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Solving for a Variable Using Addition, Subtraction, Multiplication, and Division with Fractions
> ... Math > Solving for a Variable > Solving for a Variable Using Addition, Subtraction, Multiplication, and Division with Fractions

Sometimes solving for a variable requires more than one step. This lesson demonstrates how to solve for a variable when there is addition or subtraction as well as multiplication involving fractions. It is important to understand that the concepts don’t change whether there are whole numbers, decimals, or fractions in the equation. The principles of how to solve for a variable are still the same.

Solving for a Variable on One Side Part 6-Add, Sub, Mult., and Div with Fractions

Video Source (06:53 mins) | Transcript

Equations can be much more complicated than simply having two operations. The following video will show you how to solve equations with many steps.

Example of Solving for a Variable on One Side with Many Operations Involved

Video Source (12:45 mins) | Transcript

Remember, break these large problems down step by step by the Order of Operations (PEMDAS), then undo these steps by going backward and doing the inverse operation to find out what the variable is. It is important that we do each of these operations to both sides of the equation.

Helpful Tip: Rewrite the problem after you do each step so you don’t lose track of what has already been done.

Additional Resources

Practice Problems

  1. \(\dfrac{3}{2}{\text{x}} + \dfrac{1}{4} = \dfrac{13}{4}\) (
    Solution
    x
    Solution:
    2
    )
  2. \(\dfrac{1}{3} - \dfrac{2}{3}{\text{x}} = 3\) (
    Solution
    x
    Solution: \(-4\)
    Details:
    One thing you may find helpful is to first change any subtraction to the addition of a negative number. This can help avoid losing the negative.

    This image is divided into two sides with a very light gray dotted line running vertically through the image. At the top of the left side it says, “Left-hand side.” At the top of the right side it says, “Right-hand side.” These words are in light gray indicating they are not part of the math problem. The light gray line travels down through the equal signs of two equations, one above the other. The first equation on the left-hand side shows the expression one-third minus two-thirds x. The minus or subtraction symbol is in red, bringing it to our attention. There is an equal sign to the right of this expression along the light gray line representing that the expression is equal to whatever is on the right-hand side. On the right-hand side of this equal sign is the number 3. Below this equation, the same equation has been written again, but with some variations. Instead of the minus sign, there is now a plus sign in green and the fraction negative two-thirds is in parentheses and the negative is red, showing it is the same as the subtraction sign from the previous equation. So the second expression on the left-hand side reads one-third plus (negative two-thirds)x. The negative two-thirds is all multiplied to the x. The equal sign is in the middle along the gray line, and to the right on the right-hand side is the number 3 again.

    To solve for the variable x we use the order of operations backwards, but first, let’s find our steps going forward through the order of operations assuming we had a specific number for x.

    If we were going forward, our steps would go as follows:

    • Parentheses: None this time
    • Exponents: None this time
    • Multiplication & Division: between the \(\left (-\dfrac{2}{3} \right )\) and x, so first multiply the \(\left (-\dfrac{2}{3} \right ){\text{x}}\)
    • Addition & subtraction: Add \(\dfrac{1}{3}\) to our previous answer.

    Going backwards to solve for a variable:

    Since we are solving for a variable, we do these steps in reverse.

    Step 1: Do the reverse of adding \(\dfrac{1}{3}\) to each side of the equation. In other words, subtract \(\dfrac{1}{3}\) from each side or \({\color{Red}{\text{add a negative}}}\) \(\dfrac{1}{3}\) to each side.

    This image is divided into two sides with a very light gray dotted line running vertically through the image. At the top of the left side it says, “Left-hand side.” At the top of the right side it says, “Right-hand side.” These words are in light gray indicating they are not part of the math problem. The light gray line travels down through the equal signs of an equation splitting the expression on the left from the expression on the right. The expression on the left-hand side shows negative one-third plus one-third plus (negative two-thirds)x. The negative one third is in red indicating it is a new part of the equation. To the right of this expression is an equal sign. To the right of the equal sign, under the words “Right-hand side” is the expression three plus (negative one-third). The negative one-third is in parentheses and is in red indicating it is also new to the equation.

    On the left-hand side:

    The additive inverses \(-\dfrac{1}{3}\) and \(+\dfrac{1}{3}\) add together to 0. This leaves just \(\left (-\dfrac{2}{3} \right ){\text{x}}\) on the left side.

    This image is divided into two sides with a very light gray dotted line running vertically through the image. At the top of the left side it says, “Left-hand side.” At the top of the right side it says, “Right-hand side.” These words are in light gray indicating they are not part of the math problem. The light gray line travels down through the equal signs of three equations, one above the other. The first equation on the left-hand side shows the expression negative one-third plus one-third plus (negative two-thirds) x. The negative one-third at the beginning is in red showing it is the new part being added to the left-hand side. Just below the negative one-third plus one-third is a green bracket and an arrow pointing down to the next equation under it. The arrow points to a green zero in this next equation. The left-hand side of this next equation is zero plus (negative two-thirds)x. If we look below this, still on the left-hand side, the third and final equation has just negative two-thirds x on the left-hand side. If we go back up to the top equation on the right-hand side we see three plus (negative one-third). The negative one-third is in red, indicating it is the new part of the equation being added to both sides. Below this on the right-hand side in the second equation there is still three plus (negative one-third) but this time all written in black to show that nothing has changed from the previous equation. Finally, under this on the right-hand side is the same expression of three plus (negative one-third). The right hand side is the same for all three equations in this image.

