<img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" />
Dismiss
Skip Navigation
Our Terms of Use (click here to view) have changed. By continuing to use this site, you are agreeing to our new Terms of Use.

2.6: Equations with Decimals, Fractions and Parentheses

Difficulty Level: At Grade Created by: CK-12
Turn In

Pens are $9 per dozen and pencils are $6 per dozen. Janet need to buy a half dozen of each for school. How much is the total cost of her purchase?

Watch This

Khan Academy Solving Equations with the Distributive Property

Guidance

Recall that the distributive property is a mathematical way of grouping terms. It states that the product of a number and a sum is equal to the sum of the individual products of the number and the addends. Here, you will use the distributive property with linear equations and fractions. The same rules apply. You have to multiply what is on the outside of the parentheses by what is on the inside of the parentheses and this is your first step in solving equations with one variable where there are parentheses. After you remove parentheses, you then solve the equation by combining like terms, moving constants to one side of the equals sign, variables to the other side of the equals sign, and finally isolating the variable to find the solution.

Example A

Solve: \begin{align*}\frac{2}{5}(d+4) = 6\end{align*}.

\begin{align*}\frac{2}{5}(d+4) &= 6\\ \frac{2}{5}d+\frac{8}{5} &= 6 && (\text{Apply the distributive property to remove the brackets})\end{align*}

Find the LCD for 5, 5, and 1. Since it is 5, multiply the last number by \begin{align*}\frac{5}{5}\end{align*}, to get the same denominator.

\begin{align*}\frac{2}{5}d+\frac{8}{5} &= \left({\color{red}\frac{5}{5}}\right)6\\ \frac{2}{5}d+\frac{8}{5} &= \frac{30}{5} && (\text{Simplify})\end{align*}

Since all of the denominators are the same, the equation becomes:

\begin{align*}2d+8 &= 30\\ 2d+8 {\color{red}-8} &= 30 {\color{red}-8} && (\text{Subtract} \ 8 \ \text{from both sides of the equal sign to isolate the variable})\\ 2d &= 22 && (\text{Simplify})\\ \frac{2d}{{\color{red}2}}&=\frac{22}{{\color{red}2}} && (\text{Divide by} \ 2 \ \text{to solve for the variable})\\ d &= 11\end{align*}

Therefore \begin{align*}d = 11\end{align*}.

\begin{align*}\text{Check:} &\\ \frac{2}{5}(d+4) &= 6\\ \frac{2}{5} ({\color{red}11} +4) &=6\\ \frac{2}{5}(15) &= 6\\ \frac{30}{5} &= 6\\ 6 &= 6 \ \ \end{align*}

Example B

Solve: \begin{align*}\frac{1}{4}(3x+7) =2\end{align*}.

\begin{align*}\frac{1}{4}(3x+7) &= -2\\ \frac{3}{4}x+\frac{7}{4} &= -2 && (\text{Apply the distributive property to remove the brackets})\end{align*}

Find the LCD for 4, 4, and 1. Since it is 4, multiply the last number by \begin{align*}\frac{4}{4}\end{align*}, to get the same denominator.

\begin{align*}\frac{3}{4}x+\frac{7}{4} &= \left({\color{red}\frac{4}{4}}\right)-2\\ \frac{3}{4}x+\frac{7}{4} &= \frac{-8}{4} && (\text{Simplify})\end{align*}

Since all of the denominators are the same, the equation becomes:

\begin{align*}3x+7 &= -8\\ 3x+7 {\color{red}-7} &= -8 {\color{red}-7} && (\text{Subtract} \ 7 \ \text{from both sides of the equal sign to isolate the variable})\\ 3x &= -15 && (\text{Simplify})\\ \frac{3x}{{\color{red}3}} &= \frac{-15}{{\color{red}3}} && (\text{Divide by} \ 3 \ \text{to solve for the variable})\\ x &= -5\end{align*}

Therefore \begin{align*}x = -5\end{align*}.

