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**Learn to solve equations with variables on both sides of the equal sign.**

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**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.

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**Additional Example 1A: Solving Equations with Variables on Both Sides**

Solve. 4x + 6 = x 4x + 6 = x – 4x – 4x Subtract 4x from both sides. 6 = –3x 6 –3 –3x = Divide both sides by –3. –2 = x

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Check your solution by substituting the value back into the original equation. For example, 4(-2) + 6 = -2 or -2 = -2. Helpful Hint

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**Additional Example 1B: Solving Equations with Variables on Both Sides**

Solve. 9b – 6 = 5b + 18 9b – 6 = 5b + 18 – 5b – 5b Subtract 5b from both sides. 4b – 6 = 18 Add 6 to both sides. 4b = 24 4b 4 24 = Divide both sides by 4. b = 6

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**Additional Example 1C: Solving Equations with Variables on Both Sides**

Solve. 9w + 3 = 9w + 7 9w + 3 = 9w + 7 – 9w – 9w Subtract 9w from both sides. 3 ≠ No solution. There is no number that can be substituted for the variable w to make the equation true.

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

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Check It Out: Example 1A Solve. 5x + 8 = x 5x + 8 = x – 5x – 5x Subtract 5x from both sides. 8 = –4x 8 –4 –4x = Divide both sides by –4. –2 = x

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Check It Out: Example 1B Solve. 3b – 2 = 2b + 12 3b – 2 = 2b + 12 – 2b – 2b Subtract 2b from both sides. b – 2 = Add 2 to both sides. b =

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

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To solve multi-step 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.

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**Additional Example 2: 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

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Check It Out: Example 2 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

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**Additional Example 3: Business Application**

Daisy’s Flowers sell 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.

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**Additional Example 3 Continued**

Let r represent the price of one rose. 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 services would cost the same when purchasing 9 roses.

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**Additional Example 5: Solving Literal Equations for a Variable**

The equation t = m + 10e gives the test score t for a student who answers m multiple-choice questions and e essay questions correctly. Solve this equation for e. t = m + 10e Locate e in the equation. t = m + 10e Since m is added to 10e, subtract m from both sides. –m –m t – m = e Since e is multiplied 10, divide both sides by 10. t – m = 10e 10 t – m = e 10

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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 x = 36 1 4 1 2 An orange has 45 calories. An apple has 75 calories.

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