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Backsolving

When the choices are numbers, test them in the problem instead of solving for the unknown.

When to use it

  • The question asks for a value and the four choices are numbers, listed in order.
  • Solving would take algebra you could get wrong: fractions, radicals, reciprocals, or an equation you cannot see how to start.
  • The question reads “which of the following is a solution/value…”.

How it works

  1. Start with a middle choice (B or C). Numbers are usually in order, so one test tells you whether to go bigger or smaller.
  2. Put the number into the original equation or story, not into a rearranged one.
  3. Work the arithmetic on both sides. If both sides match, you are done; if not, cross it off and move toward the side that fixes the gap.
  4. At most three tests settle any four-choice problem. Check the winner once if it was a close call.
Watch out: Testing into your own rewritten equation. A slip in rewriting makes a correct choice fail. Always test in the question as it was printed.

Time: A backsolve costs 20–30 seconds. If you can see a one-step algebra route, take it; use backsolving when you cannot.

Worked examples

Example 1

A rational equation. Cross-multiplying works, but testing is quicker and has no sign traps.

If (x + 3)/(x − 1) = 2, what is x?

  1. A−5
  2. B1
  3. C5Correct
  4. D7

Using the technique

  1. The choices are −5, 1, 5 and 7. Start in the middle with 5.

  2. Put x = 5 into (x + 3)/(x − 1): that is 8/4 = 2. It matches the right side, so 5 is the solution.

  3. Do not bother with the rest. (x = 1 would make the denominator 0, so it could never work.)

Practice this set

Example 2

Fractions on both sides. Backsolving avoids clearing denominators.

If (3x + 1)/4 = (2x − 1)/3, what is x?

  1. A−7Correct
  2. B−1
  3. C1
  4. D7

Using the technique

  1. The choices are −7, −1, 1 and 7. Try x = −1 first: (3(−1) + 1)/4 = −0.5 but (2(−1) − 1)/3 = −1. The sides differ, so −1 is out.

  2. Test −7 next (the choices are in order, and 1 and 7 are far away): (3(−7) + 1)/4 = −20/4 = −5 and (2(−7) − 1)/3 = −15/3 = −5.

  3. Both sides equal −5, so x = −7.

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Example 3

A choice here is a pair of values, so test one value that splits the pairs.

What are the solutions of x² + 6x − 7 = 0?

  1. A−1 and 7
  2. B1 and −7Correct
  3. C1 and 7
  4. D−1 and −7

Using the technique

  1. Every choice contains 1 or −1 and 7 or −7. Test x = 1: 1 + 6 − 7 = 0, so 1 is a solution. That leaves choices B (1 and −7) and C (1 and 7).

  2. Now test the second value. x = 7 gives 49 + 42 − 7 = 84, not 0. x = −7 gives 49 − 42 − 7 = 0.

  3. So the solutions are 1 and −7.

Practice this set

Practice equations in a hard-only Math set

Related lessons: Linear equations · Rational expressions · Quadratic equations