\begin{align*}\text{Check:} &\\ \frac{1}{4}(3x+7) &= -2\\ \frac{1}{4} (3 ({\color{red}-5})+7) &= -2\\ \frac{1}{4}(-15+7) &= -2\\ \frac{1}{4}(-8) &= -2\\ \frac{-8}{4} &= -2\\ -2 &= -2 \ \ \end{align*}

Example C

Solve: \begin{align*}\frac{1}{3}(x-2) = -\frac{2}{3}(2x+4)\end{align*}.

\begin{align*}\frac{1}{3}(x-2) &= -\frac{2}{3}(2x+4)\\ \frac{1}{3}x-\frac{2}{3} &= -\frac{4}{3}x-\frac{8}{3} && (\text{Apply the distributive property to remove the brackets})\end{align*}

Since all of the denominators are the same, the equation becomes:

\begin{align*}x-2 &= -4x-8\\ x {\color{red}+4x} -2 &= -4x {\color{red}+4x}-8 && (\text{Add} \ 4x \ \text{to both sides of the equal sign to combine variables})\\ 5x-2 &= -8 && (\text{Simplify})\\ 5x-2 {\color{red}+2} &= -8 {\color{red}+2} && (\text{Add} \ 2 \ \text{to both sides of the equation to isolate the variable})\\ 5x &= -6 && (\text{Simplify})\\ \frac{5x}{{\color{red}5}} &= \frac{-6}{{\color{red}5}} && (\text{Divide both sides by} \ 5 \ \text{to solve for the variable})\\ x&=\frac{-6}{5} && (\text{Simplify})\end{align*}

Therefore \begin{align*}x=\frac{-6}{5}\end{align*}.

\begin{align*}\text{Check:}&\\ \frac{1}{3}(x-2) &= -\frac{2}{3}(2x+4)\\ \frac{1}{3} \left( {\color{red}\left(\frac{-6}{5}\right)} -2 \right) &= -\frac{2}{3} \left(2 {\color{red}\left(\frac{-6}{5}\right)}+4\right)\\ 0.33(-1.2-2) &= -0.67(2(-1.2)+4)\\ 0.33(-3.2) &= -0.67(1.6)\\ -1.1 &= -1.1 \ \ \end{align*}

Concept Problem Revisited

Pens are $9 per dozen and pencils are $6 per dozen. Janet need to buy a half dozen of each for school. How much is the total cost of her purchase?

First you should write down what you know:

Let \begin{align*}x =\end{align*} total cost

Cost of pens: $9/dozen

Cost of pencils: $6/dozen

Janet needs one half dozen of each.

The total cost would therefore be:

\begin{align*}\frac{1}{2}(\$9+\$6) &= x\\ \frac{\$ 9}{2}+\frac{\$ 6}{2} &= x\\ \$4.50+\$3.00 &= x\\ \$7.50 &= x\end{align*}

Therefore Janet would need $7.50 to buy these supplies.

Vocabulary

Distributive Property
The distributive property is a mathematical way of grouping terms. It states that the product of a number and a sum is equal to the sum of the individual products of the number and the addends. For example, in the expression: \begin{align*}{\color{red}\frac{2}{3}} ({\color{blue}x + 5})\end{align*}, the distributive property states that the product of a number \begin{align*}({\color{red}\frac{2}{3}})\end{align*} and a sum \begin{align*}({\color{blue}x + 5})\end{align*} is equal to the sum of the individual products of the number \begin{align*}({\color{red}\frac{2}{3}})\end{align*} and the addends \begin{align*}({\color{blue}x}\end{align*} and \begin{align*}{\color{blue}5})\end{align*}.

