Method

How to Check Your Own Math: 6 Verification Tricks

Q: How long should checking take?

Budget about one minute of checking for every five minutes of solving. That is enough for two independent checks on almost any problem. If a check costs more than that, it is really a second solution, and you should ask whether the problem is worth solving twice.

Q: Does substituting the answer back catch every kind of error?

No, and this is the trap. Substitution verifies that your answer satisfies the equation you wrote down. If you translated the question into the wrong equation, the wrong answer will satisfy that wrong equation perfectly and the check will pass. Setup errors are caught by reading the answer back against the original story, by units, and by order-of-magnitude estimates.

Q: Why did my answer satisfy the equation but still get marked wrong?

Usually one of three reasons. You solved a correctly stated equation but answered a different question than the one asked, for example finding x when the question wanted the perimeter. You lost a second solution, which happens constantly with square roots and absolute values. Or the equation itself was a mistranslation of the word problem.

Q: Is estimating worth the time when I have a calculator?

The calculator is the reason it is worth the time. Calculators do not make arithmetic errors; they faithfully compute whatever you typed. Almost every calculator-related mistake is a data-entry error, and those produce answers that are wrong by a factor of ten, a hundred, or a sign. An estimate is the only check that reliably catches them.

Q: Which two checks should I use if the exam is nearly over?

Substitute the answer back into the original equation, and glance at the units or the plausible range. Together they take under thirty seconds per question and catch the two most common failure modes: an arithmetic slip during rearrangement, and a formula applied with a term dropped.

Six checks that engineers, actuaries, and examiners run because their mistakes are expensive. Each one is demonstrated below on work that is genuinely wrong.

By the Gauth editorial teamHow this content is produced and checkedUpdated 10 September 2026

Why re-reading your work finds nothing

Solving and checking are different processes that happen to be about the same problem. Solving runs forwards: apply rules until the unknown is alone. Checking runs backwards, or sideways, or from a completely different direction. Most people never build the second skill, which is why so many scripts contain confident, tidy, wrong working.

Re-reading is the default and it is close to worthless, because your eye supplies what you meant instead of what you wrote. A useful check has to be independent: it must reach the answer, or some property of the answer, by a route that shares as little as possible with the original solution. It also has to be cheap, because a check you will not run with four minutes left on the clock is not a check.

Independence is worth being strict about. Substituting an answer back is independent because it pushes numbers forwards through the equation as printed, while your solution pulled them backwards through a chain of rearrangements. Estimating is independent because it rounds the inputs and reorders the arithmetic. Redoing the same algebra a second time is not independent at all: the step you got wrong the first time is the step you find easy, so you will get it wrong identically and feel reassured.

There is a second question every check should answer, and it is not about correctness. It is about completeness: did you answer the question that was asked? A solution can be flawless and still lose every mark because the question wanted the perimeter and you stopped at $x$ , or wanted both roots and you reported one. Reading the question again after you have a number costs five seconds and catches a category of loss that no amount of algebraic care will.

Every technique below is demonstrated on work that is actually wrong. Watching a check confirm a correct answer teaches you nothing about what a caught error looks like.

Solve a math problem now

First solution free

1. Substitute the answer back into the original

The cheapest check there is. Take your answer, put it into the equation as originally printed — not into your rearranged version, which may already contain the error — and see whether both sides agree.

Worked example

3 (x - 4) = 2 x + 5

1
The submitted first line. The 3 reached the x but not the 4, which is the most common distribution slip there is.
$3 x - 4 = 2 x + 5$
2
From there the algebra is faultless, which is exactly why the error survives a re-read.
$x = 9$
3
Substitute 9 into the left-hand side of the original equation, brackets and all.
$3 (9 - 4) = 3 \cdot 5 = 15$
4
And into the right-hand side.
$2 (9) + 5 = 23$
5
Fifteen is not twenty-three, so 9 is not a solution. The check does not say where the mistake is, only that there is one, which is all you need to send you back to line one.
6
Redone with the 3 distributed across both terms.
$3 x - 12 = 2 x + 5 \Rightarrow x = 17$
7
Both sides now land on the same number, so the answer holds.
$3 (17 - 4) = 39, 2 (17) + 5 = 39$

Answer

x = 17

Substitution earns its keep twice over on radical and rational equations, where squaring or clearing denominators can manufacture solutions that were never there. Squaring is not reversible: it maps $2$ and $- 2$ to the same place, so the squared equation is satisfied by roots the original rejects. Checking is not good practice here, it is part of the method.

Where does this solution stop being true?

One of these lines is wrong. Click it.

One limit worth naming: substitution verifies that your answer satisfies the equation you wrote. If you mistranslated the question, the wrong answer will satisfy your wrong equation perfectly and the check will wave it through. That failure mode belongs to the next three techniques, and to the habit of reading your number back into the original sentence, covered in our guide to turning words into equations.

2. Follow the units

Units are a type system for arithmetic. Every physical quantity carries one, the units on both sides of an equation have to match, and they combine under multiplication and division exactly as the numbers do. That gives you a check that runs on the structure of a formula rather than on its numbers — so it catches whole missing factors, which substitution cannot.

Worked example

u = 0, a = 3 m s^{- 2}, t = 8 s . Find s .

