<img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" />
Skip Navigation

Single Variable Subtraction Equations

Solve one - step equations using addition.

Atoms Practice
Practice Single Variable Subtraction Equations
Practice Now
Single Variable Subtraction Equations

Let’s Think About It

The baking club realized they only have 2.7 kg of flour left from their original order. The only person to use the flour since they bought it was Sal who used .9 kg of flour to make cakes. Can you write an equation to solve for how much flour they had before Sal took some, and then solve that equation?

In this concept, you will learn to solve single variable subtraction equations.


To solve an equation in which a number is subtracted from a variable, you can use addition to isolate the variable.

You can add the same number to both sides of the equation because of the Addition Property of Equality, which states: if \begin{align*}a = b\end{align*}, then \begin{align*}a + c = b + c\end{align*}.

This means that if you add a number, \begin{align*}c\end{align*}, to one side of an equation, you must add that same number, \begin{align*}c\end{align*}, to the other side, in order to keep both sides of the equation equal.

Let’s look at an example.

Solve for \begin{align*}a\end{align*}.

\begin{align*}a- 15 = 18\end{align*}

In the equation, 15 is subtracted from \begin{align*}a\end{align*}. So, you use the addition property of equality to add 15 to both sides of the equation. This will isolate the variable, \begin{align*}a\end{align*}.

\begin{align*}\begin{array}{rcl} a- 15 &=& 18 \\ a- 15 + 15 &=& 18 + 15 \\ a + -15 + 15 &=& 33 \\ a + 0 &=& 33 \\ a &=& 33 \end{array}\end{align*}

Notice that you can rewrite \begin{align*}(-15)\end{align*} as \begin{align*}(+ -15)\end{align*}. This is a very useful strategy in solving equations. You can rewrite subtraction as adding a negative number.

The answer is \begin{align*}a = 33\end{align*}.

Here is another example.

Solve for \begin{align*}k\end{align*}.


In the equation, \begin{align*}\frac{1}{3}\end{align*} is subtracted from \begin{align*}k\end{align*}. So, you use the addition property of equality, and add \begin{align*}\frac{1}{3}\end{align*} to both sides of the equation. This isolates the variable \begin{align*}k\end{align*}.

First, add \begin{align*}\frac{1}{3}\end{align*} to both sides of the equation.

\begin{align*}\begin{array}{rcl} k-\frac{1}{3} &=& \frac{2}{3}\\ k-\frac{1}{3}+\frac{1}{3} &=& \frac{2}{3}+\frac{1}{3} \end{array}\end{align*}

Since the fractions have the same denominators you can add them together.

\begin{align*}\begin{array}{rcl} k-\frac{1}{3}+\frac{1}{3} &=& \frac{2}{3}+\frac{1}{3}\\ k &=& \frac{3}{3}\\ k &=& 1 \end{array}\end{align*}

The answer is \begin{align*}k=1\end{align*}.

Guided Practice

Harry earned $19.50 this week. That is $6.50 less than he earned last week.

Write an equation to represent \begin{align*}m\end{align*}, the amount of money, in dollars, that he earned last week. Then solve for \begin{align*}m\end{align*}.

Let \begin{align*}m\end{align*} be the amount of money Harry earned last week. Then you can write an equation.

\begin{align*}19.50 = m-6.50.\end{align*}

Next, solve the equation. Use the addition property of equality to add 6.50 to each side of the equation.

\begin{align*}\begin{array}{rcl} 19.50 &=& m- 6.50 \\ 19.50 + 6.50 &=& m- 6.50 + 6.50 \\ 26.00 &=& m + (-6.50 + 6.50) \\ 26 &=& m + 0 \\ 26 &=& m \end{array}\end{align*}

The answer is Harry earned $26.00 last week.


Solve each equation.

