3 Learn to solve equations with variables on both sides of the equal sign.
4 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.
5 Example: Solving Equations with Variables on Both Sides Solve.A. 4x + 6 = x4x + 6 = x– 4x – 4xSubtract 4x from both sides.6 = –3x6–3–3x=Divide both sides by –3.–2 = x
6 Example: Solving Equations with Variables on Both Sides Solve.B. 9b – 6 = 5b + 189b – 6 = 5b + 18– 5b – 5bSubtract 5b from both sides.4b – 6 = 18Add 6 to both sides.4b = 244b424=Divide both sides by 4.b = 6
7 Example: Solving Equations with Variables on Both Sides Solve.C. 9w + 3 = 5w w9w + 3 = 5w w9w + 3 = 9w + 7Combine like terms.– 9w – 9wSubtract 9w from both sides.3 ≠No solution. There is no number that can be substituted for the variable w to make the equation true.
8 Try ThisSolve.A. 5x + 8 = x5x + 8 = x– 5x – 5xSubtract 4x from both sides.8 = –4x8–4–4x=Divide both sides by –4.–2 = x
9 Try ThisSolve.B. 3b – 2 = 2b + 123b – 2 = 2b + 12– 2b – 2bSubtract 2b from both sides.b – 2 =Add 2 to both sides.b =
10 Try ThisSolve.C. 3w + 1 = 10w + 8 – 7w3w + 1 = 10w + 8 – 7w3w + 1 = 3w + 8Combine like terms.– 3w – 3wSubtract 3w from both sides.1 ≠No solution. There is no number that can be substituted for the variable w to make the equation true.
11 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.
12 Example: Solving Multistep Equations with Variables on Both Sides Solve.A. 10z – 15 – 4z = 8 – 2z - 1510z – 15 – 4z = 8 – 2z – 156z – 15 = –2z – 7Combine like terms.+ 2z zAdd 2z to both sides.8z – 15 = – 7Add 15 to both sides.8z = 88z 88=Divide both sides by 8.z = 1
13 Example: Solving Multistep Equations with Variables on Both Sides y53y534710B.– = y –y5343y710– = y –20( ) = 20( )y5343y710– y –Multiply by the LCD.20( ) + 20( ) – 20( )= 20(y) – 20( )y53y347104y + 12y – 15 = 20y – 1416y – 15 = 20y – 14Combine like terms.
14 Example Continued16y – 15 = 20y – 14– 16y – 16ySubtract 16y from both sides.–15 = 4y – 14Add 14 to both sides.–1 = 4y–144y=Divide both sides by 4.-14= y
15 Try ThisSolve.A. 12z – 12 – 4z = 6 – 2z + 3212z – 12 – 4z = 6 – 2z + 328z – 12 = –2z + 38Combine like terms.+ 2z zAdd 2z to both sides.10z – 12 =Add 12 to both sides.10z = 5010z10=Divide both sides by 10.z = 5
16 Try Thisy45y63468B.= y –y435y68= y –24( ) = 24( )y435y68y –Multiply by the LCD.24( ) + 24( )+ 24( )= 24(y) – 24( )y45y6386y + 20y + 18 = 24y – 1826y + 18 = 24y – 18Combine like terms.
17 Try This Continued26y + 18 = 24y – 18– 24y – 24ySubtract 24y from both sides.2y + 18 = – 18– – 18Subtract 18 from both sides.2y = –36–3622y=Divide both sides by 2.y = –18
18 Example: 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?
19 Example ContinuedFirst solve for the price of one doughnut.Let d represent the price of one doughnut.d = d– 2d – 2dSubtract 2d from both sides.= dSubtract 0.50 from both sides.– – 0.50= d0.7533d=Divide both sides by 3.The price of one doughnut is $0.25.0.25 = d
20 Example ContinuedNow find the amount of money Jamie spends each morning.Choose one of the original expressions.d(0.25) = 1.75Jamie spends $1.75 each morning.Find the number of doughnuts Jamie buys on Tuesday.Let n represent the number of doughnuts.0.25n = 1.750.25n0.251.75=Divide both sides by 0.25.n = 7; Jamie bought 7 doughnuts on Tuesday.
21 Try ThisHelene 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?
22 Try This ContinuedFirst solve for distance around the track.Let x represent the distance around the track.2x + 4 = 4x + 2– 2x – 2xSubtract 2x from both sides.4 = 2x + 2– – 2Subtract 2 from both sides.2 = 2x22x=Divide both sides by 2.The track is 1 mile around.1 = x
23 Try This ContinuedNow find the total distance Helene walks each day.Choose one of the original expressions.2x + 42(1) + 4 = 6Helene walks 6 miles each day.Find the number of laps Helene walks on Saturdays.Let n represent the number of 1-mile laps.1n = 6n = 6Helene walks 6 laps on Saturdays.
24 Lesson QuizSolve.1. 4x + 16 = 2x2. 8x – 3 = x3. 2(3x + 11) = 6x + 44. x = x – 95. 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 = –8x = 6no solution1412x = 36An orange has 45 calories. An apple has 75 calories.