How to Check Algebra Answers by Substitution

Why should you check algebra answers by substitution?

Substitution is the quickest way to know whether an answer is right without looking it up. If x = 9 solves an equation, putting 9 in for x has to make both sides equal.

It also works when you’re stuck in an exam with no answer key. A check takes less than a minute, and it catches sign slips, arithmetic errors, and distribution mistakes that you won’t spot just by rereading your working.

Rereading tends to fail because your eyes follow the same path your mind took the first time. Substitution comes at the problem from a different direction. It tests the result directly, so a mistake in the steps has nowhere to hide.

How do you check the solution to an equation?

Substitute your answer into the original equation, work out the left side, work out the right side, and compare. Do the two sides separately; don’t move anything across the equals sign.

Say you solved 3(x − 4) = x + 6 and got x = 9. The left side becomes 3(9 − 4) = 3 × 5 = 15. The right side becomes 9 + 6 = 15. They match, so x = 9 is correct.

What does it look like when an answer is wrong?

The two sides come out different. That proves the answer is wrong, and the size of the gap can hint at what went wrong.

Suppose someone solving 3(x − 4) = x + 6 distributed carelessly and wrote 3x − 4 = x + 6. That leads to x = 5. Checking in the original: 3(5 − 4) = 3, but 5 + 6 = 11. The sides disagree, so x = 5 fails.

Now look back at your first line. Comparing 3(x − 4) with 3x − 4 shows the 3 was never multiplied by the 4. Finding the first line that’s wrong is faster than redoing the whole problem.

Why must you substitute into the original problem?

Because a mistake in an early line is copied into every line after it. If you check against one of those later lines, the wrong answer will pass.

In the example above, x = 5 fits 3x − 4 = x + 6 perfectly: 11 = 11. It only fails in the original 3(x − 4) = x + 6. So copy the problem exactly as it was given, and check there.

How do you check the solutions of a quadratic equation?

Substitute each root separately. A quadratic usually has two solutions, and both have to work.

For x² − 2x − 8 = 0, the roots are x = 4 and x = −2. With 4: 16 − 8 − 8 = 0. With −2: 4 + 4 − 8 = 0. Both give 0, so both roots are right.

With irrational roots like 2 ± √5, substitution by hand gets messy. A good shortcut: the two roots of ax² + bx + c = 0 add up to −b/a. For x² − 4x − 1 = 0 that’s 4, and (2 + √5) + (2 − √5) = 4. The evaluate expression calculator can also plug a value in for you.

How do you check the answer to an inequality?

Test three numbers: the boundary value, one number inside your answer, and one number outside it. The inside number should make the inequality true, and the outside number should make it false.

For 2x − 5 > 7, the answer is x > 6. At the boundary, 2(6) − 5 = 7, so the two sides are equal there. Inside, try 7: 2(7) − 5 = 9, and 9 > 7 is true. Outside, try 0: −5 > 7 is false. Everything agrees.

If the inside number fails and the outside number passes, your inequality sign points the wrong way. That usually means a sign flip was missed or added when dividing by a negative. The article on solving inequalities explains when the sign flips.

How do you check the solution to a system of equations?

Substitute the values into every equation in the system. A solution that only fits one equation isn’t a solution.

For 2x + 3y = 7 and x − y = 1, the solution is x = 2, y = 1. In the first equation, 2(2) + 3(1) = 4 + 3 = 7. In the second, 2 − 1 = 1. Both hold, so (2, 1) is correct.

A common habit is to check only the equation you used last, which tends to pass automatically. Check the other one, because that’s where an error shows up.

Say a slip in elimination gave x = 3, y = 2. The second equation passes, since 3 − 2 = 1. The first fails: 2(3) + 3(2) = 12, not 7. Checking just one equation would have let that wrong pair through.

How do you check a factoring answer?

Expand your factors and compare the result with the original expression. If they match term for term, the factoring is right.

Say you factored x² − x − 12 as (x − 4)(x + 3). Expanding gives x² + 3x − 4x − 12, which is x² − x − 12. That matches. If you had written (x + 4)(x − 3), you’d get x² + x − 12, and the wrong middle sign would stand out at once.

This check confirms the factors are equivalent, but it doesn’t tell you whether you factored completely. Look at each factor once more to see if it breaks down further.

How do you check answers with fractions or absolute values?

The method is the same: substitute and compare each side. Fractions just need a little care with common denominators, and absolute value equations need every solution checked.

For x/3 − 1/2 = 5/6, the solution is x = 4. Substituting gives 4/3 − 1/2. Over a denominator of 6 that’s 8/6 − 3/6 = 5/6, which matches the right side.

For |2x − 3| = 7, the solutions are x = 5 and x = −2. With 5: |10 − 3| = |7| = 7. With −2: |−4 − 3| = |−7| = 7. Both work. People often find the positive case and forget the negative one, and a check makes the missing answer obvious if you test a value from each case.

When should you use a calculator to check?

After you’ve done the problem yourself. Comparing your steps with a worked solution shows you where your method went off track, which a final answer alone can’t do.

The solve for x calculator on this site shows each step for equations and inequalities, and the system of equations calculator handles two to four linear equations. Use them to find the line where your working differs, then redo a similar problem on your own.