Presentation on theme: "Solving Equations with Variables on Both Sides"— Presentation transcript:
1 Solving Equations with Variables on Both Sides 10-3Warm UpProblem of the DayLesson PresentationPre-Algebra
2 Solving Equations with Variables on Both Sides Pre-Algebra10-3Solving Equations with Variables on Both SidesWarm UpSolve.1. 2x + 9x – 3x + 8 = 162. –4 = 6x + 22 – 4x= 5– = 3x = 1x = -1327x771x = 349x162x418x = 50
3 Problem of the DayAn 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.
4 Learn to solve equations with variables on both sides of the equal sign.
5 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.
6 Additional Example 1A: 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
7 Additional Example 1B: 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
8 Additional Example 1C: 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.
9 Try This: Example 1ASolve.A. 5x + 8 = x5x + 8 = x– 5x – 5xSubtract 4x from both sides.8 = –4x8–4–4x=Divide both sides by –4.–2 = x
10 Try This: Example 1BSolve.B. 3b – 2 = 2b + 123b – 2 = 2b + 12– 2b – 2bSubtract 2b from both sides.b – 2 =Add 2 to both sides.b =
11 Try This: Example 1CSolve.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.
12 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.
13 Additional Example 2A: 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
14 Additional Example 2B: 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.
15 Additional Example 2B Continued 16y – 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
16 Try This: Example 2ASolve.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
17 Try This: Example 2By45y63468B.= 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.
18 Try This: Example 2B Continued 26y + 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
19 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?
20 Additional Example 3 Continued First 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
21 Additional Example 3 Continued Now 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.
22 Try This: Example 3Helene 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?
23 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 – 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
24 Try This: Example 3 Continued Now 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.
25 Journal: Describe the kind of solution that has no solution.
26 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.