Digital SAT Math · Algebra
Equivalent linear expressions
When two linear expressions are the same for every input, their -parts must agree and their constants must agree — separately. That comparison is the whole skill: expand, collect, then match coefficients. It turns one opaque equation into two one-step equations, and it is also how you find the constant that forces no solution or infinitely many solutions. You are not solving for a number ; you are deciding when two sides are the same object.
On the test
| Domain | Algebra (score report) |
| CB skill | Linear equations in one variable (identity and parameter items) |
| What it looks like | Two linear expressions set equal, often with a letter , , or standing for a constant; or “which expression is equivalent to…” with four linear rewrites |
| Often asked | “The equation is true for all values of . What is the value of ?”, “For what value of does the equation have no solution / infinitely many solutions?”, “Which expression is equivalent to…?”, “. What is ?” |
| Format | Multiple choice and student-produced response |
| Calculator | Allowed throughout; on this skill pencil-and-coefficient-match is almost always faster than Desmos |
Recognition cues: for all values of , where is a constant, infinitely many solutions, no solution, equivalent to, rewritten as , and parentheses with a minus sign in front of them.
What this page owns
Solving for a numeric — isolate the variable and report a number — lives on linear equations in one variable. This page is the identity and parameter half of the same CB skill: the sides are meant to match as expressions, or a constant is dialed until they do. Nonlinear rewrites (factoring a quadratic, rational cancel, completing the square) are equivalent expressions (nonlinear) in Advanced Math.
Pattern recognition
Four shapes cover essentially every item:
- Identity / for all — a letter stands in for a coefficient or constant, and the equation holds for every . Match coefficients; never solve for .
- Parameter with a solution count — “for what value of … no solution” or infinitely many. Same comparison as (1); the stem tells you which outcome to force.
- Rewrite as — expand and collect a linear expression; report , , or a combination like .
- Which is equivalent — four linear choices; expand the stem once (or expand each choice) and match.
Method
- Name the job. Identity / parameter / rewrite / which-is-equivalent. Underline the ask: a value of , a solution count, or an expression .
- Expand both sides fully. Distribute every factor onto every term inside the parentheses. A leading minus flips every sign inside, not only the first term.
- Collect like terms. One -term and one constant on each side: .
- Match coefficients.
- -coefficients:
- Constants: Those are the two equations that answer “for all ” and that set a parameter for infinitely many solutions.
- Read no-solution / infinitely many from the same comparison.
| After expand and collect | Outcome |
|---|---|
| exactly one solution (this page rarely asks you to find it — that is the one-variable solve skill) | |
| and | infinitely many solutions (the two sides are the same expression) |
| and | no solution (same -part, different constants → a false number statement) |
For a parameter item, reverse the table: to force infinitely many, set and ; to force no solution, set and check that .
| Words in the stem | What to do |
|---|---|
| for all values of / identity | expand, equate -coeffs, equate constants |
| for what value of … infinitely many | force and |
| for what value of … no solution | force , confirm |
| equivalent to / what is | expand and collect; report the combination asked |
| which expression is equivalent | expand stem (or expand choices) and match |
Worked example 1 — identity: find the constant
Stem. The equation is true for all values of , where is a constant. What is the value of ?
Step 1 — name the job. “True for all values of ” means the two sides are the same linear expression. Do not solve for .
Step 2 — expand the side with the letter.
The equation is now .
Step 3 — match coefficients.
- -coefficients:
- Constants:
Step 4 — solve the second equation as a check, not as a second unknown.
Both equations give , so the identity holds.
Check. Substitute : right side is , identical to the left. At , both sides equal ; at , both equal . Answer: .
Trap watch. Matching only the -coefficient and never checking the constants works here only because the stem was built as a true identity — on items where the two equations disagree, the constant equation is the one that exposes the impostor. Solving for (picking a number and backing out ) can produce a value that works at one point and fails at another. Dividing both sides by is illegal for an identity: it throws away .
Worked example 2 — parameter: infinitely many vs no solution
Stem. In the equation , is a constant. For what value of does the equation have infinitely many solutions? What is true for every other value of ?
Step 1 — expand the left side fully.
The equation is .
Step 2 — compare coefficients. The -coefficients already match () for every . After subtracting from both sides, only the constants remain:
Step 3 — read both outcomes from the same comparison.
- If , then (true for every ) → infinitely many solutions.
- If , then is false → no solution.
There is no value of that produces exactly one solution, because the -terms cancel before is chosen.
