where d represents distance in kilometers (km), r represents the rate in kilometers per hour (km/h), t represents the time in hours (h). Use the formula d = rt to find the indicated variables. 1)r = 60 km/h; t = 4 h; d = 2)d = 100 km; t = 2 h; r = 3)r = 55 km/h; d = 110 km; t = 4)r = 35 km/h; t = 3 h; d = 5)d = 210 km; t = 7 h; r = 6)r = 80 km/h; d = 320 km; t =

21 The formula I = prt is used to find the amount of simple interest on a given amount, where I is the interest; p is the principal amount; r is the rate of percent; and t is the time in years. Thurman borrowed $13,500 from his brother for 4 years at an annual percentage rate of 6%. How much interest will he pay if he pays the entire loan off at the end of the fourth year? What is the total amount he will repay?

22 Formulas are used everyday to solve problems, whether you are computing gas mileage for your car (mpg = m ÷ g) or changing degrees Celsius to Fahrenheit (F = 9/5C + 32), or even solving the Pythagorean Theorem (a² + b² = c²) to find distance.


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