Skip to main content
K12 LibreTexts

2.4.5: Multi-Step Equations with Decimals, Fractions, and Parentheses

  1. Last updated
  2. Save as PDF
  • Page ID
    4360

    • \( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \) \( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)\(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\) \(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\)\(\newcommand{\AA}{\unicode[.8,0]{x212B}}\)

    Equations with Decimals, Fractions and Parentheses

    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.

    Let's solve the following equations:

    1. (2/5)(d+4)=6.

    (2/5)(d+4)=6

    (2/5)d+(8/5)=6 (Apply the distributive property to remove the parentheses)

    Find the LCD for 5, 5, and 1. Since it is 5, multiply the last number by 5/5, to get the same denominator.

    (25)d+(85)=(55)6

    (2/5)d+(8/5)=(30/5) (Simplify)

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

    2d+8=30

    2d+8−8=30-8

    2d=22

    2d/2=22/2

    d=11 (Subtract 8 from both sides of the equals sign)(Simplify)(Divide by 2 to solve for the variable)

    Therefore d=11.

    Check: (2/5)(d+4)=6

    (2/5)(11+4)=6

    (2/5)(15)=6

    30/5=6

    6=6

    1. 1/4(3x+7)=-2.

    1/4(3x+7)=-2

    (3/4)x+(7/4)=−2 (Apply the distributive property)

    Find the LCD for 4, 4, and 1. Since it is 4, multiply the last number by 4/4, to get the same denominator.

    (3/4)x+(7/4)=(4/4)−2

    (3/4)x+(7/4)=−8/4 (Simplify)

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

    3x+7=-8

    3x+7−7=-8-7 (Subtract 7 from both sides of the equals sign)

    3x=-15 (Simplify)

    3x/3=-15/3 (Divide by 3 to solve for the variable)

    x=−5

    Therefore x=−5.

    Check: (1/4)(3x+7)

    1/4(3(−5)+7)=−2

    1/4(−15+7)=−2

    1/4(−8)=−2

    −8/4=−2

    −2=−2

    1. 1/3(x−2)=−2/3(2x+4).

    1/3(x−2)=−2/3(2x+4)

    (1/3)x−(2/3)=(−4/3)x−(8/3) (Apply the distributive property)

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

    x−2=−4x−8

    x+4x−2=−4x+4x−8 (Add 4x to both sides of the equals sign)

    5x−2=−8 (Simplify)

    5x−2+2=−8+2 (Add 2 to both sides of the equation)

    5x=−6 (Simplify)

    5x/5=−6/5 (Divide both sides by 5 to solve for the variable)

    x=−6/5 (Simplify)

    Therefore x=−6/5.

    Check: 1/3(x−2)=−2/3(2x+4)

    1/3((−6/5)−2)=−2/3(2(−6/5)+4)

    0.33(−1.2−2)=−0.67(2(−1.2)+4)

    0.33(−3.2)=−0.67(1.6)

    −1.1=−1.1

    Examples

    Example 2.4.5.1

    Earlier, you were told that pens are $9 per dozen and pencils are $6 per dozen and that Janet needs to buy a half dozen of each for school. How much is the total cost of her purchase?

    Solution

    First you should write down what you know:

    Let x= 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:

    12($9+$6)$92+$62$4.50+$3.00$7.50=x=x=x=x

    Therefore Janet would need $7.50 to buy these supplies.

    Example 2.4.5.2

    Solve for x: 12(5x+3)=1.

    Solution

    12(5x+3)52x+32=1=1(Apply the distributive property)

    Find the LCD for 2, 2, and 1. Since it is 2, multiply the last number by 22, to get the same denominator.

    52x+3252x+32=1(22)=22(Simplify)

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

    5x+35x+3−35x5x5x=2=2−3=−1=−15=−15(Subtract 3 from both sides of the equals sign)(Simplify)(Divide both sides by 5)(Simplify)

    Therefore x=−15.

    Example 2.4.5.3

    Solve for x: 23(9x−6)=2.

    Solution

    23(9x−6)183x−123=2=2(Apply the distributive property)

    Find the LCD for 3, 3, and 1. Since it is 3, multiply the last number by 33, to get the same denominator.

    183x−123183x−123=2(33)=63(Simplify)

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

    18x−1218x−12+1218x18x18x=6=6+12=18=1818=1(Add 12 to both sides of the equals sign)(Simplify)(Divide both sides by 18)(Simplify)

    Therefore x=1.

    Example 2.4.5.4

    Solve for x: 23(3x+9)=14(2x+5)

    Solution

    23(3x+9)63x+183=14(2x+5)=24x+54(Apply the distributive property)

    Find the LCD for 3, 3, and 4, 4. Since it is 12, multiply the first two fractions by 44 and the second two fractions by 33, to get the same denominator.

    (44)63x+(44)1832412x+7212=(33)24x+(33)54=612x+1512(Simplify)

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

    24x+7224x+72−7224x24x−6x18x18x18x=6x+15=6x+15−72=6x−57=6x−6x−57=−57=−5718=−5718(Subtract 72 from both sides of the equals sign)(Simplify)(Subtract 6x from both sides of the equals sign)(Simplify)(Divide both sides by 18)(Simplify)

    Review

    Solve for the variable in each of the following equations.

    1. 1/2(x+5)=6
    2. 1/4(g+2)=8
    3. 0.4(b+2)=2
    4. 0.5(r−12)=4
    5. 1/4(x−16)=7
    6. 26.5−k=0.5(50−k)
    7. 2(1.5c+4)=−1
    8. −1/2(3x−5)=7
    9. 0.35+0.10(m−1)=5.45
    10. (1/4)+(2/3)(t+1)=1/2
    11. (1/2)x−3(x+4)=2/3
    12. (−5/8)x+x=1/8
    13. 0.4(12−d)=18
    14. 0.25(x+3)=0.4(x−5)
    15. 2/3(t−2)=3/4(t+2)

    Review (Answers)

    To see the Review answers, open this PDF file and look for section 2.6.

    Vocabulary

    Term Definition
    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.

    Additional Resources

    Video:

    Practice: Multi-Step Equations with Decimals, Fractions, and Parentheses

    This page titled 2.4.5: Multi-Step Equations with Decimals, Fractions, and Parentheses is shared under a CC BY-NC license and was authored, remixed, and/or curated by CK-12 Foundation via source content that was edited to the style and standards of the LibreTexts platform; a detailed edit history is available upon request.