8.EE.7: Solve Linear Equations in One Variable
Also written as 8.EE.C.7 and CCSS.MATH.CONTENT.8.EE.C.7 · Cluster: Analyze and solve linear equations and pairs of simultaneous linear equations.
The Official Wording
Solve linear equations in one variable.
Common Core State Standards, Grade 8, Expressions & Equations · Read it at the source · Checked October 14, 2026
What 8.EE.7 Means in Plain English
Eighth graders solve equations with the variable on both sides, distributing and combining like terms first: 2(x + 3) = 5x - 6 becomes 2x + 6 = 5x - 6, so 12 = 3x and x = 4. Coefficients may be fractions or decimals.
Part a adds the other outcomes: simplifying 2x + 3 = 2x + 7 ends at 3 = 7, so no number works, while ending at 4 = 4 means every number does. High school solves the same equations with letters as coefficients (HSA-REI.3).
8.EE.7 I Can Statements
- I can solve equations with variables on both sides.
- I can distribute and combine like terms before solving.
- I can tell when an equation has one, no or infinitely many solutions.
8.EE.7 Worked Example
One equation that needs the distributive property, then two small changes that give no solution and infinitely many.
Solve (1/2)(6x - 8) + 2x = 3(x + 2). Then say what happens if the right side is changed to 5x + 6, or to 5x - 4.
- Distribute on both sides. Half of 6x is 3x and half of -8 is -4; 3 times x + 2 is 3x + 6.3x - 4 + 2x = 3x + 6
- Collect like terms on the left.5x - 4 = 3x + 6
- Subtract 3x from both sides to get the variable on one side, then add 4 to both sides.2x - 4 = 6, so 2x = 10
- Divide both sides by 2. The equation has reached the form x = a: one solution.x = 5
- With 5x + 6 on the right, subtracting 5x from both sides leaves a false statement.5x - 4 = 5x + 6 becomes -4 = 6: no solution
- With 5x - 4 on the right, the two sides are identical, so every x makes it true.5x - 4 = 5x - 4 becomes -4 = -4: infinitely many solutions
Check it: Substitute x = 5 into the original: (1/2)(30 - 8) + 10 = 11 + 10 = 21, and 3(5 + 2) = 21. Both sides match.
Common Mistakes with 8.EE.7
Distributing to the first term only
3(x + 2) = 3x + 2
Try this: Draw arrows from the 3 to each term inside the parentheses, so 3(x + 2) becomes 3x + 6.
Reading 0 = 0 as x = 0
“5x - 4 = 5x - 4 gives 0 = 0, so x = 0.”
Try this: Read what is left: no x remains and the statement is true, so every value of x is a solution.
Dropping the sign when subtracting a group
5x - (x - 9) = 4x - 9
Try this: Rewrite the minus as -1 times the group: -(x - 9) = -x + 9, so the result is 4x + 9.
Where 8.EE.7 Fits
Practice 8.EE.7
Quick Check
- Solve 4(x - 3) = 2x + 6.
- How many solutions does 3(2x + 1) = 6x + 3 have?
- Solve (2/3)x - 1 = (1/3)x + 4.