math class 678 4-in-1 skill sheets
8.EE.C.7a 8th Grade Expressions & Equations

One, None, or Infinite Solutions

Give examples of linear equations in one variable with one solution, infinitely many solutions, or no solutions.

How to explain it

At this standard, students classify the solution type of a linear equation in one variable — one solution, no solution, or infinitely many solutions — by solving or simplifying until variables cancel or a unique value is found.

The anchor students hold onto: Solve normally. If variables cancel and the residual is TRUE (e.g. 5 = 5), infinite solutions; if FALSE (e.g. 3 = 7), no solution; otherwise x = a.

Students need this for 8.EE.C.8 systems of linear equations — systems can have one solution (intersecting), no solution (parallel lines), or infinitely many (coincident lines).

Worked examples

Example 1 One Solution
5x − 3 = 2x + 9
Step 1Subtract 2x: 3x − 3 = 9.
Step 2Add 3: 3x = 12.
Step 3Divide by 3: x = 4.
Step 4One solution: x = 4.
Answerx = 4
Example 2 No Solution
2x + 3 = 2x + 7
Step 1Subtract 2x: 3 = 7.
Step 23 = 7 is FALSE.
Step 3No solution.
AnswerNo solution
Example 3 Infinitely Many
3(x + 2) = 3x + 6
Step 1Distribute: 3x + 6 = 3x + 6.
Step 2Subtract 3x: 6 = 6.
Step 36 = 6 is TRUE.
Step 4Infinitely many solutions.
AnswerInfinitely many solutions

Common mistakes

What students write Variables cancel to 15 = 14, and a student writes x = 15 − 14 = 1 as the answer.
The fix When variable terms cancel, there is no x to isolate. The remaining statement is 15 ≠ 14 (false), so the answer is No Solution — not x = 1.
Try this Alex solved 5(x + 3) = 5x + 14. [Alex's work] 5x + 15 = 5x + 14 15 = 14 x = 15 − 14 = 1 Describe Alex's error and state the correct answer.
What students write Distributing 3(x + 2) as 3x + 2 (only first term) and reaching the wrong solution type.
The fix Distribute to EVERY term inside: 3(x + 2) = 3x + 6, not 3x + 2. An incorrect distribution changes the solution type entirely.

Teacher tip

Head off the two predictable errors before they happen. First: When variable terms cancel, there is no x to isolate. The remaining statement is 15 ≠ 14 (false), so the answer is No Solution — not x = 1. Second: Distribute to EVERY term inside: 3(x + 2) = 3x + 6, not 3x + 2. An incorrect distribution changes the solution type entirely.