The two correct answers are a. −6x + 15 and c. 15 − 6x. Distributing −3 across (2x − 5) gives −3·2x = −6x and −3·(−5) = +15, so the simplified form is −6x + 15. Because addition is commutative, 15 − 6x names the same value, which is why exactly two of the five options qualify.
The distribution mechanism behind −3(2x − 5)
The expression −3(2x − 5) is a product of a factor and a binomial. The distributive property rewrites it as the sum of two separate products: multiply −3 by each term inside the parentheses, then combine.
Work the signs carefully. The first product is −3 · 2x = −6x, a negative coefficient on x. The second product is −3 · (−5), and multiplying two negatives yields a positive 15. That second sign is where most errors happen, and it is the hinge the whole problem turns on.
So the expression expands to −6x + 15. This is not a new number; it is the same quantity written differently. For any value of x, the original and the expanded form return identical results, which is the definition of equivalence.
A quick numeric check confirms it. Let x = 1: the original gives −3(2 − 5) = −3(−3) = 9, and −6(1) + 15 = 9. Let x = 0: the original gives −3(−5) = 15, and −6(0) + 15 = 15. Both forms agree at every test point, so they are equivalent expressions.
What each answer choice evaluates to
Each option is a candidate rewrite, and only some preserve the value. Some differ by a sign, and others rearrange terms without changing the total. The table below runs every choice through a spot check at two values of x.
| Choice | Expression | Check at x = 1 | Check at x = 0 | Verdict |
|---|---|---|---|---|
| a | −6x + 15 | 9 | 15 | Equivalent |
| b | −6x − 15 | −21 | −15 | Not equivalent (sign of constant) |
| c | 15 − 6x | 9 | 15 | Equivalent |
| d | −6x − 5 | −11 | −5 | Not equivalent (constant and middle term) |
| e | 6x − 15 | −9 | −15 | Not equivalent (both signs flipped) |
Choice a matches the expansion exactly: −6x + 15. Choice c is the same two terms in reverse order, since 15 − 6x equals −6x + 15. Choice b keeps the −6x but makes the constant negative, flipping one sign. Choice d mishandles the multiplication of −3 · (−5), treating it as −5 instead of +15. Choice e reverses the sign of the entire expression, producing the negative of the correct answer.
Only a and c preserve the value at every x. That is the two-answer requirement satisfied, and it explains why the question asks for two rather than one.
Sign rules and the commutative property
Two algebraic principles govern this item, and naming them makes the reasoning portable to harder problems.
The first is the rule for multiplying signed numbers. A negative times a positive is negative, giving −6x. A negative times a negative is positive, giving +15. The phrase "minus times minus is plus" is the entire reason the constant is +15 and not −15, so choice d fails on this single step.
The second is the commutative property of addition. The expression −6x + 15 can be written 15 + (−6x), which appears as 15 − 6x. Order does not change the sum, so these are identical quantities. Choice c exploits this freedom of arrangement.
A useful analogy is debt and credit. A −6x term is like a recurring debt that grows with x, while +15 is a fixed credit. Whether you state the credit first or last, the net position is unchanged. Affirming that arithmetic order is safe is what lets choice c stand alongside choice a.
Edge cases: when answers look different but are equal
Students often mark c as wrong because it "looks different" from a. The surface arrangement differs, but the underlying value does not. Equivalence is about value for every input, not about visual matching.
A related trap is treating −6x + 15 and 15 − 6x as if the subtraction sign attaches to the wrong thing. In 15 − 6x, only the 6x is subtracted, so the term is −6x. Confusing this with −(15 − 6x), which would be −15 + 6x, is the error behind choice e.
Fractions and negative x values behave the same way. Test x = ½: the original gives −3(1 − 5) = 12, and −6(0.5) + 15 = 12. Test x = −2: the original gives −3(−4 − 5) = 27, and −6(−2) + 15 = 27. The equivalence holds across the entire real line, not just convenient integers.
Common misconceptions that produce wrong answers
Most wrong answers trace to one of a few predictable mistakes. Recognizing them turns a guessing problem into a diagnostic one.
Forgetting to multiply the constant term, leaving −6x then copying −5 unchanged. This produces choice d. Dropping the negative sign on the second product, writing −3 · (−5) = −15. This produces choice b. Distributing only partway and stopping before simplifying signs. Believing a rearranged expression must be wrong because it does not match the expanded form visually. This wrongly eliminates choice c. Flipping the sign of the whole expression by factoring out a negative incorrectly. This produces choice e. Applying the distributive property only to the x term and treating the parentheses as a grouping symbol that must remain.
The counterintuitive detail: choice c is the correct answer many students eliminate fastest, precisely because it looks least like their expanded result. Ordered differently but equal in value, it tests whether you understand equivalence as a numerical property rather than a spelling test. Professionals check equivalence by substituting a value, not by eye.
Practical actions for checking equivalence reliably
A repeatable procedure removes the doubt. Use these steps on any "select all equivalent expressions" item.
Distribute fully, multiplying the outside factor by every term inside the parentheses, including the constant. Simplify signs before combining, so −3 · (−5) becomes +15 immediately. Substitute two test values, ideally x = 0 and x = 1, into both the original and each candidate. Equal outputs at both are strong evidence of equivalence. Accept rearrangements that preserve value, such as 15 − 6x, and reject sign changes such as 6x − 15. Match your expanded form against the choices and count, confirming exactly two survive.
Substituting x = 0 is especially efficient. At x = 0 every x term vanishes, so the output equals the constant term. The original gives 15, which immediately eliminates choices b, d, and e, whose constants are −15, −5, and −15. That single substitution narrows five options to two in seconds.
This method generalizes to harder items with three or more correct answers. The constant-term test still screens for sign errors on constants, and a second substitution at x = 1 catches coefficient errors on x.
Limits of the shortcut and what it cannot prove
Substitution is a screening tool, not a proof. Matching outputs at two points is strong evidence but not a guarantee of equivalence for all x; a genuine proof distributes and simplifies symbolically. For multiple-choice items with linear expressions, two well-chosen test points reliably separate the correct answers, but a rigorous answer justifies the equivalence by algebra.
The problem is also specific to linear expressions. Nonlinear expressions can agree at several points and still differ elsewhere, so substitution becomes riskier without symbolic confirmation. When in doubt, expand fully and compare coefficients term by term, which is the definitive equivalence test.
One more caveat: equivalence is about value, not about matching "simplified form" as a convention. Two expressions can be equivalent while only one is fully simplified. Accept equivalent arrangements like 15 − 6x even though −6x + 15 is the conventional order, because the question tests value, not formatting preference.
This is a general educational explanation of algebraic equivalence, not a diagnostic or tutoring assessment, and no test outcome is promised. For help with a specific course or placement requirement, consult a qualified local instructor.