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Solving for a Variable on One Side Using Multiplication, Addition, and Subtraction
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Not all problems only have one thing happening at a time, we often have multiplication and addition. We still add the additive inverse to both sides and multiply both sides by the multiplicative inverse, but which comes first? The following video will show this process:

Solving for a Variable on One Side Part 3-Multiplication, addition, and subtraction

Video Source (07:18 mins) | Transcript

Remember, when solving for a variable, we use the Order of Operations (PEMDAS), but we go backward instead. In your practice problems, you’re just dealing with Multiplication and Addition/Subtraction, so first you’ll add the additive inverse, then you multiply by the multiplicative inverse.

Additional Resources

Practice Problems

Solve for the variable:
  1. \(5{\text{M}} + 2 = 12\) (
    Solution
    x
    Solution: \(2\)
    Details:
    Where there are several operations within an equation, it is helpful to identify the operations and the order we should do them in.

    In this example, there is addition and multiplication.
    This is a picture of the equation 5m+2=12. 'Multiplication' is written with an arrow pointing to 5M. 'Addition' is written with an arrow pointing to the addition symbol.

    When solving for a variable, we do the order of operations backward.

    The order of operations is as follows:
    1. Parentheses
    2. Exponents
    3. Multiplication & Division
    4. Addition and Subtraction

    Step 1: Going backward we start by undoing the addition and subtraction:

    In the equation \(5{\text{M}} + 2 = 12\) we need to remove the addition of the \(+2\) from the left-hand side. We do this by adding the additive inverse to both sides. The additive inverse of \(+2\) is \(-2\).
    This is a picture of the equation 5M+2=12. There is a vertical dashed line through the equal sign. “Left-hand side” is written on the left side of the dashed line with “Right-hand side” written on the right. +-2 is written below the +2 on the left and below 12 on the right.

    The \(+2\) and \(-2\) add to zero, so we are left with \(5{\text{M}}\) on the left side.

    The \(12\) and \(-2\) add to \(10\) on the right side.
    This is a picture of the equation 5M+2=12. There is a vertical dashed line through the equal sign. “Left-hand side” is written on the left side of the dashed line with “Right-hand side” written on the right. “+-2” is written below the +2 on the left and below 12 on the right. There is a line beneath that indicating that addition has been performed. Below that, 5M+0=10 has been written.

    Step 2: Undo any multiplication or division

    The \(5\) is currently being multiplied to the \({\text{M}}\). We need to remove this to get \({\text{M}}\) all by itself. We do this by multiplying both sides of the equation by the multiplicative inverse of \(5\).

    The multiplicative inverse of \(5\) is \(\dfrac{1}{5}\).

    \(5{\text{M}} = 10\)

    \({\color{Cyan} \dfrac{1}{5}}\left ( 5{\text{M}} \right ) = \left ( 10 \right ){\color{Cyan} \dfrac{1}{5}}\)

    Since \(\dfrac{1}{5} \times 5 = 1\), we are left with \(1{\text{M}}\). Anything multiplied by \(1\) is still itself, so \(1{\text{M}} = {\text{M}}\).

    \({\color{Cyan} \dfrac{1}{5}}\left ( {\color{DarkGreen} 5}{\text{M}} \right ) = \left ( 10 \right ){\color{Cyan} \dfrac{1}{5}}\)

    \({\color{Cyan} \dfrac{1}{5}}\cdot{\color{DarkGreen} \dfrac{5}{1}}\left ( {\text{M}} \right ) = \left ( 10 \right ){\color{Cyan} \dfrac{1}{5}}\)

    \(\dfrac{{\color{Cyan} 1}*{\color{DarkGreen} 5}}{{\color{Cyan} 5}*{\color{DarkGreen} 1}}\left ( {\text{M}} \right ) = \left ( 10 \right ){\color{Cyan} \dfrac{1}{5}}\)

    \(\dfrac{{\color{Orange} 5}}{{\color{Orange} 5}}\left ( {\text{M}} \right ) = \left ( 10 \right ){\color{Cyan} \dfrac{1}{5}}\)

    \({\color{Orange} 1}{\text{M}} = \left ( 10 \right ){\color{Cyan} \dfrac{1}{5}}\)

    In the previous image, multiplication is written in several different ways.
    - A number and a variable right next to each other without any other operations between them means they are being multiplied together. Ex: \(5{\text{M}}\)
    - Parentheses are used to show multiplication between the numbers inside the parentheses and the numbers outside the parentheses.
    - A dot is used between the fractions showing they are being multiplied together.
    - The \(*\) symbol is used within the fraction to show multiplication between the numbers in the fraction.