Check. With , both sides are ; , , and all work. With , the equation becomes , so , false for every . Answer: gives infinitely many solutions; every other value of gives no solution.
Trap watch. Swapping the labels — calling “no solution” because “the cancelled” — is the classic mix-up; cancellation only says you are in the special-case branch, and the constants decide which special case. Distributing as (the 5 hits only the first term) produces the impostor . Reporting a value of instead of answers a question nobody asked.
Worked example 3 — rewrite with a leading minus
Stem. The expression is equivalent to , where and are constants. What is the value of ?
Step 1 — distribute the factor onto every term inside.
(The second term is , not : a minus times a minus.)
Step 2 — combine with the leading 6.
So and .
Step 3 — finish the ask.
Check. At : original ; rewrite . At : both sides equal . Answer: .
Trap watch. Distributing the over only the first term — writing — is the named partial-negative trap; then , which will be a listed choice. Flipping only the and leaving the alone gives the same way. Reporting or alone answers half the question when the stem asked for .
Practice
Answer before you open the explanation. Items will mix identities, parameter solution-counts, and linear rewrites; two will be student-produced response (type the number, no choices). Every wrong choice is a specific named trap from this page.
Question 1 Warm-up
Which of the following expressions is equivalent to 3(2x + 5) - 4?
Show the answer Choice C
Why it is right
Distribute the 3 onto both terms inside the parentheses: 3(2x + 5) = 6x + 15. The expression is then 6x + 15 - 4, and combining the two constants gives 6x + 11. Check the rewrite at two inputs: at x = 1 the original is 3(7) - 4 = 17 and the rewrite is 6 + 11 = 17; at x = 0 both are 11.
Why each other choice fails
- Choice A
- 2x + 11 multiplies the 3 onto the constant 5 only and leaves the 2x untouched: 2x + 15 - 4. A factor outside parentheses multiplies every term inside, the x-term included.
- Choice B
- 6x + 1 distributes the 3 over the first term only, writing 6x + 5 - 4. The 5 has to be multiplied by 3 as well, which turns it into 15 before the 4 is subtracted.
- Choice D
- 6x + 19 adds the 4 instead of subtracting it: 6x + 15 + 4. The sign in front of a term travels with it, so the trailing - 4 is still a subtraction after the distribution.
Question 2 Standard
The equation 6x + 4 = a(x + 2) - 8 is true for all values of x, where a is a constant. What is the value of a?
Show the answer Choice D
Why it is right
Expand the right side: a(x + 2) - 8 = ax + 2a - 8. Because the equation holds for every x, the two sides are the same linear expression, so the x-coefficients must agree and the constants must agree separately. From the x-terms, a = 6. From the constants, 4 = 2a - 8, so 2a = 12 and a = 6 as well. Substituting a = 6 gives 6(x + 2) - 8 = 6x + 12 - 8 = 6x + 4, identical to the left side.
Why each other choice fails
- Choice A
- -2 comes from a sign error moving the -8 across the equals sign: writing 4 - 8 = 2a instead of 4 + 8 = 2a gives 2a = -4 and a = -2. The -8 is subtracted on the right, so it is added to both sides to undo it.
- Choice B
- 2 matches 4 with 2a and drops the -8 entirely. Every constant on the right belongs to the constant equation; the -8 sits outside the parentheses but is still part of the right side, so the comparison is 4 = 2a - 8.
- Choice C
- 3 reads the coefficient of x on the right as 2a, as though the 2 inside the parentheses multiplied x as well. Expanding a(x + 2) gives ax + 2a, so the 2 lands on the constant term and the x-comparison is 6 = a, not 6 = 2a.
Question 3 Standard
The expression 8 - 4(x - 3) is equivalent to ax + b, where a and b are constants. What is the value of a + b?
Show the answer Choice C
Why it is right
Distribute -4 onto both terms inside the parentheses: -4(x - 3) = -4x + 12, because a negative times a negative is positive. The expression becomes 8 - 4x + 12 = -4x + 20, so a = -4 and b = 20 and a + b = 16. Check at x = 1: the original is 8 - 4(-2) = 16 and the rewrite is -4 + 20 = 16.
Why each other choice fails
- Choice A
- -8 keeps the second term negative, writing -4(x - 3) = -4x - 12 and so 8 - 4x - 12 = -4x - 4, giving a + b = -4 + (-4) = -8. Multiplying -4 by -3 gives +12, since two negatives make a positive product.
