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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. Solve. A. 4 x + 6 = x Additional Example 1A: Solving Equations with Variables on Both Sides 4 x + 6 = x
- 4 x โ 4 x 6 = โ 3 x Subtract 4x from both sides. Divide both sides by โ 3.
- 2 = x
- 3 x = โ 3 Solve. B. 9 b โ 6 = 5 b + 18 Additional Example 1B: Solving Equations with Variables on Both Sides 9 b โ 6 = 5 b + 18
- 5 b โ 5 b 4 b โ 6 = 18 4 b 4
Subtract 5b from both sides. Divide both sides by 4. b = 6
4 b = 24 Add 6 to both sides.
Solve. C. 9 w + 3 = 5 w + 7 + 4 w Additional Example 1C: Solving Equations with Variables on Both Sides 9 w + 3 = 5 w + 7 + 4 w 3 โ 7 9 w + 3 = 9 w + 7 Combine like terms.
- 9 w โ 9 w Subtract 9w from both sides. No solution. There is no number that can be substituted for the variable w to make the equation true. Solve. A. 5 x + 8 = x Try This: Example 1A 5 x + 8 = x
- 5 x โ 5 x 8 = โ 4 x Subtract 4x from both sides. Divide both sides by โ 4.
- 2 = x
- 4 x = โ 4 Solve. B. 3 b โ 2 = 2 b + 12 3 b โ 2 = 2 b + 12
- 2 b โ 2 b b โ 2 = 12 Subtract 2b from both sides.
- 2 + 2 b = 14 Add 2 to both sides. Try This: Example 1B Solve. C. 3 w + 1 = 10 w + 8 โ 7 w 3 w + 1 = 10 w + 8 โ 7 w 1 โ 8 3 w + 1 = 3 w + 8 Combine like terms.
- 3 w โ 3 w 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
Solve. A. 12 z โ 12 โ 4 z = 6 โ 2 z + 32 Try This: Example 2A 12 z โ 12 โ 4 z = 6 โ 2 z + 32
- 12 + 8 z โ 12 = โ 2 z + 38 Combine like terms.
- 2 z + 2 z (^) Add 2z to both sides. 10 z โ 12 = + 38 10 z = 50 z = 5 Add 12 to both sides. 1010 z (^) = (^5010) Divide both sides by 10.
B.
Multiply by the LCD. 6 y + 20 y + 18 = 24 y โ 18 26 y + 18 = 24 y โ 18 Combine like terms. y 4
5 y 6
+ + = y โ 8 y 4
5 y 6
y 4
5 y 6
24 ( ) + 24( )+ 24( )= 24( y ) โ 24 ( )
y 4 5 y 6
Try This: Example 2B Subtract 18 from both sides. 2 y + 18 = โ 18 2 y = โ 36
2 y 2 =^ Divide both sides by 2. y = โ 18 26 y + 18 = 24 y โ 18
- 24 y โ 24 y (^) Subtract 24y from both sides. Try This: Example 2B Continued (^) 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. 1.25 + 2 d = 0.50 + 5 d Let d represent the price of one doughnut.
- 2 d โ 2 d 1.25 = 0.50 + 3 d Subtract 2d from both sides.
- 0.50 โ 0. Subtract 0.50 from both sides. 0.75 = 3 d
3 3 d = (^3) Divide both sides by 3. 0.25 = d The price of one doughnut is $0.25. Additional Example 3 Continued Now find the amount of money Jamie spends each morning. 1.25 + 2 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. 0.25 n = 1. n = 7; Jamie bought 7 doughnuts on Tuesday. Divide both sides by 0.25. 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. 2 x + 4 = 4 x + 2 Let x represent the distance around the track.
- 2 x โ 2 x 4 = 2 x + 2 Subtract 2x from both sides.
- 2 โ 2 Subtract 2 from both sides. 2 = 2 x 2 2 2 x = 2 Divide both sides by 2. 1 = x The track is 1 mile around.
