The idea
A linear equation in one variable contains only first-degree terms in that variable after simplification. A solution makes its two sides equal. Adding the same quantity to both sides, or multiplying or dividing both sides by the same nonzero number, preserves its solutions.
Compare the conditions
After expanding a linear equation into Ax + B = Cx + D, compare both the coefficients and the constants.
- A ≠ C
- Exactly one solution: collect the x-terms and divide by the nonzero coefficient.
- A = C and B = D
- Infinitely many solutions: every x makes the equation true.
- A = C and B ≠ D
- No solution: the x-terms cancel, leaving a false constant statement.
A step-by-step method
- Distribute multiplication across every term inside parentheses.
- Combine like terms, then add or subtract the same quantity on both sides to collect the variable terms.
- If the remaining variable coefficient is nonzero, divide by it and check the solution. If the variable terms cancel, a true constant statement gives infinitely many solutions and a false one gives no solution.
- When a linear equation with unknown constants must have infinitely many solutions, match both the variable coefficients and constant terms after expansion.
Worked example
Attempt this original question, then open the reasoning to compare your approach.
For constants a and b, the equation a(3x − 2) = 12x + b has infinitely many solutions for x. What is a + b?
- −12
- −4
- 4
- 12
Show answer and reasoning
Answer: B · −4
Expand the left side: 3ax − 2a = 12x + b. Infinitely many solutions require the two linear expressions to be identical for every x, so both the x-coefficients and the constant terms must match.
Compare coefficients: 3a = 12, so a = 4. Compare constants: b = −2a = −8.
Therefore a + b = −4. Substitution gives 12x − 8 = 12x − 8, an identity true for every x. Checking just one chosen x would not establish infinitely many solutions. By contrast, equal x-coefficients with different constant terms would give no solution.
Why each option works or fails
- −12 is a difference (−8 − 4), not the requested sum a + b.
- Correct: a = 4 and b = −8, so a + b = −4.
- 4 is a alone. The constant term must also match.
- 12 results from using b = +8; distributing a gives the constant −2a, not +2a.
Try a new question
Work independently before opening the explanation. Check what changed and whether the same rule still applies.
For the equation 3x + 12 = k(2x + 8) + 5, k is a constant. Which value of k makes the equation have no solution for x?
- 0
- 1
- 3/2
- 3
Show answer and reasoning
Answer: C · 3/2
Expand and collect terms: (3 − 2k)x = 8k − 7. A nonzero x-coefficient gives exactly one solution, so no solution requires 3 − 2k = 0.
This gives k = 3/2. The right side is then 8(3/2) − 7 = 5, so the equation becomes 0x = 5, which is impossible.
Equivalently, substitution in the original equation gives 3x + 12 = 3x + 17. Matching x-coefficients creates no solution only when the constant terms differ; equal constants would instead give infinitely many solutions.
Why each option works or fails
- For k = 0, the equation becomes 3x + 12 = 5 and has the solution x = −7/3.
- For k = 1, it becomes 3x + 12 = 2x + 13, so x = 1.
- Correct: k = 3/2 gives equal x-coefficients but unequal constants, leaving a contradiction.
- For k = 3, it becomes 3x + 12 = 6x + 29, so x = −17/3.
Common mistakes
- Infinitely many solutions does not mean that you should find one particular value of x. It requires an identity after simplification.
- Matching only the x-coefficient is insufficient: the constant terms must also match.
- Checking one or a few chosen values is not a general proof. Matching the coefficients of these linear expressions establishes the identity.
Sources and related guides
This guide was prepared with AI assistance. Its rules and worked examples were checked against the references below; no independent expert review is claimed. Examples are original and are not official SAT questions.
Revision: Original examples compare identities with equations that have no solution.