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

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

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.

Learn to solve equations with variables on both sides of the equal sign.

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.

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

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

Additional Example 1C: Solving Equations with Variables on Both Sides Solve. C. 9w + 3 = 5w + 7 + 4w 9w + 3 = 5w + 7 + 4w 9w + 3 = 9w + 7 Combine like terms. – 9w – 9w Subtract 9w from both sides. 3 ≠ 7 No solution. There is no number that can be substituted for the variable w to make the equation true.

Try This: Example 1A Solve. A. 5x + 8 = x 5x + 8 = x – 5x – 5x Subtract 4x from both sides. 8 = –4x 8 –4 –4x = Divide both sides by –4. –2 = x

Try This: Example 1B Solve. B. 3b – 2 = 2b + 12 3b – 2 = 2b + 12 – 2b – 2b Subtract 2b from both sides. b – 2 = 12 + 2 + 2 Add 2 to both sides. b = 14

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

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.

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

Additional Example 2B: Solving Multistep Equations with Variables on Both Sides y 5 3y 5 3 4 7 10 B. + – = 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(y) – 20( ) y 5 3y 3 4 7 10 4y + 12y – 15 = 20y – 14 16y – 15 = 20y – 14 Combine like terms.

Additional Example 2B Continued 16y – 15 = 20y – 14 – 16y – 16y Subtract 16y from both sides. –15 = 4y – 14 + 14 + 14 Add 14 to both sides. –1 = 4y –1 4 4y = Divide both sides by 4. -1 4 = y

Try This: Example 2A Solve. A. 12z – 12 – 4z = 6 – 2z + 32 12z – 12 – 4z = 6 – 2z + 32 8z – 12 = –2z + 38 Combine like terms. + 2z + 2z Add 2z to both sides. 10z – 12 = + 38 + 12 +12 Add 12 to both sides. 10z = 50 10z 50 10 = Divide both sides by 10. z = 5

Try This: Example 2B y 4 5y 6 3 4 6 8 B. + + = 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(y) – 24( ) y 4 5y 6 3 8 6y + 20y + 18 = 24y – 18 26y + 18 = 24y – 18 Combine like terms.

Try This: Example 2B Continued 26y + 18 = 24y – 18 – 24y – 24y Subtract 24y from both sides. 2y + 18 = – 18 – 18 – 18 Subtract 18 from both sides. 2y = –36 –36 2 2y = Divide both sides by 2. y = –18

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?

Additional Example 3 Continued First solve for the price of one doughnut. Let d represent the price of one doughnut. 1.25 + 2d = 0.50 + 5d – 2d – 2d Subtract 2d from both sides. 1.25 = 0.50 + 3d Subtract 0.50 from both sides. – 0.50 – 0.50 0.75 = 3d 0.75 3 3d = Divide both sides by 3. The price of one doughnut is $0.25. 0.25 = d

Additional Example 3 Continued Now find the amount of money Jamie spends each morning. Choose one of the original expressions. 1.25 + 2d 1.25 + 2(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.

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?

Try This: Example 3 Continued First solve for distance around the track. Let x represent the distance around the track. 2x + 4 = 4x + 2 – 2x – 2x Subtract 2x from both sides. 4 = 2x + 2 – 2 – 2 Subtract 2 from both sides. 2 = 2x 2 2x = Divide both sides by 2. The track is 1 mile around. 1 = x

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

Journal: Describe the kind of solution that has no solution.

Lesson Quiz Solve. 1. 4x + 16 = 2x 2. 8x – 3 = 15 + 5x 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.