- Choice B
- 1 distributes the -4 over the first term only: 8 - 4x - 3 = -4x + 5, so a + b = -4 + 5 = 1. The factor in front of the parentheses multiplies every term inside, not just the one nearest to it.
- Choice D
- 24 drops the sign on the x-term, reading the rewrite as 4x + 20 and adding 4 + 20. The subtraction in front of 4(x - 3) makes the coefficient of x negative, so a = -4.
Question 4 Standard
In the equation 3(2x + a) = 6x + 21, a is a constant. For what value of a does the equation have infinitely many solutions?
Show the answer Choice A
Why it is right
Expand the left side: 3(2x + a) = 6x + 3a. The x-coefficients are already equal (6 = 6) whatever a is, so the solution count is decided by the constants alone. Infinitely many solutions means the two sides are the same expression, which requires 3a = 21, so a = 7. With a = 7 the equation reads 6x + 21 = 6x + 21, true for every x; any other value of a leaves a false numerical statement and therefore no solution.
Why each other choice fails
- Choice B
- 18 subtracts 3 from 21 instead of dividing, treating 3a as 3 + a. The 3 multiplies a after the distribution, so it is undone by division: a = 21/3 = 7.
- Choice C
- 21 distributes the 3 over the first term only, reading 3(2x + a) as 6x + a, and then matches a directly with 21. The factor outside multiplies both terms inside, so the constant on the left is 3a, not a.
- Choice D
- 63 multiplies 21 by 3 instead of dividing by it. That reverses the operation needed to undo 3a = 21; substituting a = 63 makes the left side 6x + 189, nowhere near the right side.
Question 5 Standard
How many solutions does the equation 3(2x + 5) = 6x + 15 have?
Show the answer Choice D
Why it is right
Expand the left side: 3(2x + 5) = 6x + 15. The equation becomes 6x + 15 = 6x + 15, so the two sides are the same linear expression. The x-coefficients agree (6 = 6) and the constants agree (15 = 15), which is exactly the condition for an identity. Subtracting 6x from both sides leaves 15 = 15, a statement that is true no matter what x is, so every real number is a solution.
Why each other choice fails
- Choice A
- No solution swaps the two special cases. The x-terms cancelling only says the equation is in the special-case branch; the constants decide which case it is. Here the leftover statement 15 = 15 is true, and no solution would require the leftover constants to disagree.
- Choice B
- Exactly one solution assumes that any equation containing x must pin x down to a single value. Once both sides carry the same x-term there is nothing left to isolate, so the count can only be none or infinitely many.
- Choice C
- Exactly two solutions counts the two coefficient comparisons, x-terms and constants, as though each produced a solution. Those are two conditions on the expressions, not two values of x, and a linear equation can never have exactly two solutions.
Question 6 Harder
In the equation a(x - 4) = 5x + 12 - 2x, a is a constant. For what value of a does the equation have no solution?
Show the answer Choice B
Why it is right
Collect the right side first: 5x + 12 - 2x = 3x + 12. Expanding the left gives ax - 4a. No solution means the x-terms match while the constants disagree, so a = 3 is required. With a = 3 the constant on the left is -4(3) = -12 and the constant on the right is 12, and -12 does not equal 12, so the equation reduces to the false statement -12 = 12 and no value of x satisfies it.
Why each other choice fails
- Choice A
- -3 matches the constants instead of the x-coefficients, solving -4a = 12. That is the condition for the constant terms to agree, which is the wrong half of the comparison: no solution needs the x-coefficients equal and the constants different.
- Choice C
- 5 reads the coefficient of x on the right as 5 without collecting the -2x that trails the constant. The right side is 5x - 2x + 12 = 3x + 12, so the coefficient to match is 3, not 5.
- Choice D
- 7 collects 5x and -2x as 7x, adding the coefficients instead of subtracting. The -2x is subtracted, so the right side's x-coefficient is 5 - 2 = 3.
Question 7 Harder Student-produced response
The expression 5(2x - 3) - 3(x - 4) is equivalent to ax + b, where a and b are constants. What is the value of a + b?
Show the answer 4
Why it is right
Distribute each factor onto every term inside its parentheses: 5(2x - 3) = 10x - 15, and -3(x - 4) = -3x + 12 because -3 times -4 is +12. Adding the two results gives 10x - 15 - 3x + 12 = 7x - 3, so a = 7 and b = -3 and a + b = 4. Check at x = 2: the original is 5(1) - 3(-2) = 11 and the rewrite is 7(2) - 3 = 11.
