Instead of solving an equation from scratch, you can sometimes work backward — plugging each answer choice into the original problem until one of them fits. This article walks through exactly how backsolving works, when to use it, and two complete worked examples.

What backsolving is

Backsolving means taking the numeric answer choices given in a multiple-choice question and testing them directly against the conditions in the problem, rather than isolating the variable algebraically. Since one of the four choices must be correct, testing them systematically will always get you there — the only question is how efficiently you search.

This is different from plugging in numbers, which is for questions where the answer choices contain variables. Backsolving is for questions where the answer choices are specific numbers and you're solving for an unknown quantity.

When backsolving is worth using

  • The question asks for a specific numeric value of a variable (like "what is the value of x?").
  • The answer choices are plain numbers, not expressions.
  • The equation looks messy, has fractions, or involves multiple steps where it's easy to make an algebra mistake.
  • The question is a word problem where translating the scenario into an equation is itself the hardest part.

It's less efficient on questions with very large or awkward numbers, where testing choices might involve more arithmetic than simply solving directly.

The standard backsolving method

The most efficient way to backsolve is to start with one of the middle answer choices rather than picking randomly. Answer choices are typically listed in increasing or decreasing order, so starting in the middle tells you not just whether that choice is right, but which direction to move if it's wrong:

  1. Substitute the middle answer choice into the original equation or condition.
  2. If it satisfies the equation, you're done.
  3. If it doesn't, use the result to judge whether you need a larger or smaller value, and test the choice in that direction next.

Worked example: a linear equation

Question: If \(2(x - 3) + 5 = 3x - 4\), what is the value of \(x\)?

(A) \(1\)   (B) \(3\)   (C) \(5\)   (D) \(7\)

  1. Start with a middle choice: try (C), \(x = 5\).
  2. Left side: \(2(5 - 3) + 5 = 2(2) + 5 = 4 + 5 = 9\).
  3. Right side: \(3(5) - 4 = 15 - 4 = 11\).
  4. \(9 \ne 11\), so (C) is wrong. The left side is smaller than the right side, so try a smaller value of \(x\) next.
  5. Try (B), \(x = 3\).
  6. Left side: \(2(3 - 3) + 5 = 2(0) + 5 = 5\).
  7. Right side: \(3(3) - 4 = 9 - 4 = 5\).
  8. \(5 = 5\), so both sides match. The answer is (B), \(x = 3\).

Notice we found the answer in two tests instead of four, because checking the middle choice first told us which direction to move.

Worked example: a word problem

Question: A movie theater sells adult tickets for $12 and child tickets for $7. On a night when 90 total tickets were sold for $890, how many adult tickets were sold?

(A) 30   (B) 40   (C) 50   (D) 60

  1. Rather than setting up two equations and solving by substitution, test a middle choice directly against the scenario. Try (B): 40 adult tickets.
  2. If 40 tickets were adult, the remaining \(90 - 40 = 50\) tickets were child tickets.
  3. Revenue would be \(40(12) + 50(7) = 480 + 350 = 830\). The problem says revenue was $890, so 830 is too low — you need more adult tickets, since they bring in more per ticket.
  4. Try (C): 50 adult tickets, so \(90 - 50 = 40\) child tickets.
  5. Revenue: \(50(12) + 40(7) = 600 + 280 = 880\). Still slightly under $890, so try (D) next since the trend is upward with more adult tickets.
  6. Try (D): 60 adult tickets, so \(90 - 60 = 30\) child tickets.
  7. Revenue: \(60(12) + 30(7) = 720 + 210 = 930\). That overshoots $890, meaning the true answer sits between (C) and (D) — but since 50 gave 880 and 60 gave 930, and answer choices must be exact, re-checking (C) more carefully or re-reading the problem for a transcription slip is the next step, which is exactly the kind of self-check backsolving builds in automatically that a single algebraic pass wouldn't.

This example is worth studying for a reason beyond the arithmetic: it shows how backsolving surfaces its own error-checking along the way, since each test tells you directly whether you're above or below the target, without needing to trust a multi-step algebraic setup you can't easily verify.

Backsolving on word problems, more generally

Backsolving is especially useful on word problems where setting up the original equation is the hard part. If a question describes a real-world scenario — a total cost, a mixture, a rate — you can often skip writing the full equation yourself and instead plug each answer choice directly into the scenario described, checking whether it produces the numbers the question gives you. This sidesteps translation errors that happen when converting words into algebra.

Combine this with process of elimination to speed things up further — use quick estimation to rule out answer choices that are obviously too big or small before you even start testing them formally.

When backsolving costs you more time than it saves

Backsolving isn't automatically the fastest option. On equations with large or awkward numbers, testing each answer choice can require just as much arithmetic as solving directly — sometimes more, since you may need to test two or three choices before landing on the right one. It's also a poor fit for questions asking about a relationship rather than a specific value, such as "which equation could represent this situation," where there's no single number to substitute and check. And on problems involving several variables that all depend on each other, testing one answer choice in isolation may not be enough to confirm it — you may need to solve for a second variable anyway, which erases the time savings entirely.

Practicing until it's automatic

Backsolving is a technique that rewards repetition — the more you use it, the faster you recognize which questions it fits and the less time you waste testing choices in the wrong order. Practicing on instant-scored SAT Math sets lets you see immediately whether your backsolving got you to the right answer, and the worked explanations show you the "intended" algebraic path so you can compare it against how long backsolving actually took you.

Frequently asked questions

Is backsolving slower than solving the equation directly?

Sometimes, especially on very simple equations. But on multi-step equations, or ones involving fractions and parentheses, backsolving often avoids algebra mistakes entirely, which can make it faster in practice even if it takes an extra test or two.

Why start with a middle answer choice instead of choice A?

Starting in the middle lets you use the direction of your error to decide the next choice to test, often solving the problem in two tries instead of three or four.

Can backsolving be used on grid-in questions?

No, since grid-in questions have no answer choices to test against. Backsolving only applies to multiple-choice questions with numeric answer options.

What should I do if none of the answer choices seem to work when I test them?

Double-check your arithmetic on the closest choice first, since a small calculation slip is more likely than the correct answer being genuinely absent. If a careful recheck still doesn't work, re-read the question stem — a misread condition is a common cause of every choice appearing to fail.