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Pre-Algebra 10-3 Solving Equations with Variables on Both Sides 10-3 Solving Equations with Variables on Both Sides Pre-Algebra Warm Up Warm Up Problem of the Day Problem of the Day Lesson Presentation Lesson Presentation

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Pre-Algebra 10-3 Solving Equations with Variables on Both Sides Warm Up Solve. 1. 2x + 9x – 3x + 8 = – 4 = 6x + 22 – 4x 3. + = 5 4. – = 3 x = 1 x = -13 x = 34 Pre-Algebra 10-3 Solving Equations with Variables on Both Sides 2 7 x x9x 16 2x2x x = 50

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Pre-Algebra 10-3 Solving Equations with Variables on Both Sides Problem of the Day An equilateral triangle and a regular pentagon have the same perimeter. Each side of the pentagon is 3 inches shorter than each side of the triangle. What is the perimeter of the triangle? 22.5 in.

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Pre-Algebra 10-3 Solving Equations with Variables on Both Sides Learn to solve equations with variables on both sides of the equal sign.

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Pre-Algebra 10-3 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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Pre-Algebra 10-3 Solving Equations with Variables on Both Sides Solve. A. 4x + 6 = x Additional Example 1A: Solving Equations with Variables on Both Sides 4x + 6 = x – 4x 6 = –3x Subtract 4x from both sides. Divide both sides by –3. –2 = x 6 –3 –3x –3 =

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Pre-Algebra 10-3 Solving Equations with Variables on Both Sides Solve. B. 9b – 6 = 5b + 18 Additional Example 1B: Solving Equations with Variables on Both Sides 9b – 6 = 5b + 18 – 5b 4b – 6 = 18 4b4b = Subtract 5b from both sides. Divide both sides by 4. b = b = 24 Add 6 to both sides.

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Pre-Algebra 10-3 Solving Equations with Variables on Both Sides Solve. C. 9w + 3 = 5w w Additional Example 1C: Solving Equations with Variables on Both Sides 9w + 3 = 5w w 3 ≠ 7 9w + 3 = 9w + 7Combine like terms. – 9w Subtract 9w from both sides. No solution. There is no number that can be substituted for the variable w to make the equation true.

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Pre-Algebra 10-3 Solving Equations with Variables on Both Sides Solve. A. 5x + 8 = x Try This: Example 1A 5x + 8 = x – 5x 8 = –4x Subtract 4x from both sides. Divide both sides by –4. –2 = x 8 –4 –4x –4 =

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Pre-Algebra 10-3 Solving Equations with Variables on Both Sides Solve. B. 3b – 2 = 2b b – 2 = 2b + 12 – 2b b – 2 = 12 Subtract 2b from both sides. + 2 b = 14 Add 2 to both sides. Try This: Example 1B

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Pre-Algebra 10-3 Solving Equations with Variables on Both Sides Solve. C. 3w + 1 = 10w + 8 – 7w 3w + 1 = 10w + 8 – 7w 1 ≠ 8 3w + 1 = 3w + 8Combine like terms. – 3w Subtract 3w from both sides. No solution. There is no number that can be substituted for the variable w to make the equation true. Try This: Example 1C

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Pre-Algebra 10-3 Solving Equations with Variables on Both Sides To solve multistep 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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Pre-Algebra 10-3 Solving Equations with Variables on Both Sides Solve. A. 10z – 15 – 4z = 8 – 2z - 15 Additional Example 2A: Solving Multistep Equations with Variables on Both Sides 10z – 15 – 4z = 8 – 2z – z – 15 = –2z – 7Combine like terms. + 2z Add 2z to both sides. 8z – 15 = – 7 8z = 8 z = 1 Add 15 to both sides. Divide both sides by 8. 8z =

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Pre-Algebra 10-3 Solving Equations with Variables on Both Sides B. Additional Example 2B: Solving Multistep Equations with Variables on Both Sides Multiply by the LCD. 4y + 12y – 15 = 20y – 14 16y – 15 = 20y – 14Combine like terms. y5y y53y – = y – y5y y53y – = y – 20 ( ) = 20 ( ) y5y y53y – y – 20 ( ) + 20 ( ) – 20 ( ) = 20(y) – 20 ( ) y5y5 3y53y

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Pre-Algebra 10-3 Solving Equations with Variables on Both Sides Additional Example 2B Continued Add 14 to both sides. –15 = 4y – 14 –1 = 4y + 14 –1 4 4y4y 4 = Divide both sides by 4. 4 = y 16y – 15 = 20y – 14 – 16y Subtract 16y from both sides.

