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# 2.6: Equations with Decimals, Fractions and Parentheses

Difficulty Level: Advanced Created by: CK-12
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Practice Equations with Decimals, Fractions, and Parentheses

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

### Equations with Decimals, Fractions and Parentheses

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 equations that contain decimals and/or fractions. The same rules apply. If the equation has parentheses, your first step is to multiply what is on the outside of the parentheses by what is on the inside of the parentheses. After you remove parentheses, you then solve the equation by combining like terms, moving constants to one side of the equals sign and variables to the other side of the equals sign, and finally isolating the variable to find the solution.

#### Solve for d

Solve: 25(d+4)=6\begin{align*}\frac{2}{5}(d+4) = 6\end{align*}.

25(d+4)25d+85=6=6(Apply the distributive property to remove the parentheses)\begin{align*}\frac{2}{5}(d+4) &= 6\\ \frac{2}{5}d+\frac{8}{5} &= 6 && (\text{Apply the distributive property to remove the parentheses})\end{align*}

Find the LCD for 5, 5, and 1. Since it is 5, multiply the last number by 55\begin{align*}\frac{5}{5}\end{align*}, to get the same denominator.

25d+8525d+85=(55)6=305(Simplify)\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:

2d+82d+882d2d2d=30=308=22=222=11(Subtract 8 from both sides of the equals sign)(Simplify)(Divide by 2 to solve for the variable)\begin{align*}2d+8 &= 30\\ 2d+8 {\color{red}-8} &= 30 {\color{red}-8} && (\text{Subtract} \ 8 \ \text{from both sides of the equals sign})\\ 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 d=11\begin{align*}d = 11\end{align*}.

Check:25(d+4)25(11+4)25(15)3056=6=6=6=6=6  \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*}

#### Solve for x

Solve: 14(3x+7)=2\begin{align*}\frac{1}{4}(3x+7) =2\end{align*}.

14(3x+7)34x+74=2=2(Apply the distributive property)\begin{align*}\frac{1}{4}(3x+7) &= -2\\ \frac{3}{4}x+\frac{7}{4} &= -2 && (\text{Apply the distributive property})\end{align*}

Find the LCD for 4, 4, and 1. Since it is 4, multiply the last number by 44\begin{align*}\frac{4}{4}\end{align*}, to get the same denominator.

34x+7434x+74=(44)2=84(Simplify)\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:

3x+73x+773x3x3x=8=87=15=153=5(Subtract 7 from both sides of the equals sign)(Simplify)(Divide by 3 to solve for the variable)\begin{align*}3x+7 &= -8\\ 3x+7 {\color{red}-7} &= -8 {\color{red}-7} && (\text{Subtract} \ 7 \ \text{from both sides of the equals sign})\\ 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 x=5\begin{align*}x = -5\end{align*}.

Check:14(3x+7)14(3(5)+7)14(15+7)14(8)842=2=2=2=2=2=2  \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*}

#### Solve for x

Solve: 13(x2)=23(2x+4)\begin{align*}\frac{1}{3}(x-2) = -\frac{2}{3}(2x+4)\end{align*}.

13(x2)13x23=23(2x+4)=43x83(Apply the distributive property)\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})\end{align*}

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

x2x+4x25x25x2+25x5x5x=4x8=4x+4x8=8=8+2=6=65=65(Add 4x to both sides of the equals sign)(Simplify)(Add 2 to both sides of the equation)(Simplify)(Divide both sides by 5 to solve for the variable)(Simplify)\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 equals sign})\\ 5x-2 &= -8 && (\text{Simplify})\\ 5x-2 {\color{red}+2} &= -8 {\color{red}+2} && (\text{Add} \ 2 \ \text{to both sides of the equation})\\ 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 x=65\begin{align*}x=\frac{-6}{5}\end{align*}.

Check:13(x2)13((65)2)0.33(1.22)0.33(3.2)1.1=23(2x+4)=23(2(65)+4)=0.67(2(1.2)+4)=0.67(1.6)=1.1  \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*}

### Examples

#### Example 1

Earlier, you were given a problem about Janet's purchase.

Pens are $9 per dozen and pencils are$6 per dozen. Janet needs 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.

#### Example 2

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

\begin{align*}\frac{1}{2} (5x+3) &=1 \\ \frac{5}{2}x+\frac{3}{2}&=1 && (\text{Apply the distributive property})\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 equals sign})\\ 5x &= -1 && (\text{Simplify})\\ \frac{5x}{{\color{red}5}} &= \frac{-1}{{\color{red}5}} && (\text{Divide both sides by 5})\\ x &= \frac{-1}{5} && (\text{Simplify})\end{align*}

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

#### Example 3

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

\begin{align*}\frac{2}{3} (9x-6) &= 2\\ \frac{18}{3}x-\frac{12}{3} &= 2 && (\text{Apply the distributive property})\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 equals sign})\\ 18x &= 18 && (\text{Simplify})\\ \frac{18x}{{\color{red}18}} &= \frac{18}{{\color{red}18}} && (\text{Divide both sides by 18})\\ x &= 1 && (\text{Simplify})\end{align*}

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

#### Example 4

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

\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})\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 equals sign})\\ 24x &= 6x-57 && (\text{Simplify})\\ 24x {\color{red}-6x} &= 6x {\color{red}-6x} - 57 && (\text{Subtract} \ 6x \ \text{from both sides of the equals sign})\\ 18x &= -57 && (\text{Simplify})\\ \frac{18x}{{\color{red}18}} &= \frac{-57}{{\color{red}18}} && (\text{Divide both sides by 18})\\ x &= \frac{-57}{18} && (\text{Simplify})\end{align*}

### Review

Solve for the variable in each of the following equations.

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*}
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*}
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*}

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### Vocabulary Language: English

distributive property

The distributive property states that the product of an expression and a sum is equal to the sum of the products of the expression and each term in the sum. For example, $a(b + c) = ab + ac$.

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