    On the right-hand side:

    We have \(3+\left (-\dfrac{1}{3} \right )\)

    Adding fractions requires common denominators. The number 3 is a whole number, so it has an invisible denominator of 1. We can rewrite it as \(\dfrac{3}{1}\)

    The greatest common denominator between 1 and 3 is 3. Multiply \(\dfrac{3}{1}\) by \(\dfrac{3}{3}\) to get the common denominator 3.

    This image is divided into two sides with a very light gray dotted line running vertically through the image. At the top of the left side it says, “Left-hand side.” At the top of the right side it says, “Right-hand side.” These words are in light gray indicating they are not part of the math problem. The light gray line travels down through the equal signs of three equations, one above the other. The first equation on the left-hand side shows the expression negative two-thirds x. The left-hand side expressions all stay the same in this image, but the right-hand side changes, showing the steps for solving the right-hand side. The top expression on the right-hand side is three plus (negative one-third). The three is in blue indicating that something is going to happen to it in the next step. The next line on the right-hand side of the equation shows is (three-over-one) times three-over-three plus (negative one-third). The three-over-one is blue showing it is the same as the three in the previous equation. The three-over-three is green to emphasize that it is new. Below the (three-over-one) times three-over-three is a green bracket pointing down to the next version of the equation. Just below the bracket is the fraction nine-over-three. This is added to (negative one-third). The fraction nine-over-three is also green to represent that it is the product of the blue three-over-one and the green three-over-three.

    When adding or subtracting fractions with common denominators, we simply add or subtract the numbers in the numerator. The denominator stays the same because that is the part of the fraction that tells us what sizes the pieces are.

    \(\left ( \dfrac{9}{3} \right )+\left ( -\dfrac{1}{3} \right )=\dfrac{8}{3}\)

    We subtract \(9-1\). The negative on the \(\left ( \dfrac{1}{3} \right )\) is the same as adding a negative 1 or subtracting 1.

    This leaves us with \(\left ( -\dfrac{2}{3} \right ){\text{x}}\) on the left and \(\dfrac{8}{3}\) on the right.

    This image is divided into two sides with a very light gray dotted line running vertically through the image. At the top of the left side it says, “Left-hand side.” At the top of the right side it says, “Right-hand side.” These words are in light gray indicating they are not part of the math problem. The light gray line travels down through the equal signs of three equations, one above the other. The first equation on the left-hand side shows the expression negative two-thirds x. It is equal to the expression on the right-hand side that says nine-thirds plus (negative one-third). The nine-thirds is in green to show we are working with it. The negative on the one-third is in red to highlight that it is negative. The next equation down also has negative two-thirds x on the left-hand side. It is equal to the expression on the right-hand side which shows a fraction where the numerator is a green nine and then a red subtraction symbol followed by a one. The green nine comes from the numerator of the nine-thirds in the previous equation. The red subtraction symbol comes from the negative symbol in the previous equation. The one comes from the one-third in the previous equation. The denominator of this fraction is 3. The last equation at the bottom of this image has negative two-thirds x on the left-hand side. To the right of this is the equal sign and the fraction eight-thirds. The eight is colored blue to represent it is the result of nine minus 1 in the previous equation.

    Step 2: Multiply both sides by the multiplicative inverse of \(-\dfrac{2}{3}\).

    This will leave just 1x on the left-hand side.

    This image is divided into two sides with a very light gray dotted line running vertically through the image. At the top of the left side it says, “Left-hand side.” At the top of the right side it says, “Right-hand side.” These words are in light gray indicating they are not part of the math problem. The light gray line travels down through the equal signs of three equations, one above the other. The first equation on the left-hand side shows the expression (negative three-halves)(negative two-thirds) x. The parentheses express that the negative three-halves is being multiplied to the negative two-thirds. All of it is multiplied to x. The negative three-halves is in blue to show it is new. The negative is in red to highlight it. To the right is the equal sign. On the right-hand side of the equal sign is the expression (negative three-halves) eight-thirds. The parentheses next to the eight-thirds indicate the two fractions are being multiplied together. The negative is red to emphasize it and the three-halves is blue to show it is new. Below this equation, on the left-hand side is 1x. The equal sign is between the left and right-hand sides. On the right-hand side is the fraction negative twenty-four-sixths meaning twenty-four in the numerator and six in the denominator. The negative is red to highlight it and the fraction twenty-four-sixths is green to show it is the solution of the previous right-hand side expression. The equation at the bottom is x on the left-hand side equal to negative 4 on the right-hand side. The negative is red to highlight it and the four is green to indicate it is the solution of the previous expression.