    We now have \({\text{M}}\) all by itself on one side of the equal sign. The next step is to simplify the other side of the equal sign.
    This is a picture of the equation \(1M=10\dfrac{1}{5}\). Next line is: \(M=\dfrac{10}{1} \times \dfrac{1}{5}=\dfrac{10 \times 1}{1 \times 5}=\dfrac{10}{5}=2\)

    \(\left ( 10 \right )\left ( \dfrac{1}{5} \right )\) is the same as \(\dfrac{10}{5}=2\).

    Our final solution: \({\text{M}} = 2\)
    )
  2. \(-2{\text{Q}} + 1 = 5\) (
    Solution
    x
    Solution:
    \(-2\)
    )
  3. \(2{\text{n}} + 9 = -5\) (
    Solution
    x
    Solution:
    \(-7\)
    )
  4. \(8{\text{k}} {-} 7 = 17\) (
    Solution
    x
    Solution:
    \(3\)
    Details:
    In this example problem, there are two operations going on: multiplication and the addition of a negative number.

    We need to unravel our equation in order to get the variable \({\text{k}}\) all by itself on one side of the equal sign. We do this by doing the order of operations backward.

    Step 1: Do the inverse of any addition or subtraction
    In this example, \(-7\) is being added to \(8{\text{k}}\).

    \(8{\text{k}} {-} 7 = 8{\text{k}} + (-7)\)

    This is a picture of the equation \(8K-7=17\). There is a vertical dashed line through the equal sign. “Left-hand side” is written on the left side of the dashed line with “Right-hand side” written on the right. \(8k+-7=17\) is written below the original equation, with a horizontal bracket indicating that subtracting 7 is equivalent to adding negative \(-7\).

    In order to remove the \(-7\), we add the additive inverse to both sides.

    The additive inverse of \(-7\) is \(+7\)

    On the left-hand side:
    \(8{\text{k}} {-} 7 + 7 = 8{\text{k}} + 0 = 8{\text{k}}\)

    On the right-hand side:
    \(17 + 7 = 24\)

    This is a picture of the equation \(8K-7+7=17+7\). There is a vertical dashed line through the equal sign. “Left-hand side” is written on the left side of the dashed line with “Right-hand side” written on the right. \(8K+0=24\) is written on the next line. There are horizontal brackets between the two equations indicating that \(-7+7=0\) and \(17+7=24\).

    Step 2: Now the only operation on \({\text{k}}\) is the multiplication of \(8\). We isolate \({\text{k}}\) by multiplying both sides of the equation by the multiplicative inverse of \(8\).

    \(8\left ( \dfrac{1}{8} \right ) = 1\)

    Therefore, the multiplicative inverse of \(8\) is \(\dfrac{1}{8}\).

    On the left-hand side of the equal sign:
    \(\left ( \dfrac{1}{8} \right )\left ( 8{\text{k}} \right ) = 1{\text{k}} = {\text{k}}\)

    On the right-hand side of the equal sign:
    \(\left ( \dfrac{1}{8} \right )24 = \left ( \dfrac{1}{8} \right )\left ( \dfrac{24}{1} \right ) = \left ( \dfrac{24}{8} \right ) = 3\)

    \({\color{DarkGreen} 8}{\text{k}}={\color{Orange} 24}\)

    \({\color{Cyan} \left ( \dfrac{1}{8} \right )} 8{\color{DarkGreen} {\text{k}}} = {\color{Cyan} \left ( \dfrac{1}{8} \right )}{\color{Orange} 24}\)

    \(\left ( \dfrac{{\color{Cyan} 1}*{\color{DarkGreen} 8}}{{\color{Cyan} 8}*{\color{DarkGreen} 1}} \right ){\text{k}} = \dfrac{{\color{Cyan} 1}*{\color{Orange} 24}}{{\color{Cyan} 8}*{\color{Orange} 1}}\)

    \(1{\text{k}}=\dfrac{{\color{DarkOrange} 24}}{{\color{Cyan} 8}}\)

    \({\text{k}}=3\)

    Our final solution: \({\text{k}} = 3\)
    )
  5. \(-3{\text{x}} + 14 = -7\) (
    Solution
    x
    Solution:
    \(7\)
    )
  6. \(-5{\text{q}} {-} 29 = 41\) (
    Video Solution
    x
    Solution: \(-14\)
    Details:

    (Video Source | Transcript)
    )
  7. \(10{\text{f}} {-} 6 = 24\) (
    Video Solution
    x
    Solution: \(3\)
    Details:

    (Video Source | Transcript)
    )

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