3Extension of AF4.1 Solve two-step linear equations and inequalities in one variable over the rational numbers, interpret the solution or solutions in the context from which they arose, and verify the reasonableness of the results.Also covered: AF1.1CaliforniaStandards
4Additional Example 1A: Solving Equations with Variables on Both Sides Solve.4x + 6 = x4x + 6 = xTo collect the variable terms on one side, subtract 4x from both sides.– 4x – 4x6 = –3x6–3–3x=Since x is multiplied by -3, divide both sides by –3.–2 = x
5You can always check your solution by substituting the value back into the original equation. Helpful Hint
6Additional Example 1B: Solving Equations with Variables on Both Sides Solve.9b – 6 = 5b + 189b – 6 = 5b + 18To collect the variable terms on one side, subtract 5b from both sides.– 5b – 5b4b – 6 = 18Since 6 is subtracted from 4b, add 6 to both sides.4b = 244b424=Since b is multiplied by 4, divide both sides by 4.b = 6
7Additional Example 1C: Solving Equations with Variables on Both Sides Solve.9w + 3 = 9w + 79w + 3 = 9w + 7– 9w – 9wTo collect the variable terms on one side, subtract 9w from both sides.3 ≠There is no solution. There is no number that can be substituted for the variable w to make the equation true.
8if the variables in an equation are eliminated and the resulting statement is false, the equation has no solution.Helpful Hint
9Check It Out! Example 1ASolve.5x + 8 = x5x + 8 = xTo collect the variable terms on one side, subtract 5x from both sides.– 5x – 5x8 = –4x8–4–4x=Since x is multiplied by –4, divide both sides by –4.–2 = x
10Check It Out! Example 1BSolve.3b – 2 = 2b + 123b – 2 = 2b + 12To collect the variable terms on one side, subtract 2b from both sides.– 2b – 2bb – 2 =Since 2 is subtracted from b, add 2 to both sides.b =
11Check It Out! Example 1CSolve.3w + 1 = 3w + 83w + 1 = 3w + 8– 3w – 3wTo collect the variable terms on one side, subtract 3w from both sides.1 ≠No solution. There is no number that can be substituted for the variable w to make the equation true.
12To solve more complicated equations, you may need to first simplify by combining like terms or clearing fractions. Then add or subtract to collect variable terms on one side of the equation. Finally, use properties of equality to isolate the variable.
13Additional Example 2A: Solving Multi-Step Equations with Variables on Both Sides Solve.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
14Additional Example 2B: Solving Multi-Step Equations with Variables on Both Sides y53y534710– = y –y5343y710– = y –20( ) = 20( )y5343y710– y –Multiply by the LCD, 20.20( ) + 20( ) – 20( )= 20(y) – 20( )y53y347104y + 12y – 15 = 20y – 1416y – 15 = 20y – 14Combine like terms.
15Additional 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
16Check It Out! Example 2ASolve.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
17Check It Out! Example 2By45y63468= y –y435y68= y –24( ) = 24( )y435y68y –Multiply by the LCD, 24.24( ) + 24( )+ 24( )= 24(y) – 24( )y45y6386y + 20y + 18 = 24y – 1826y + 18 = 24y – 18Combine like terms.
18Check It Out! 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
19Additional Example 3: Business Application Daisy’s Flowers sells a rose bouquet for $39.95 plus $2.95 for every rose. A competing florist sells a similar bouquet for $26.00 plus $4.50 for every rose. Find the number of roses that would make both florists' bouquets cost the same price. What is the price?Write an equation for each service. Let c represent the total cost and r represent the number of roses.total cost is flat fee plus cost for each roseDaisy’s: c = rOther: c = r
20Additional Example 3 Continued Now write an equation showing that the costs are equal.r = rSubtract 2.95r from both sides.– 2.95r – 2.95r= rSubtract from both sides.– – 26.00= r13.951.551.55r 1.55=Divide both sides by 1.55.9 = rThe two bouquets from either florist would cost the same when purchasing 9 roses.
21Additional Example 3 Continued To find the cost, substitute 9 for r into either equation.Daisy’s:Other florist:c = rc = rc = (9)c = (9)c =c =c = 66.5c = 66.5The cost for a bouquet with 9 roses at either florist is $66.50.
22Check It Out! Example 3Marla’s Gift Baskets sells a muffin basket for $22.00 plus $2.25 for every balloon. A competing service sells a similar muffin basket for $16.00 plus $3.00 for every balloon. Find the number of balloons that would make both baskets cost the same price.Write an equation for each service. Let c represent the total cost and b represent the number of balloons.total cost is flat fee plus cost for each balloonMarla’s: c = bOther: c = b
23Check It Out! Example 3 Continued Now write an equation showing that the costs are equal.b = bSubtract 2.25b from both sides.– 2.25b – 2.25b= bSubtract from both sides.– – 16.00= b6.000.750.75b 0.75=Divide both sides by 0.75.8 = bThe two services would cost the same when purchasing a muffin basket with 8 balloons.
24Lesson 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.