Example 1

Solve for \begin{align*}x\end{align*}.

\begin{align*}x- 44 = 22\end{align*}

Use the addition property of equality and add 44 to both sides of the equation.

\begin{align*}\begin{array}{rcl} x- 44 &=& 22 \\ x- 44+44 &=& 22+44 \\ x &=& 66 \end{array}\end{align*}

The answer is \begin{align*}x = 66\end{align*}.

Example 2

Solve for \begin{align*}x\end{align*}.

\begin{align*}x-1.3 = 5.6\end{align*}

Use the addition property of equality and add 1.3 to both sides of the equation.

\begin{align*}\begin{array}{rcl} x- 1.3 &=& 5.6 \\ x- 1.3+1.3 &=& 5.6+1.3 \\ x &=& 6.9 \end{array}\end{align*}

The answer is \begin{align*}x = 6.9\end{align*}.

Example 3

Solve for \begin{align*}y\end{align*}.


Use the addition property of equality and add \begin{align*}\frac{1}{4}\end{align*} to both sides of the equation.

\begin{align*} \begin{array}{rcl} y-\frac{1}{4} &=& \frac{1}{2} \\ y-\frac{1}{4}+\frac{1}{4} &=& \frac{1}{2}+\frac{1}{4} \\ y &=& \frac{1}{2}+\frac{1}{4} \end{array}\end{align*}

To add \begin{align*}\frac{1}{2}+\frac{1}{4}\end{align*} you need common denominators. Write \begin{align*}\frac{1}{2}\end{align*} as the equivalent fraction \begin{align*}\frac{2}{4}\end{align*} and then add.

\begin{align*}\begin{array}{rcl} y &=& \frac{2}{4}+\frac{1}{4} \\ y &=& \frac{3}{4} \end{array}\end{align*}

The answer is \begin{align*}y=\frac{3}{4}\end{align*}.

Follow Up

Remember the baking club and Sal who used their flour to make cakes? They need to know how much flour they originally had so they can pay their supplier.

The club started with some flour, but Sal took .9 kg of this flour to make cake. They had 2.7 kg left. You have to find how much they had to begin with.

First, you need to write an equation that represents this information. Let \begin{align*}x\end{align*}, represent the amount of flour the group originally had. This amount, minus the .9 kg Sal took equals 2.7 kg.


Use the addition property of equality and add .9 to both sides of the equation and solve.

\begin{align*}\begin{array}{rcl} x-.9+.9 &=& 2.7+.9 \\ x &=& 3.6 \end{array}\end{align*}

The answer is they originally had 3.6 kg of flour.

Video Review


Explore More

Solve each single-variable subtraction equation.

  1. \begin{align*}x-8 = 9\end{align*}
  2. \begin{align*}x- 18 = 29\end{align*}
  3. \begin{align*}a- 9 = 29\end{align*}
  4. \begin{align*}a-4 = 30\end{align*}
  5. \begin{align*}b-14 = 27\end{align*}
  6. \begin{align*}b-13 = 50\end{align*}
  7. \begin{align*}y-23 = 57\end{align*}
  8. \begin{align*}y- 15 = 27\end{align*}
  9. \begin{align*}x-9 = 32\end{align*}
  10. \begin{align*}c- 19 = 32\end{align*}
  11. \begin{align*}x-1.9 = 3.2\end{align*}
  12. \begin{align*}y-2.9 = 4.5\end{align*}
  13. \begin{align*}c-6.7 = 8.9\end{align*}
  14. \begin{align*}c-1.23 = 3.54\end{align*}
  15. \begin{align*}c-5.67 = 8.97\end{align*}




The result of a subtraction operation is called a difference.


An expression is a mathematical phrase containing variables, operations and/or numbers. Expressions do not include comparative operators such as equal signs or inequality symbols.


To simplify means to rewrite an expression to make it as "simple" as possible. You can simplify by removing parentheses, combining like terms, or reducing fractions.


The sum is the result after two or more amounts have been added together.

Image Attributions


Please wait...
Please wait...

Original text