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Course 3 11-3 Solving Equations with Variables on Both Sides Learn to solve equations with variables on both sides of the equal sign.

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**Solving Equations with Variables on Both Sides**

Course 3 11-3 Solving Equations with Variables on Both Sides Some problems produce equations that have variables on both sides of the equal sign. Solving an equation with variables on both sides is similar to solving an equation with a variable on only one side. You can add or subtract a term containing a variable on both sides of an equation.

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**Additional Example 1A: Solving Equations with Variables on Both Sides**

Course 3 11-3 Solving Equations with Variables on Both Sides Additional Example 1A: Solving Equations with Variables on Both Sides Solve. 4x + 6 = x 4x + 6 = x – 4x – 4x Subtract 4x from both sides. 6 = –3x 6 –3 –3x = Divide both sides by –3. –2 = x

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**Solving Equations with Variables on Both Sides**

Course 3 11-3 Solving Equations with Variables on Both Sides Check your solution by substituting the value back into the original equation. For example, 4(-2) + 6 = -2 or -2 = -2. Helpful Hint

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**Additional Example 1B: Solving Equations with Variables on Both Sides**

Course 3 11-3 Solving Equations with Variables on Both Sides Additional Example 1B: Solving Equations with Variables on Both Sides Solve. 9b – 6 = 5b + 18 9b – 6 = 5b + 18 – 5b – 5b Subtract 5b from both sides. 4b – 6 = 18 Add 6 to both sides. 4b = 24 4b 4 24 = Divide both sides by 4. b = 6

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**Additional Example 1C: Solving Equations with Variables on Both Sides**

Course 3 11-3 Solving Equations with Variables on Both Sides Additional Example 1C: Solving Equations with Variables on Both Sides Solve. 9w + 3 = 9w + 7 9w + 3 = 9w + 7 – 9w – 9w Subtract 9w from both sides. 3 ≠ No solution. There is no number that can be substituted for the variable w to make the equation true.

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**Solving Equations with Variables on Both Sides**

Course 3 11-3 Solving Equations with Variables on Both Sides if the variables in an equation are eliminated and the resulting statement is false, the equation has no solution. Helpful Hint

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**Solving Equations with Variables on Both Sides**

Course 3 11-3 Solving Equations with Variables on Both Sides Check It Out: Example 1A Solve. 5x + 8 = x 5x + 8 = x – 5x – 5x Subtract 5x from both sides. 8 = –4x 8 –4 –4x = Divide both sides by –4. –2 = x

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**Solving Equations with Variables on Both Sides**

Course 3 11-3 Solving Equations with Variables on Both Sides Check It Out: Example 1B Solve. 3b – 2 = 2b + 12 3b – 2 = 2b + 12 – 2b – 2b Subtract 2b from both sides. b – 2 = Add 2 to both sides. b =

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**Solving Equations with Variables on Both Sides**

Course 3 11-3 Solving Equations with Variables on Both Sides Check It Out: Example 1C Solve. 3w + 1 = 3w + 8 3w + 1 = 3w + 8 – 3w – 3w Subtract 3w from both sides. 1 ≠ No solution. There is no number that can be substituted for the variable w to make the equation true.

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**Solving Equations with Variables on Both Sides**

Course 3 11-3 Solving Equations with Variables on Both Sides To solve multi-step equations with variables on both sides, first combine like terms and clear fractions. Then add or subtract variable terms to both sides so that the variable occurs on only one side of the equation. Then use properties of equality to isolate the variable.

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**Solving Equations with Variables on Both Sides**

Course 3 11-3 Solving Equations with Variables on Both Sides Additional Example 2A: Solving Multi-Step Equations with Variables on Both Sides Solve. 10z – 15 – 4z = 8 – 2z - 15 10z – 15 – 4z = 8 – 2z – 15 6z – 15 = –2z – 7 Combine like terms. + 2z z Add 2z to both sides. 8z – 15 = – 7 Add 15 to both sides. 8z = 8 8z 8 8 = Divide both sides by 8. z = 1

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**Solving Equations with Variables on Both Sides**

Course 3 11-3 Solving Equations with Variables on Both Sides Additional Example 2B: Solving Multi-Step Equations with Variables on Both Sides y 5 3y 5 3 4 7 10 – = y – y 5 3 4 3y 7 10 – = y – 20( ) = 20( ) y 5 3 4 3y 7 10 – y – Multiply by the LCD, 20. 20( ) + 20( ) – 20( )= 20(y) – 20( ) y 5 3y 3 4 7 10 4y + 12y – 15 = 20y – 14 16y – 15 = 20y – 14 Combine like terms.

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**Additional Example 2B Continued**

Course 3 11-3 Solving Equations with Variables on Both Sides Additional Example 2B Continued 16y – 15 = 20y – 14 – 16y – 16y Subtract 16y from both sides. –15 = 4y – 14 Add 14 to both sides. –1 = 4y –1 4 4y = Divide both sides by 4. –1 4 = y

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**Solving Equations with Variables on Both Sides**

Course 3 11-3 Solving Equations with Variables on Both Sides Check It Out: Example 2A Solve. 12z – 12 – 4z = 6 – 2z + 32 12z – 12 – 4z = 6 – 2z + 32 8z – 12 = –2z + 38 Combine like terms. + 2z z Add 2z to both sides. 10z – 12 = Add 12 to both sides. 10z = 50 10z 10 = Divide both sides by 10. z = 5

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**Solving Equations with Variables on Both Sides**

Course 3 11-3 Solving Equations with Variables on Both Sides Check It Out: Example 2B y 4 5y 6 3 4 6 8 = y – y 4 3 5y 6 8 = y – 24( ) = 24( ) y 4 3 5y 6 8 y – Multiply by the LCD, 24. 24( ) + 24( )+ 24( )= 24(y) – 24( ) y 4 5y 6 3 8 6y + 20y + 18 = 24y – 18 26y + 18 = 24y – 18 Combine like terms.

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**Check It Out: Example 2B Continued**

Course 3 11-3 Solving Equations with Variables on Both Sides Check It Out: Example 2B Continued 26y + 18 = 24y – 18 – 24y – 24y Subtract 24y from both sides. 2y + 18 = – 18 – – 18 Subtract 18 from both sides. 2y = –36 –36 2 2y = Divide both sides by 2. y = –18

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**Additional Example 4 Continued**

Course 3 11-3 Solving Equations with Variables on Both Sides Additional Example 4 Continued Now find the amount of money Jamie spends each morning. Choose one of the original expressions. d (0.25) = 1.75 Jamie spends $1.75 each morning. Find the number of doughnuts Jamie buys on Tuesday. Let n represent the number of doughnuts. 0.25n = 1.75 0.25n 0.25 1.75 = Divide both sides by 0.25. n = 7; Jamie bought 7 doughnuts on Tuesday.

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**Insert Lesson Title Here**

Course 3 11-3 Solving Equations with Variables on Both Sides Insert Lesson Title Here Lesson Quiz Solve. 1. 4x + 16 = 2x 2. 8x – 3 = x 3. 2(3x + 11) = 6x + 4 4. x = x – 9 5. An apple has about 30 calories more than an orange. Five oranges have about as many calories as 3 apples. How many calories are in each? x = –8 x = 6 no solution 1 4 1 2 x = 36 An orange has 45 calories. An apple has 75 calories.

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