Guided Practice

1. Solve for x: \begin{align*}\frac{1}{2}(5x+3)=1\end{align*}.

2. Solve for x: \begin{align*}\frac{2}{3}(9x-6)=2\end{align*}.

3. Solve for x: \begin{align*}\frac{2}{3}(3x+9)=\frac{1}{4}(2x+5)\end{align*}.

Answers:

1.

\begin{align*}\frac{1}{2} (5x+3) &=1 \\ \frac{5}{2}x+\frac{3}{2}&=1 && (\text{Apply the distributive property to remove the brackets})\end{align*}

Find the LCD for 2, 2, and 1. Since it is 2, multiply the last number by \begin{align*}\frac{2}{2}\end{align*}, to get the same denominator.

\begin{align*}\frac{5}{2}x+\frac{3}{2} &= 1 {\color{red}\left(\frac{2}{2}\right)}\\ \frac{5}{2}x+\frac{3}{2} &= \frac{2}{2} && (\text{Simplify})\end{align*}

Since all of the denominators are the same, the equation becomes:

\begin{align*}5x+3 &= 2\\ 5x+3 {\color{red}-3} &= 2 {\color{red}-3} && (\text{Subtract} \ 3 \ \text{from both sides of the equal sign to isolate the variable})\\ 5x &= -1 && (\text{Simplify})\\ \frac{5x}{{\color{red}5}} &= \frac{-1}{{\color{red}5}} && (\text{Divide both sides by the} \ 5 \ \text{to solve for the variable})\\ x &= \frac{-1}{5} && (\text{Simplify})\end{align*}

Therefore \begin{align*}x=\frac{-1}{5}\end{align*}.

\begin{align*}\text{Check:} &\\ \frac{1}{2} (5x+3) &= 1\\ \frac{1}{2} \left(5 \left({\color{red}\frac{-1}{5}}\right)+3\right) &= 1\\ \frac{1}{2}(-1+3) &= 1\\ \frac{1}{2}(2) &= 1\\ 1 &= 1 \ \ \end{align*}

2.

\begin{align*}\frac{2}{3} (9x-6) &= 2\\ \frac{18}{3}x-\frac{12}{3} &= 2 && (\text{Apply the distributive property to remove the brackets})\end{align*}

Find the LCD for 3, 3, and 1. Since it is 3, multiply the last number by \begin{align*}\frac{3}{3}\end{align*}, to get the same denominator.

\begin{align*}\frac{18}{3}x-\frac{12}{3} &= 2 {\color{red}\left(\frac{3}{3}\right)}\\ \frac{18}{3}x-\frac{12}{3} &= \frac{6}{3} && (\text{Simplify})\end{align*}

Since all of the denominators are the same, the equation becomes:

\begin{align*}18x-12 &= 6\\ 18x-12 {\color{red}+12} &= 6 {\color{red}+12} && (\text{Add} \ 12 \ \text{to both sides of the equal sign to isolate the variable})\\ 18x &= 18 && (\text{Simplify})\\ \frac{18x}{{\color{red}18}} &= \frac{18}{{\color{red}18}} && (\text{Divide both sides by the} \ 18 \ \text{to solve for the variable})\\ x &= 1 && (\text{Simplify})\end{align*}

Therefore \begin{align*}x=1\end{align*}.

\begin{align*}\text{Check:} &\\ \frac{2}{3}(9x-6) &= 2\\ \frac{2}{3}(9({\color{red}1})-6) &= 2\\ \frac{2}{3}(3) &= 2\\ 2 &= 2 \ \ \end{align*}

3.

\begin{align*}\frac{2}{3}(3x+9) &= \frac{1}{4}(2x+5)\\ \frac{6}{3}x+\frac{18}{3} &= \frac{2}{4}x+\frac{5}{4} && (\text{Apply the distributive property to remove the brackets})\end{align*}

Find the LCD for 3, 3, and 4, 4. Since it is 12, multiply the first two fractions by \begin{align*}\frac{4}{4}\end{align*} and the second two fractions by \begin{align*}\frac{3}{3}\end{align*}, to get the same denominator.

