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**Solve Multi-Step Equations**

Review of Chapter 2

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**Steps to Solve Equations with Variables on Both Sides**

1) Simplify each side Get rid of double Negatives Distribute Combine Like Terms 2) Move variables to same side “Smaller to the bigger” 3) Solve by using INVERSE Operations

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**Use Steps to Solve Equation: 2(x + 7) + 3 = 5x - 1**

1) distribute 2x + 17 = 5x - 1 2) Combine like terms -2x x 3) Get variables on same side – use inverse operation 17 = 3x - 1 4) Solve 2 step equation 18 = 3x x = 6 bkevil

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**Check - replace “x” with solution 2(x + 7) + 3 = 5x - 1**

2(6 + 7) + 3 =5(6) - 1 1) Replace X = 6 2(13) + 3 =5(6) - 1 2) Follow order of operations on both sides of equation = 29 = 29 3) Checks X = 6 is the solution to the equation bkevil

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**Use Steps to Solve Equation:**

-3x + 4 = 5x – 8 Get variables on same side of equation – use inverse operation (add 3x) +3x x 4 = 8x - 8 Solve 2 step equation 12 = 8x x = 3/2 bkevil

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**Use Steps to Solve Equation: 4(1 – 2x) = 4 – 6x**

Get rid of ( ) -- distribute +8x x Get variables on same side – use inverse operation (add 6x) 4 = 4 + 2x Solve 2 step equation 0 = 2x Undo by using inverse -2 + 4 undo 2nd undo 1st x = 0 bkevil

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**Use Steps to Solve Equation:**

9 + 5x = 5x + 9 Get variables on same side of equation – use inverse operation (subtract 5x) -5x -5x 9 = 9 When solving, if you get a TRUE STATEMENT, then that means that any real number works. Infinite Solutions bkevil

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**Use Steps to Solve Equation:**

6x – 1 = 6x – 8 Get variables on same side of equation – use inverse operation (subtract 6x) -6x x -1 = - 8 The variables zeroed out and remaining is a false statement where a number is equal to a different number, so there will be no number that will work in the equation. x = no solutions The solution is no real numbers or empty set bkevil

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**Review Steps to Solve Equations with Variables on Both Sides**

1) Simplify each side Get rid of double Negatives Distribute Combine Like Terms 2) Move variables to same side “Smaller to the bigger” 3) Solve by using INVERSE Operations bkevil

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Solve the equation. 1. 2m – 6 + 4m = 12 ANSWER 3 2. 6a – 5(a – 1) = 11 ANSWER 6

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Solve the equation. 3. A charter bus company charges $11.25 per ticket plus a handling charge of $.50 per ticket, and a $15 fee for booking the bus. If a group pays $297 to charter a bus, how many tickets did they buy? ANSWER 24 tickets

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Independent Practice Solve the equation. g – 2 + g = 16 ANSWER 2 b + 2(b – 4) = 47 11 ANSWER 3. –6 + 4(2c + 1) = –34 –4 ANSWER

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Independent Practice (x – 6) = 12 2 3 24 ANSWER 5. Joe drove 405 miles in 7 hours. He drove at a rate of 55 miles per hour during the first part of the trip and 60 miles per hour during the second part. How many hours did he drive at a rate of 55 miles per hour? 3 h ANSWER

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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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