To check a solution, substitute your answer into the original equation, evaluate the left side and the right side separately, and confirm they are equal. This habit turns a doubtful answer into a verified one and gives you confidence in any algebra question.
It closes the route through equations and formulas and pairs with every lesson before it.
What is the idea behind it?
A solution is a value that makes the original relationship true. So the only test that matters is whether the original relationship is true when that value is put in.
Checking is not repeating the solution. It is a different calculation that approaches the answer from the other direction, which is why it catches mistakes that redoing the same steps would miss.
How to do it, step by step
- Copy the original equation (not a later line).
- Substitute your value for every occurrence of the letter, using brackets round negatives.
- Evaluate the left side completely on its own.
- Evaluate the right side completely on its own.
- Compare. Equal means correct. Different means go back.
- For a word problem, also ask whether the answer fits the story.
Worked example
Solve 7 − 2(x − 1) = x + 12 and verify.
Step 1, expand: 7 − 2x + 2 = x + 12, so 9 − 2x = x + 12.
Step 2, collect: subtract x and 12: −3 = 3x, so x = −1.
Check in the original: left side 7 − 2(−1 − 1) = 7 − 2 × (−2) = 7 + 4 = 11. Right side −1 + 12 = 11. Both equal 11. ✓
The mistake to watch for
The trap is checking in a line you already rearranged.
Mistaken working: 7 − 2(x − 1) becomes 7 − 2x − 2 (the sign of the second term is wrong). Then 5 − 2x = x + 12, so −7 = 3x and x = −7/3.
Checking in 5 − 2x = x + 12 passes: 5 + 14/3 = 29/3 and −7/3 + 12 = 29/3.
That check agrees because the error is built into the line being tested. In the original: left side 7 − 2(−7/3 − 1) = 7 + 20/3 = 41/3, while the right side is 29/3. They differ, so the answer is wrong.
The correction is to always scroll back up to the original question and copy that line, nothing else.
Check yourself
1. A student solves 4(x + 3) = 2x + 20 and gets x = 2. Is this right? If not, find the correct value.
Show answer
Check x = 2 in the original: left side 4 × 5 = 20, right side 2 × 2 + 20 = 24. These differ, so it is wrong.
Solve: 4x + 12 = 2x + 20, so 2x = 8 and x = 4. Check: 4 × 7 = 28 and 8 + 20 = 28. ✓ The correct value is x = 4.
2. Solve 3(x + 5) = 2 − x and check your answer.
Show answer
Expand: 3x + 15 = 2 − x. Add x: 4x + 15 = 2. Subtract 15: 4x = −13. So x = −13/4.
Check: left side 3(−13/4 + 5) = 3 × 7/4 = 21/4. Right side 2 + 13/4 = 21/4. ✓
3. A student rearranges P = 2(l + w) and writes w = P − 2l. Test the formula with l = 5 and w = 3. Is it correct?
Show answer
With l = 5 and w = 3, P = 2 × 8 = 16. The student’s formula gives w = 16 − 10 = 6, which is not 3. So it is wrong.
The correct rearrangement is w = P/2 − l. Test: 16/2 − 5 = 3. ✓
Where this leads next
Put all five lessons together in the equations and formulas practice set, and record any slips in the mistake log and retest queue. When you meet two unknowns, simultaneous relationships uses the same checking habit on two equations at once.
If you want someone to read your checking line by line, that is what online one-to-one Mathematics tuition can offer.