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**2.5 Solving equations with variables on both sides**

You will solve equations with variables on both sides. Essential Question: How do you solve equations with variables on both sides?

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**Solve an equation with variables on both sides**

EXAMPLE 1 Solve an equation with variables on both sides Solve 7 – 8x = 4x – 17. 7 – 8x = 4x – 17 Write original equation. 7 – 8x + 8x = 4x – x Add 8x to each side. 7 = 12x – 17 Simplify each side. 24 = 12x Add 17 to each side. 2 = x Divide each side by 12. ANSWER The solution is 2. Check by substituting 2 for x in the original equation.

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**Solve an equation with variables on both sides**

EXAMPLE 1 Solve an equation with variables on both sides CHECK 7 – 8x = 4x – 17 Write original equation. 7 – 8(2) = 4(2) – 17 ? Substitute 2 for x. –9 = 4(2) – 17 ? Simplify left side. –9 = –9 Simplify right side. Solution checks.

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**Solve an equation with grouping symbols**

EXAMPLE 2 Solve an equation with grouping symbols 9x – 5 = 1 4 (16x + 60). Solve 1 4 (16x + 60) 9x – 5 = Write original equation. 9x – 5 = 4x + 15 Distributive property 5x – 5 = 15 Subtract 4x from each side. 5x = 20 Add 5 to each side. x = 4 Divide each side by 5.

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GUIDED PRACTICE for Examples 1 and 2 Solve the equation. Check your solution. – 3m = 5m 3 ANSWER

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GUIDED PRACTICE for Examples 1 and 2 Solve the equation. Check your solution. c = 4c – 7 ANSWER 9

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GUIDED PRACTICE for Examples 1 and 2 Solve the equation. Check your solution. – 3k = 17k – 2k ANSWER –8

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GUIDED PRACTICE for Examples 1 and 2 Solve the equation. Check your solution. z – 2 = 2(3z – 4) ANSWER 6

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GUIDED PRACTICE for Examples 1 and 2 Solve the equation. Check your solution. – 4a = 5(a – 3) ANSWER 2

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GUIDED PRACTICE for Examples 1 and 2 Solve the equation. Check your solution. 8y – 6 = 2 3 (6y + 15) 6. ANSWER 4

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EXAMPLE 3 Solve a real-world problem CAR SALES A car dealership sold 78 new cars and 67 used cars this year. The number of new cars sold by the dealership has been increasing by 6 cars each year. The number of used cars sold by the dealership has been decreasing by 4 cars each year. If these trends continue, in how many years will the number of new cars sold be twice the number of used cars sold?

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EXAMPLE 3 Solve a real-world problem SOLUTION Let x represent the number of years from now. So, 6x represents the increase in the number of new cars sold over x years and –4x represents the decrease in the number of used cars sold over x years. Write a verbal model. 67 78 + 6x = 2 ( (– 4 x) )

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**Solve a real-world problem**

EXAMPLE 3 Solve a real-world problem 78 + 6x = 2(67 – 4x) Write equation. 78 + 6x = 134 – 8x Distributive property x = 134 Add 8x to each side. 14x = 56 Subtract 78 from each side. x = 4 Divide each side by 14. ANSWER The number of new cars sold will be twice the number of used cars sold in 4 years.

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EXAMPLE 3 Solve a real-world problem CHECK You can use a table to check your answer. YEAR 1 2 3 4 Used car sold 67 63 59 55 51 New car sold 78 84 90 96 102

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GUIDED PRACTICE for Example 3 7. WHAT IF? In Example 3, suppose the car dealership sold 50 new cars this year instead of 78. In how many years will the number of new cars sold be twice the number of used cars sold? ANSWER 6 yr

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**Identify the number of solutions of an equation**

EXAMPLE 4 Identify the number of solutions of an equation Solve the equation, if possible. a. 3x = 3(x + 4) b. 2x + 10 = 2(x + 5) SOLUTION a. 3x = 3(x + 4) Original equation 3x = 3x + 12 Distributive property The equation 3x = 3x + 12 is not true because the number 3x cannot be equal to 12 more than itself. So, the equation has no solution. This can be demonstrated by continuing to solve the equation.

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**Identify the number of solutions of an equation**

EXAMPLE 4 Identify the number of solutions of an equation 3x – 3x = 3x + 12 – 3x Subtract 3x from each side. 0 = 12 Simplify. ANSWER The statement 0 = 12 is not true, so the equation has no solution.

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**Identify the number of solutions of an equation**

EXAMPLE 1 EXAMPLE 4 Identify the number of solutions of an equation b. 2x + 10 = 2(x + 5) Original equation 2x + 10 = 2x + 10 Distributive property ANSWER Notice that the statement 2x + 10 = 2x + 10 is true for all values of x. So, the equation is an identity, and the solution is all real numbers.

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GUIDED PRACTICE for Example 4 Solve the equation, if possible. z + 12 = 9(z + 3) ANSWER no solution

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GUIDED PRACTICE for Example 4 Solve the equation, if possible. w + 1 = 8w + 1 ANSWER

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GUIDED PRACTICE for Example 4 Solve the equation, if possible. (2a + 2) = 2(3a + 3) ANSWER identity

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Daily Homework Quiz Solve the equation, if possible. 3(3x + 6) = 9(x + 2) 1. ANSWER The equation is an identity. 7(h – 4) = 2h + 17 2. ANSWER 9 8 – 2w = 6w – 8 3. ANSWER 2

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Daily Homework Quiz 4g + 3 = 2(2g + 3) 4. ANSWER The equation has no solution. Bryson is looking for a repair service for general household maintenance. One service charges $75 to join the service and $30 per hour. Another service charge $45 per hour. After how many hours of service is the total cost for the two services the same? 5. ANSWER 5 h

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**You will solve equations with variables on both sides.**

Essential Question: How do you solve equations with variables on both sides? You will solve equations with variables on both sides. To solve equations with variables on both sides, collect the variable terms on one side and the constant terms on the other. • Some equations, called identities, are true for all values of the variable. Other equations have no solutions. To solve equations with variables on both sides, first simplify the expressions on each side of the equation by using the distributive property to remove grouping symbols and then combining like terms. Next, use properties of equality to collect variable terms on one side of the equation and constants on the other. Then solve the equation by isolating the variable.

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