Answers students type instead
- 7
- Reports a alone. The rewrite is correct, but the question asks for the sum a + b, so b = -3 still has to be added.
- -12
- Distributes the -3 over the first term only: 10x - 15 - 3x - 4 = 7x - 19, so a + b = -12. The -3 multiplies the -4 as well, and that product is +12.
- -3
- Reports b alone, the constant term of the rewrite, rather than the requested sum a + b.
Question 8 Harder
The equation 4(x + 3) + x + 3 = 2x + k(x + 5) is true for all values of x, where k is a constant. What is the value of k?
Show the answer Choice A
Why it is right
Collect each side into the form (number)x + (number). The left side is 4x + 12 + x + 3 = 5x + 15. The right side is 2x + kx + 5k = (2 + k)x + 5k. For the equation to hold for every x the x-coefficients must agree and the constants must agree. From the x-terms, 5 = 2 + k, so k = 3. From the constants, 15 = 5k, so k = 3 as well; both comparisons give the same value, which confirms the identity.
Why each other choice fails
- Choice B
- 5 matches the left coefficient 5 with k directly and ignores the 2x already sitting on the right. The right side's x-coefficient is 2 + k, so the comparison is 5 = 2 + k and k = 3.
- Choice C
- 15 matches the left constant 15 with k and forgets that k is multiplied by the 5 inside the parentheses. Expanding k(x + 5) gives kx + 5k, so the constant comparison is 15 = 5k.
- Choice D
- 75 multiplies 15 by 5 instead of dividing when undoing 15 = 5k. Because 5k means 5 times k, k is recovered by dividing: k = 15/5 = 3.
Question 9 Harder
In the equation 6x + c = 2(3x + 4), c is a constant. Which of the following statements about the solutions of the equation is true?
Show the answer Choice A
Why it is right
Expand the right side: 2(3x + 4) = 6x + 8. The equation is 6x + c = 6x + 8, and the x-coefficients are equal for every c, so subtracting 6x from both sides leaves c = 8. When c = 8 that leftover statement is true for every x, so the equation has infinitely many solutions; when c is anything else the statement is false for every x, so the equation has no solution. No value of c gives exactly one solution, because the x-terms cancel before c is chosen.
Why each other choice fails
- Choice B
- This swaps the two outcomes. Matching x-coefficients puts the equation into the special-case branch, and the constants decide which case: equal constants give infinitely many solutions and unequal constants give none, not the other way round.
- Choice C
- c = 4 comes from distributing the 2 over the first term only, reading 2(3x + 4) as 6x + 4. The 2 multiplies the 4 as well, so the constant on the right is 8 and the identity value of c is 8.
- Choice D
- Exactly one solution would be right if the two sides had different x-coefficients, but both sides carry exactly 6x. The variable cancels, so no single value of x can ever be isolated for any c.
Question 10 Hardest
The equation 3(2x - 5) + 4x = a(x + 2) + b is true for all values of x, where a and b are constants. What is the value of a + b?
Show the answer Choice B
Why it is right
Collect the left side: 3(2x - 5) + 4x = 6x - 15 + 4x = 10x - 15. Expand the right side: a(x + 2) + b = ax + 2a + b. Matching x-coefficients gives a = 10. Matching constants gives 2a + b = -15, and with a = 10 that is 20 + b = -15, so b = -35. Therefore a + b = 10 + (-35) = -25. Check: 10(x + 2) - 35 = 10x + 20 - 35 = 10x - 15, identical to the left side.
Why each other choice fails
- Choice A
- -35 reports b alone. Both constants were found correctly, but the question asks for the sum a + b, so the coefficient a = 10 still has to be added to -35.
- Choice C
- -5 takes b to be the left-hand constant -15 and adds it to a = 10. The constant on the right is 2a + b, not b by itself, so the -15 must first be reduced by 2a = 20 to isolate b.
- Choice D
- 10 reports a alone, the x-coefficient. That is only half of the rewrite; the constant b = -35 is part of the requested sum a + b.
Question 11 Hardest Student-produced response
In the equation a(4x + 6) = 10x + 15, a is a constant. For what value of a does the equation have infinitely many solutions?
