# Preview Warm Up California Standards Lesson Presentation.

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Preview Warm Up California Standards Lesson Presentation

Warm Up Solve. 1. 2x + 9x – 3x + 8 = 16 2. –4 = 6x + 22 – 4x 3. + = 5
= 5 – = 3 x = 1 x = –13 2 7 x 7 7 1 x = 34 9x 16 2x 4 1 8 x = 50

Extension 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: AF1.1 California Standards

Additional Example 1A: Solving Equations with Variables on Both Sides
Solve. 4x + 6 = x 4x + 6 = x To collect the variable terms on one side, subtract 4x from both sides. – 4x – 4x 6 = –3x 6 –3 –3x = Since x is multiplied by -3, divide both sides by –3. –2 = x

You can always check your solution by substituting the value back into the original equation.

Additional Example 1B: Solving Equations with Variables on Both Sides
Solve. 9b – 6 = 5b + 18 9b – 6 = 5b + 18 To collect the variable terms on one side, subtract 5b from both sides. – 5b – 5b 4b – 6 = 18 Since 6 is subtracted from 4b, add 6 to both sides. 4b = 24 4b 4 24 = Since b is multiplied by 4, divide both sides by 4. b = 6

Additional Example 1C: Solving Equations with Variables on Both Sides
Solve. 9w + 3 = 9w + 7 9w + 3 = 9w + 7 – 9w – 9w To 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.

if the variables in an equation are eliminated and the resulting statement is false, the equation has no solution. Helpful Hint

Check It Out! Example 1A Solve. 5x + 8 = x 5x + 8 = x To collect the variable terms on one side, subtract 5x from both sides. – 5x – 5x 8 = –4x 8 –4 –4x = Since x is multiplied by –4, divide both sides by –4. –2 = x

Check It Out! Example 1B Solve. 3b – 2 = 2b + 12 3b – 2 = 2b + 12 To collect the variable terms on one side, subtract 2b from both sides. – 2b – 2b b – 2 = Since 2 is subtracted from b, add 2 to both sides. b =

Check It Out! Example 1C Solve. 3w + 1 = 3w + 8 3w + 1 = 3w + 8 – 3w – 3w To 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.

To 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.

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

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

16y – 15 = 20y – 14 – 16y – 16y Subtract 16y from both sides. –15 = 4y – 14 Add 14 to both sides. –1 = 4y –1 4 4y = Divide both sides by 4. –1 4 = y

Check It Out! Example 2A Solve. 12z – 12 – 4z = 6 – 2z + 32 12z – 12 – 4z = 6 – 2z + 32 8z – 12 = –2z + 38 Combine like terms. + 2z z Add 2z to both sides. 10z – 12 = Add 12 to both sides. 10z = 50 10z 10 = Divide both sides by 10. z = 5

Check It Out! Example 2B y 4 5y 6 3 4 6 8 = 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( )= 24(y) – 24( ) y 4 5y 6 3 8 6y + 20y + 18 = 24y – 18 26y + 18 = 24y – 18 Combine like terms.

Check It Out! Example 2B Continued
26y + 18 = 24y – 18 – 24y – 24y Subtract 24y from both sides. 2y + 18 = – 18 – – 18 Subtract 18 from both sides. 2y = –36 –36 2 2y = Divide both sides by 2. y = –18

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 rose Daisy’s: c = r Other: c = r

Now write an equation showing that the costs are equal. r = r Subtract 2.95r from both sides. – 2.95r – 2.95r = r Subtract from both sides. – – 26.00 = r 13.95 1.55 1.55r 1.55 = Divide both sides by 1.55. 9 = r The two bouquets from either florist would cost the same when purchasing 9 roses.

To find the cost, substitute 9 for r into either equation. Daisy’s: Other florist: c = r c = r c = (9) c = (9) c = c = c = 66.5 c = 66.5 The cost for a bouquet with 9 roses at either florist is \$66.50.

Check It Out! Example 3 Marla’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 balloon Marla’s: c = b Other: c = b

Check It Out! Example 3 Continued
Now write an equation showing that the costs are equal. b = b Subtract 2.25b from both sides. – 2.25b – 2.25b = b Subtract from both sides. – – 16.00 = b 6.00 0.75 0.75b 0.75 = Divide both sides by 0.75. 8 = b The two services would cost the same when purchasing a muffin basket with 8 balloons.

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