One Step Equations - Math Steps, Examples & Questions

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One step equations

Here you will learn about one step equations, including how to solve them and how to solve problems using them.

Students first learn about algebraic expressions and equations in the $6$ th grade with their work with expressions and equations. They expand their knowledge of equations as they move through middle school math, Pre-algebra and Algebra $1.$

What is one step equations?

One step equations are algebraic equations made up of variables and numbers. An equation is a mathematical sentence that shows two expressions are equal. One step equations can be solved in one step using either addition, subtraction, multiplication, or division.

For example,

x+5=8

In order to solve an algebraic equation, you will need to isolate the variable (get it alone) on one side of the equation by doing the inverse operation.

Let’s look at examples of solving one-step equations.

What is one step equations?

Common Core State Standards

How does this relate to $6$ th grade math, $7$ th grade math, and $8$ th grade math?

Grade 6 Number System (6.NS.C.6.c)
Find and position integers and other rational numbers on a horizontal or vertical number line diagram; find and position pairs of integers and other rational numbers on a coordinate plane.

Grade 7 – Ratios and Proportional Relationships (7.RP.A.2.b)
Identify the constant of proportionality (unit rate) in tables, graphs, equations, diagrams, and verbal descriptions of proportional relationships.

Grade 8 – Expressions and Equations (8.EE.B.5)
Graph proportional relationships, interpreting the unit rate as the slope of the graph.

[FREE] One Step Equations Worksheet (Grade 6 to High School)

Use this worksheet to check your grade 6 to high school students’ understanding of one step equations. 15 questions with answers to identify areas of strength and support!

DOWNLOAD FREE

[FREE] One Step Equations Worksheet (Grade 6 to High School)

Use this worksheet to check your grade 6 to high school students’ understanding of one step equations. 15 questions with answers to identify areas of strength and support!

DOWNLOAD FREE

How to solve one step equations

In order to solve one step equations using inverse operations:

Identify the inverse operation to use.
Do the inverse operation on both sides of the equation.
Solve for the unknown variable.
Check the answer.

One step equation examples

Example 1: solve the equation using addition/subtraction

Solve the equation for $x.$

x-8=2

Identify the inverse operation to use.

$x-8=2 \to$ inverse operation is addition because addition undoes subtraction

2Do the inverse operation on both sides of the equation.

\begin{aligned} & x-8=2 \\\\ & +8=+8 \end{aligned}

3Solve for the unknown variable.

On the left side of the equation $-8+8=0$

On the right side of the equation $2+8=10$

\begin{aligned} x-8&=2 \\\\ +8&=+8 \\\\ x+0&=10 \\\\ x&=10 \end{aligned}

4Check the answer.

Example 2: solve for the variable using addition/subtraction

Solve the equation for $x.$

x+6=-7

Identify the inverse operation to use.

$x+6=-7 \to$ inverse operation is subtraction because subtraction undoes addition.

Do the inverse operation on both sides of the equation.

\begin{aligned} x+6 & =-7 \\\\-6 & =-6 \end{aligned}

Solve for the unknown variable.

On the left side of the equation $+6-6=0$

On the right side of the equation $-7-6=-13$

$\begin{aligned} x+6&=-7 \\\\ -6&=-6 \\\\ x+0&=-13 \\\\ x&=-13 \end{aligned}$

Check the answer.

Example 3: solve the equation using multiplication/division

Solve the equation, $\cfrac{1}{2} x=20,$ for $x.$

Identify the inverse operation to use.

$\cfrac{1}{2} x=20 →$ inverse operation of multiplication is division

Do the inverse operation on both sides of the equation.

\begin{aligned} & \cfrac{1}{2} x=20 \\\\& \cfrac{1}{2} \div \cfrac{1}{2} x=20 \div \cfrac{1}{2} \end{aligned}

Solve for the unknown variable.

One the left side of the equation $\cfrac{1}{2} \div \cfrac{1}{2}=\cfrac{1}{2} \times \cfrac{2}{1}=\cfrac{2}{2}=1$

On the right side of the equation $20 \div \cfrac{1}{2}=20 \times \cfrac{2}{1}=\cfrac{40}{1}=40$

$\begin{aligned} \cfrac{1}{2} x&=20 \\\\ \cfrac{1}{2} \div \cfrac{1}{2} x&=20 \div \cfrac{1}{2} \\\\ 1 x&=40 \\\\ x&=40 \end{aligned}$

Check the answer.

Example 4: solve the equation using multiplication/division

Solve the equation, $\cfrac{x}{-3}=-4$ for $x.$

Identify the inverse operation to use.