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Pre-Algebra 10-3 Solving Equations with Variables on Both Sides Solve. A. 12z – 12 – 4z = 6 – 2z + 32 Try This: Example 2A 12z – 12 – 4z = 6 – 2z z – 12 = –2z + 38Combine like terms. + 2z Add 2z to both sides. 10z – 12 = z = 50 z = 5 Add 12 to both sides. Divide both sides by z =

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Pre-Algebra 10-3 Solving Equations with Variables on Both Sides B. Multiply by the LCD. 6y + 20y + 18 = 24y – 18 26y + 18 = 24y – 18Combine like terms. y4y y65y = y – y4y y65y ( ) = 24 ( ) y4y y65y y – 24 ( ) + 24 ( ) + 24 ( ) = 24(y) – 24 ( ) y4y4 5y65y Try This: Example 2B

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Pre-Algebra 10-3 Solving Equations with Variables on Both Sides Subtract 18 from both sides. 2y + 18 = – 18 2y = –36 – 18 –36 2 2y2y 2 = Divide both sides by 2. y = –18 26y + 18 = 24y – 18 – 24y Subtract 24y from both sides. Try This: Example 2B Continued

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Pre-Algebra 10-3 Solving Equations with Variables on Both Sides Additional Example 3: Consumer Application Jamie spends the same amount of money each morning. On Sunday, he bought a newspaper for $1.25 and also bought two doughnuts. On Monday, he bought a newspaper for fifty cents and bought five doughnuts. On Tuesday, he spent the same amount of money and bought just doughnuts. How many doughnuts did he buy on Tuesday?

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Pre-Algebra 10-3 Solving Equations with Variables on Both Sides Additional Example 3 Continued First solve for the price of one doughnut d = d Let d represent the price of one doughnut. – 2d 1.25 = d Subtract 2d from both sides. – 0.50 Subtract 0.50 from both sides = 3d d3d 3 = Divide both sides by = d The price of one doughnut is $0.25.

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Pre-Algebra 10-3 Solving Equations with Variables on Both Sides Additional Example 3 Continued Now find the amount of money Jamie spends each morning d Choose one of the original expressions. Jamie spends $1.75 each morning (0.25) = n = Let n represent the number of doughnuts. Find the number of doughnuts Jamie buys on Tuesday. 0.25n = 1.75 n = 7; Jamie bought 7 doughnuts on Tuesday. Divide both sides by 0.25.

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Pre-Algebra 10-3 Solving Equations with Variables on Both Sides Try This: Example 3 Helene walks the same distance every day. On Tuesdays and Thursdays, she walks 2 laps on the track, and then walks 4 miles. On Mondays, Wednesdays, and Fridays, she walks 4 laps on the track and then walks 2 miles. On Saturdays, she just walks laps. How many laps does she walk on Saturdays?

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Pre-Algebra 10-3 Solving Equations with Variables on Both Sides Try This: Example 3 Continued First solve for distance around the track. 2x + 4 = 4x + 2 Let x represent the distance around the track. – 2x 4 = 2x + 2 Subtract 2x from both sides. – 2 Subtract 2 from both sides. 2 = 2x 2 2 2x2x 2 = Divide both sides by 2. 1 = x The track is 1 mile around.

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Pre-Algebra 10-3 Solving Equations with Variables on Both Sides Try This: Example 3 Continued Now find the total distance Helene walks each day. 2x + 4 Choose one of the original expressions. Helene walks 6 miles each day.2(1) + 4 = 6 Let n represent the number of 1-mile laps. Find the number of laps Helene walks on Saturdays. 1n = 6 Helene walks 6 laps on Saturdays. n = 6

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Pre-Algebra 10-3 Solving Equations with Variables on Both Sides Journal: Describe the kind of solution that has no solution.

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Pre-Algebra 10-3 Solving Equations with Variables on Both Sides Lesson Quiz Solve. 1. 4x + 16 = 2x 2. 8x – 3 = x 3. 2(3x + 11) = 6x 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 = 6 x = –8 no solution x = An orange has 45 calories. An apple has 75 calories.

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