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You are reading an older version of this FlexBook® textbook: CK-12 Algebra I Concepts Go to the latest version.

3.2: One-Step Equations Transformed by Multiplication/Division

Difficulty Level: At Grade Created by: CK-12
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Practice One-Step Equations Transformed by Multiplication/Division
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What if you had an algebraic equation involving multiplication or division like -5x = 3 ? How could you solve it for the unknown variable x ? After completing this Concept, you'll be able to solve equations like this one.

Watch This

CK-12 Foundation: 0302S Solving Equations with Multiplication and Division (H264)

Guidance

Suppose you are selling pizza for $1.50 a slice and you can get eight slices out of a single pizza. How much money do you get for a single pizza? It shouldn’t take you long to figure out that you get 8 \times \$1.50 = \$12.00 . You solved this problem by multiplying. Here’s how to do the same thing algebraically, using x to stand for the cost in dollars of the whole pizza.

Example A

Solve  \frac{1}{8} \cdot x = 1.5 .

Our x is being multiplied by one-eighth. To cancel that out and get x by itself, we have to multiply by the reciprocal, 8. Don’t forget to multiply both sides of the equation.

8 \left ( \frac{1}{8} \cdot x \right ) &= 8(1.5)\\x &= 12

Example B

Solve 0.25x = 5.25 .

0.25 is the decimal equivalent of one fourth, so to cancel out the 0.25 factor we would multiply by 4.

4(0.25x) &= 4(5.25)\\x &= 21

Solving by division is another way to isolate x . Suppose you buy five identical candy bars, and you are charged $3.25. How much did each candy bar cost? You might just divide $3.25 by 5, but let’s see how this problem looks in algebra.

Example C

Solve 5x = 3.25 .

To cancel the 5, we divide both sides by 5.

 \frac{5x}{5} &= \frac{3.25}{5}\\x &= 0.65

Example D

Solve 1.375x = 1.2 .

Divide by 1.375

 x &= \frac{1.2}{1.375}\\x &= 0.8 \overline{72}

Notice the bar above the final two decimals; it means that those digits recur, or repeat. The full answer is 0.872727272727272....

To see more examples of one - and two-step equation solving, watch the Khan Academy video series starting at http://www.youtube.com/watch?v=bAerID24QJ0 .

Watch this video for help with the Examples above.

CK-12 Foundation: Solving Equations with Multiplication and Division

Vocabulary

  • An equation in which each term is either a constant or the product of a constant and a single variable is a linear equation .
  • We can add, subtract, multiply, or divide both sides of an equation by the same value and still have an equivalent equation .
  • To solve an equation, isolate the unknown variable on one side of the equation by applying one or more arithmetic operations to both sides.

Guided Practice

Solve:

a)  \frac{9x}{5} = 5 .

b) 7x = \frac{5}{11} .

Solutions:

a) \frac{9x}{5} is equivalent to  \frac{9}{5} \cdot x , so to cancel out that  \frac{9}{5} , we multiply by the reciprocal,  \frac{5}{9} .

 \frac{5}{9} \left ( \frac{9x}{5} \right ) &= \frac{5}{9}(5)\\x &= \frac{25}{9}

b) Divide both sides by 7.

x &= \frac{5}{11.7}\\x &= \frac{5}{77}

Practice

For 1-5, solve the following equations for x .

  1. 7x = 21
  2. 4x = 1
  3. \frac{5x}{12} = \frac{2}{3}
  4. 0.01x = 11
  5. \frac{-2x}{9} = \frac{10}{3}

For 6-10, solve the following equations for the unknown variable.

  1. 21s = 3
  2. -7a = -5
  3. \frac{7f}{11} = \frac{7}{11}
  4. 6r = \frac{3}{8}
  5. \frac{9b}{16} = \frac{3}{8}

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Difficulty Level:

At Grade

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Date Created:

Aug 13, 2012

Last Modified:

Feb 19, 2015
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