$\cfrac{x}{-3}=-4 →$ inverse operation of division is multiplication

Do the inverse operation on both sides of the equation.

\cfrac{x}{-3} \cdot-3=-4 \cdot-3

Solve for the unknown variable.

On the left side of the equation $\cfrac{x}{-3} \bullet-3=\cfrac{-3 x}{-3}=1 x$

On the right side of the equation $-4 \cdot-3=12$

$\begin{aligned} \cfrac{x}{-3} \cdot-3 & =-4 \cdot-3 \\\\1 x & =12 \\\\x & =12 \end{aligned}$

Check the answer.

\begin{aligned} & \cfrac{x}{-3}=-4 \\\\& \cfrac{12}{-3}=-4 \end{aligned}

Example 5: solve the equation using multiplication/division

Solve the equation for $x.$

-x=9

Identify the inverse operation to use.

$-x$ means $-1x$

$-x=9 \to$ The inverse operation of multiplication is division

Do the inverse operation on both sides of the equation.

\cfrac{-x}{-1}=\cfrac{9}{-1}

Solve for the unknown variable.

On the left side of the equation $\cfrac{-x}{-1}=1 x$

On the right side of the equation $\cfrac{9}{-1}=-9$

$\begin{aligned} \cfrac{-x}{-1}&=\cfrac{9}{-1} \\\\ x&=-9 \end{aligned}$

Check the answer.

How to solve one step equations word problems

In order to solve one step equation word problems:

Identify the variable and write the equation.
Identify the inverse operation to use.
Do the inverse operation on both sides of the equation.
Solve for the unknown variable.
Check the answer.

Example 6: problem solving

The product of a number and $-5$ is $100.$ Find the number.

Identify the variable and write the equation.

“A number” is represented by $x.$

Product means multiplication. So the equation is: $-5x=100$

Identify the inverse operation to use.

$-5 x=100 \to$ The inverse operation of multiplication is division.

Do the inverse operation on both sides of the equation.

\cfrac{-5 x}{-5}=\cfrac{100}{-5}

Solve for the unknown variable.

On the left side of the equation $\cfrac{-5 x}{-5}=1 x$

On the right side of the equation $\cfrac{100}{-5}=-20$

$\begin{aligned} \cfrac{-5 x}{-5}&=\cfrac{100}{-5} \\\\ 1 x&=-20 \\\\ x&=-20 \end{aligned}$

The number is $-20.$

Check the answer.

Example 7: problem solving

The sum of a number and $-7$ is $18.$ Find the number.

Identify the variable and write the equation.

“A number” is represented by $x.$

Sum means to add. So the equation is: $x+(-7)=18$

Identify the inverse operation to use.

The inverse operation of addition is subtraction.

Do the inverse operation on both sides of the equation.

\begin{aligned} & x+(-7)=18 \\\\& x+(-7)-(-7)=18-(-7) \end{aligned}

Solve for the unknown variable.

\begin{aligned} x+(-7)&=18 \\\\ x+(-7)-(-7)&=18-(-7) \\\\ x&=18-(-7) \\\\ x&=18+7 \\\\ x&=25 \end{aligned}

Check the answer.

Teaching tips for one step equations

Use manipulatives such as algebra tiles to help students build conceptual understanding of the abstract concept.

Reinforce additive inverse and multiplicative inverse, along with the concept that $0+x = x$ and $1x=x.$

Although practice math worksheets and equation worksheets help students to review skills and procedures, visual models provide an opportunity for students to formulate an understanding of abstract algebraic concepts.

Use digital platforms that incorporate game playing for students to review skills and procedures.

Have students always check the value they get for the variable to make sure their answer makes sense and is correct.

Easy mistakes to make

Using the incorrect inverse operation
For example, $x+6=9$ using addition to solve instead of subtraction.

Confusing integer rules
For example,

Practice one step equations

x=-18

x=-2

x=2

x=18

$x-8=-10 \to$ The inverse operation is addition, so add $8$ to both sides of the equation.

\begin{aligned} x-8&=-10 \\\\ +8&=+8 \end{aligned}

On the left side of the equation $-8+8=0.$ So it is, $0+x = x$

On the right side of the equation $-10+8=-2$

\begin{aligned} x-8&=-10 \\\\ 8&=+8 \\\\ x+0&=-2 \\\\ x&=-2 \end{aligned}

Check:

One Step Equations 15 US

x=14

x=6

x=-14

x=-6

$x+10=-4 \to$ The inverse operation is subtraction, so subtract $10$ to both sides of the equation.