    \(-\dfrac{24}{6}=-4\)

    Our final solution: \({\text{x}} = -4\)
    )
  3. \(3=-2{\text{D}}+\dfrac{2}{3}\) (
    Solution
    x
    Solution:
    \(-\dfrac{7}{6}\)
    )
  4. \(-\dfrac{2}{3}=\dfrac{2}{3}{\text{U}}{-}\dfrac{1}{2}\) (
    Video Solution
    x
    Solution: \(-\dfrac{1}{4}\)
    Details:

    (Video Source | Transcript)
    )
  5. \(8 = (5{\text{x}} {-} 2) - 5\) (
    Solution
    x
    Solution:
    3
    )
  6. \(2\left ( \dfrac{{\text{F}}+1}{-2}\right )-8=-4\) (
    Video Solution
    x
    Solution: \(-5\)
    Details:

    (Video Source | Transcript)
    )
  7. \(1.3{\text{d}} + 5.2 = -2.6\) (
    Solution
    x
    Solution: \(-6\)
    Details:
    We use the order of operations backwards to unravel this equation and solve for the variable d. This example uses decimal numbers. We treat them the same as integers or fractions.

    Step 1: Do the opposite of any addition or subtraction (S&A in PEMDAS)

    In this example, we need to add the additive inverse of positive 5.2 to both sides.

    The additive inverse is \(-5.2\), so we need to add \(-5.2\) to both sides of the equation.

    This image shows two equations, one on top of the other. The top equation is one-point-three d plus five-point-two equals negative two-point-six. The five-point-two is in blue to show we are doing something with it in the next step. The next equation is one-point-three d plus five-point-two plus negative five-point-two equals negative two-point-six plus negative five-point-two. The negative five-point-two on both sides of the equal sign is orange to indicate it is new on both sides.

    The left side of the equal sign:

    \(5.2 + -5.2 = 0\) leaving \(1.3{\text{d}}\) on the left.

    The right side of the equal sign:

    \(-2.6 + -5.2 = -7.8\)

    Remember the rules of adding a negative and a negative. They just become more negative.

    This image shows three equations, one on top of the other. The first equation at the top is one-point-three d plus five-point-two plus negative five-point-two equals negative two-point-six plus negative five-point-two. The negative five-point-two on both sides of the equal sign is orange to indicate it is new on both sides. The positive five-point-two on the left-hand side of the equal sign is in blue. Under the blue five-point-two plus the orange negative five-point-two is a green bracket with the point of the bracket pointing down to a green zero in the next equation. This indicates that the sum of the blue five-point-two and the orange negative five-point-two is zero. On the right-hand side, under the negative two-point-six plus the orange negative five-point-two is another green bracket whose point points down to the next equation at the number negative seven-point-eight. The negative is red and the seven-point-eight is green indicating it is the sum of the numbers negative two-point-six and negative five-point-two. Thus, the whole equation under the first is one-point-three d plus green zero equals negative green seven-point-eight. The last equation at the bottom of the image is one-point-three d equals negative seven-point-eight.

    Step 2: Do the opposite of any multiplication or division (D&M in PEMDAS)

    In this example, we need to multiply both sides by the multiplicative inverse of \(1.3\). The multiplicative inverse of a decimal is found the same way as the multiplicative inverse of an integer: \(\dfrac{1}{\text{decimal}}\)

    Multiply both sides by \(\dfrac{1}{1.3}\)

    This image shows two equations, one above the other. The top equation is one-point-three d equals (negative seven-point-eight). The negative is red to bring attention to it. The bottom equation is the same as the top equation except that both sides of the equal sign have the fraction (one-over-one-point-three) in blue multiplied what was originally there. In other words, (one-over-one-point-three) times one-point-three d equals (one-over-one-point-three) times (negative seven-point-eight).

    The left side of the equation:

    \(\left ( \dfrac{1}{1.3} \right )\left ( -1.3 \right ) = 1\) so we are left with 1d.

    The right side of the equation:

    \(\left ( \dfrac{1}{1.3} \right )\left ( -7.8 \right ) = \dfrac{\left ( -7.8 \right )}{\left ( 1.3 \right )}=-6\)

    We can solve this using longhand or a calculator.

    This image shows two equations, one over the other. The top equation is the same as the bottom equation of the previous image. It is (one-over-one-point-three) times one-point-three d equals (one-over-one-point-three) times (negative seven-point-eight). The fraction (one-over-one-point-three) on each side of the equal sign is blue. The bottom equation on the left-hand side shows a fraction whose numerator is one-point-three and denominator is a blue one-point-three to help show where the numbers went in the multiplication. This fraction is all multiplied to the variable d. The left-hand side is equal to the fraction negative seven-point-eight over one-point-three. Again the one-point-three is blue help identify where it came from.

    Our final solution: \({\text{d}} = -6\)

    Be sure not to lose the negative in the final solution.
    )

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    Solving for a Variable on One Side Using Multiplication and Division with Fractions
    Solving Variables on Both Sides of the Equation