Show the answer 5/2
Why it is right
Expand the left side: a(4x + 6) = 4ax + 6a. Infinitely many solutions means the two sides are the same expression, so the x-coefficients must agree and the constants must agree. From the x-terms, 4a = 10, so a = 10/4 = 5/2. From the constants, 6a = 15, so a = 15/6 = 5/2 as well. Both comparisons give the same value, so a = 5/2 makes the left side 10x + 15, identical to the right side.
Answers students type instead
- 6
- From 4a = 10, subtracting 4 instead of dividing by it gives 6. The 4 multiplies a after the distribution, so it is undone by division.
- 10
- Reads the x-coefficient on the left as a rather than 4a, matching a straight to 10. Distributing a over 4x gives 4ax, so the comparison is 4a = 10.
- 0.4
- Inverts the constant comparison, computing 6/15 = 0.4 instead of 15/6. The equation from the constants is 6a = 15, so a = 15/6.
Question 12 Hardest
In the equation p(x + 4) = 6x + q, p and q are constants. Which of the following statements is true?
Show the answer Choice D
Why it is right
Expanding the left side gives px + 4p, so the equation is px + 4p = 6x + q. The two sides are the same expression exactly when the x-coefficients agree, p = 6, and the constants agree, 4p = q, which forces q = 24. With p = 6 and q = 24 both sides read 6x + 24, so every value of x satisfies the equation and there are infinitely many solutions. Any other pair either leaves different x-coefficients, giving exactly one solution, or equal x-coefficients with different constants, giving none.
Why each other choice fails
- Choice A
- This matches the x-coefficients and stops there. With p = 6 and q = 10 the equation is 6x + 24 = 6x + 10, whose constants disagree, so it has no solution rather than infinitely many; an identity needs both comparisons to hold.
- Choice B
- This swaps the two constants. With p = 24 and q = 6 the equation is 24x + 96 = 6x + 6, and the x-coefficients differ, so it has exactly one solution, x = -5, not infinitely many.
- Choice C
- This swaps the two outcomes for the correct pair. With p = 6 and q = 24 both sides are 6x + 24, so the leftover statement 24 = 24 is true for every x. Matching constants mean infinitely many solutions; no solution is the case where the constants disagree.
Common mistakes
- Distributing a negative over only the first term — written as instead of . The factor outside multiplies every term inside; a leading minus flips every sign.
- Equating constants but forgetting the -coefficient — setting the plain numbers equal and never checking that the coefficients of also match (or the reverse). An identity needs both.
- Swapping “no solution” and “infinitely many” — both require matching -coefficients; the constants decide. Same constants → infinitely many; different constants → none.
- Dividing by a variable expression — cancelling from both sides of an identity and losing the requirement that the sides agree at . Expand and match instead.
- Solving for on an identity item — producing a number for the variable when the stem asked for the constant , or when every is supposed to work.
- Reporting when the ask was (or the reverse) — finishing the expand correctly and then answering the wrong combination.
- Stopping after one of the two coefficient equations — matching -terms, writing down , and never confirming the constants agree (the place the test hides a disagreement).
- Treating “the variable cancelled” as automatic no solution — cancellation only puts you in the special-case branch; you still have to read whether the remaining statement is true or false.
FAQ
How is this different from “linear equations in one variable”? That page is about finishing a solve: isolate (or an expression built from ) and report a number. This page is about identities and parameters: the sides are meant to be the same expression for every , or a constant is chosen so that the solution count is none or infinitely many. The algebra of distributing and collecting is shared; the ask is not.
How is this different from Advanced Math “equivalent expressions”? Here the expressions stay linear — degree 1 after expand. There you factor quadratics, cancel rational factors, complete the square, and rewrite radicals. Matching coefficients appears in both places; on this page there is only an -term and a constant to match.
Can I pick a number for and backsolve for ? On multiple choice, substituting one convenient can eliminate choices, but a single point never proves an identity — two different linear expressions can agree at one and disagree everywhere else. Safer: expand and match both coefficients. On student-produced response there is nothing to test against, so the coefficient match is required.
What if the two coefficient equations disagree? Then no value of the constant makes the equation true for all . On a “for all ” item that is a signal you distributed wrong. On a solution-count item it simply means that value is not the infinite-solutions case — check whether it still produces no solution (same -coefficient, different constants).
Do I need Desmos? Optional check only. Graph both sides or slider-drag the constant to confirm. The coefficient comparison is finished before the graph loads, and it is the method that works when the answer is a grid-in for .
Fraction or integer for a grid-in? Enter the exact value. If matching coefficients produces a fraction, enter the fraction; do not round.