\begin{align*}\left({\color{red}\frac{4}{4}}\right) \frac{6}{3}x+\left({\color{red}\frac{4}{4}}\right) \frac{18}{3} &= \left({\color{red}\frac{3}{3}}\right) \frac{2}{4}x+\left({\color{red}\frac{3}{3}}\right) \frac{5}{4}\\ \frac{24}{12}x+\frac{72}{12} &= \frac{6}{12}x+\frac{15}{12} && (\text{Simplify})\end{align*}

Since all of the denominators are the same, the equation becomes:

\begin{align*}24x+72 &= 6x+15\\ 24x+72 {\color{red}-72} &= 6x+15 {\color{red}-72} && (\text{Subtract} \ 72 \ \text{from both sides of the equal sign to isolate the variable})\\ 24x &= 6x-57 && (\text{Simplify})\\ 24x {\color{red}-6x} &= 6x {\color{red}-6x} - 57 && (\text{Subtract} \ 6x \ \text{from both sides of the equal sign to get variables on same side})\\ 18x &= -57 && (\text{Simplify})\\ \frac{18x}{{\color{red}18}} &= \frac{-57}{{\color{red}18}} && (\text{Divide both sides by} \ 18 \ \text{to solve for the variable})\\ x &= \frac{-57}{18} && (\text{Simplify})\end{align*}

Therefore \begin{align*}x=\frac{-57}{18}\end{align*}.

\begin{align*}\text{Check:} &\\ \frac{2}{3} (3x+9) &= \frac{1}{4}(2x+5)\\\ \frac{2}{3} \left(3 \left( {\color{red}\frac{-57}{18}}\right)+9\right) &= \frac{1}{2} \left(2 \left({\color{red}\frac{-57}{18}} \right)+5\right)\\ \frac{2}{3}(-9.5+9) &= \frac{1}{4}(-6.33+5)\\ \frac{2}{3}(-0.5) &= \frac{1}{4}(-1.33)\\ -0.33 &= -0.33 \ \ \end{align*}

Practice

Solve for the variable in each of the following problems.

  1. \begin{align*}\frac{1}{2} (x+5)=6\end{align*}
  2. \begin{align*}\frac{1}{4}(g+2)=8\end{align*}
  3. \begin{align*}0.4(b+2)=2\end{align*}
  4. \begin{align*}0.5(r-12)=4\end{align*}
  5. \begin{align*}\frac{1}{4}(x-16)=7\end{align*}

Solve for the variable in each of the following problems.

  1. \begin{align*}26.5-k=0.5(50-k)\end{align*}
  2. \begin{align*}2(1.5c+4)=-1\end{align*}
  3. \begin{align*}-\frac{1}{2}(3x-5)=7\end{align*}
  4. \begin{align*}0.35+0.10(m-1)=5.45\end{align*}
  5. \begin{align*}\frac{1}{4}+\frac{2}{3}(t+1)=\frac{1}{2}\end{align*}

Solve for the variable in each of the following problems.

  1. \begin{align*}\frac{1}{2}x-3 (x+4)=\frac{2}{3}\end{align*}
  2. \begin{align*}-\frac{5}{8}x+x=\frac{1}{8}\end{align*}
  3. \begin{align*}0.4(12-d)=18\end{align*}
  4. \begin{align*}0.25(x+3)=0.4(x-5)\end{align*}
  5. \begin{align*}\frac{2}{3}(t-2)=\frac{3}{4}(t+2)\end{align*}

Notes/Highlights Having trouble? Report an issue.

Color Highlighted Text Notes
Show More

Image Attributions

Show Hide Details
Description
Difficulty Level:
At Grade
Grades:
Date Created:
Dec 19, 2012
Last Modified:
Nov 04, 2015

We need you!

At the moment, we do not have exercises for Equations with Decimals, Fractions and Parentheses.

Save or share your relevant files like activites, homework and worksheet.
To add resources, you must be the owner of the Modality. Click Customize to make your own copy.
Please wait...
Please wait...
Image Detail
Sizes: Medium | Original
 
Here