\begin{aligned} x+10 & =-4 \\\\ -10 & =-10 \end{aligned}

On the left side of the equation $10-10=0.$ So it is, $0+x = x$

On the right side of the equation $-4-10 =-14$

\begin{aligned} x+10&=-4 \\\\ -10&=-10 \\\\ x+0&=-14 \\\\ x&=-14 \end{aligned}

Check:

One Step Equations 16 US

x=6

x=-6

x=\cfrac{1}{6}

x=-\cfrac{1}{6}

$-3x=18 \to$ The inverse operation is division, so divide both sides of the equation by $-3.$

\cfrac{-3 x}{-3}=\cfrac{18}{-3}

On the left side of the equation $\cfrac{-3 x}{-3}=1 x$ and $1 x=x$

On the right side of the equation $\cfrac{18}{-3}=-6$

\begin{aligned} \cfrac{-3 x}{-3}&=\cfrac{18}{-3} \\\\ x&=-6 \end{aligned}

Check:

One Step Equations 17 US

x=-\cfrac{1}{3}

x=-3

x=\frac{2}{3}

x=\cfrac{1}{3}

$\cfrac{1}{3}+x=\cfrac{2}{3} →$ The inverse operation is subtraction, so subtract both sides of the equation by $\cfrac{1}{3}.$

\cfrac{1}{3}-\cfrac{1}{3}+x=\cfrac{2}{3}-\cfrac{1}{3}

On the left side of the equation $\cfrac{1}{3}-\cfrac{1}{3}+x=0+x$ and $0+x=x$

On the right side of the equation $\cfrac{2}{3}-\cfrac{1}{3}=\cfrac{1}{3}$

\begin{aligned} \cfrac{1}{3}-\cfrac{1}{3}+x&=\cfrac{2}{3}-\cfrac{1}{3} \\\\ 0+x&=\frac{1}{3} \\\\ x&=\cfrac{1}{3} \end{aligned}

Check:

One Step Equations 18 US

x=12

x=26

x=-12

x=-26

“A number” is represented by $x.$ Sum means to use the operation of addition.

The equation is: $x+7=19$

$x+7=19 \to$ The inverse operation of addition is subtraction, so subtract $7$ from both sides of the equation.

x+7-7=19-7

On the left side of the equation, $x+7-7 = x+0$ and $x+0$ is $x$

On the right side of the equation, $19-7=12$

\begin{aligned} x+7-7&=19-7 \\\\ x+0&=12 \\\\ x&=12 \end{aligned}

The number is $12.$

Check:

One Step Equations 19 US

-4

4

3

-3

“A number” is represented by $x.$ Quotient means to use the operation of division.

The equation is: $\cfrac{x}{3}=8$

$\cfrac{x}{3}=8 →$ The inverse operation of division is multiplication, so multiply both sides of the equation by $3.$

\begin{aligned} \cfrac{x}{3}&=8 \\\\ \cfrac{x}{3} \cdot 3&=8 \cdot 3 \\\\ \cfrac{x}{3} \cdot \cfrac{3}{1}&=8 \cdot 3 \end{aligned}

On the left side of the equation $\cfrac{3 x}{3}=1 x, 1 x=x$

On the right side of the equation, $8 \cdot 3=24$

\begin{aligned} \cfrac{x}{3} \cdot \cfrac{3}{1}&=8 \cdot 3 \\\\ 1 x&=24 \\\\ x&=24 \end{aligned}

The number is $24.$

Check:

One Step Equations 20 US

-5

5

43

-43

“A number” is represented by $x.$ Difference means to use the operation of subtraction.

The equation is: $x-(-19)=24$

$x-(-19)=24 \to$ The inverse operation to subtraction is addition

x-(-19)=24

\begin{aligned} x-(-19)+(-19)&=24+(-19) \\\\ x&=24+(-19) \\\\ x&=5 \end{aligned}

Check:

\begin{aligned} x-(-19)&=24 \\\\ 5-(-19)&=24 \\\\ 5+19&=24 \end{aligned}

One step equations FAQs

Is the variable always $“x”$ in equations?

No, the variable can be any letter (typically lowercase). The most common variable as you get into higher levels of mathematics is $x.$

When solving equations, do you only use one of the four basic arithmetic operations?

No, as you progress through middle and high school, you will use more than one operation and can also use square root or cube root to solve for the equation.

How do you solve two-step equations and multi-step equations?

The strategies used to solve two-step equations and multi-step equations build upon the strategies used to solve one-step equations. You will use inverse operations (opposite operations) and order of operations in the reverse order.

For example, to find the value of the variable, $x,$ in the equation $2x-1=11,$ you first must move the $-1$ to the opposite side of the equation. Then to get the variable to have a coefficient of $1,$ divide by $2.$

The next lessons are

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