1
The written line drops the square from the time. It looks close enough to the remembered formula to pass unnoticed.
$s = \frac{1}{2} a t = \frac{1}{2} (3) (8) = 12$
2
Run the units through the same expression, ignoring the numbers entirely. Acceleration is metres per second squared; time is seconds.
$m s^{- 2} \times s = m s^{- 1}$
3
That is a speed. The question asked for a distance, so whatever is on the page is not the right formula — and you reached that conclusion without remembering the right one.
4
To land on metres, the seconds have to cancel twice, which means a second factor of time.
$m s^{- 2} \times s^{2} = m$
5
Redo with the square restored. The units now agree with the quantity being asked for.
$s = \frac{1}{2} (3) (8^{2}) = \frac{1}{2} (3) (64) = 96 m$

Answer

s = 96 m

The same check settles the question that stalls more students than any other in applied problems: do I multiply by the rate or divide by it? A pump moves 4.5 litres per minute and you need to fill 1,080 litres. Multiplying gives

1080 L \times 4.5 L mi n^{- 1} = 4860 L^{2} mi n^{- 1}

which is not a unit of anything. Dividing cancels the litres and leaves minutes, which is what the question wanted:

\frac{1080 L}{4.5 L mi n ^{- 1}} = 240 min = 4 hours

Prefixes deserve their own moment of suspicion, because they hide factors of a thousand inside a single letter. A rate given in millilitres and a volume given in litres differ by $1 0^{3}$ ; a capacitance in microfarads and a charge in coulombs differ by $1 0^{6}$ . Write every quantity in base units before the first line of algebra. Converting at the end means converting a number you can no longer sanity-check.

Pure algebra has no units, but it has an equivalent: degree. Multiplying a linear factor by a quadratic must produce a cubic, and the constant term of the product must be the product of the two constant terms. Expand $(x + 3) (x^{2} - 2 x + 5)$ and you should get $x^{3} + x^{2} - x + 15$ . If your answer is quadratic, you dropped a term; if the constant is not 15, the last multiplication went wrong. Both checks take about two seconds and neither requires redoing the expansion.

3. Estimate the order of magnitude before you compute

Calculators do not make arithmetic errors. They compute exactly what you typed, which means almost every calculator-related mistake is a data-entry error — a misplaced decimal, a missing bracket, a digit transposed. Those errors do not produce answers that are slightly off; they produce answers that are off by factors of ten. An estimate made before you touch the keypad is the only check that reliably catches them, and it costs about twenty seconds.

Worked example

$2, 400 invested at 6% compounded annually for 12 years.

1
Estimate first. The rule of 72 says money doubles in roughly 72 divided by the percentage rate — here, twelve years, which is exactly the term.
$\frac{72}{6} = 12 years to double$
2
So the answer should sit close to twice the principal. Write that number down before computing anything.
$expect \approx 4800$
3
The keyed-in calculation, with 1.6 typed where 1.06 was meant. Nothing about the digits on screen looks alarming in isolation.
$2400 \times 1. 6^{12} \approx 675, 540$
4
That is roughly 140 times the estimate. Method errors are usually wrong by a small factor; being wrong by two orders of magnitude points almost always at data entry, so you retype rather than rethink.
5
Retyped correctly. The growth factor is just above 2, exactly as the rule of 72 predicted.
$1.0 6^{12} = 2.0122 \dots \Rightarrow 2400 \times 1.0 6^{12} = 4829.27$

Answer

$4, 829.27

Estimation also comes with free bounds you rarely have to derive. A mean has to lie between the smallest and largest value in the data. A probability has to lie between 0 and 1. A correlation coefficient has to lie between −1 and 1. A standard deviation for a modestly sized sample is usually near a quarter of the range, so a value ten times that size means a squaring or a division went astray — the arithmetic is walked through slowly in standard deviation by hand. Each of these is an order-of-magnitude check where somebody has already worked out the bounds for you.

4. Push the answer to a limiting case

The most powerful check on the list and the least used. Take your answer — especially a symbolic one — and send a variable somewhere you already know the result: to zero, to infinity, to equality with another variable, to a value that makes the situation trivial. If the formula does not do the obvious thing at the extreme, it is wrong in the middle too.

Worked example

A finishes a job alone in 6 h; B finishes it alone in 3 h. Together?

1
The submitted answer averages the two times, which is the instinctive move and the wrong one.
$\frac{6 + 3}{2} = 4.5 hours$
2
Before checking the arithmetic, establish a bound you can argue for without solving anything: a second worker cannot make a job take longer than the faster worker alone.
3
B alone finishes in 3 hours, so any correct answer has to be under 3 hours. 4.5 fails that test, and so does the other popular wrong answer of 9.
$t_{together} < 3 hours$
4
The bound also tells you what kind of thing to add. Times do not add here; rates do. A completes one sixth of the job per hour, B one third.
$\frac{1}{6} + \frac{1}{3} = \frac{1}{6} + \frac{2}{6} = \frac{1}{2}$
5
Half a job per hour means one job in two hours, comfortably inside the bound the limiting case gave you.
$t = \frac{1}{\frac{1}{2}} = 2 hours$

Answer

2 hours

The same move works on formulas you have derived rather than looked up. Suppose a projectile launched at speed $v_{0}$ and angle $θ$ comes out with maximum height

h = \frac{v _{0}^{2} sin ^{2} θ}{g} .

Set $θ = 9 0^{\circ}$ , so the projectile goes straight up and the problem becomes one you can solve by energy alone. The formula predicts $h = v_{0}^{2} / g$ , but conservation of energy says

\frac{1}{2} m v_{0}^{2} = m g h ⟹ h = \frac{v _{0}^{2}}{2 g},

so the derived formula is too large by a factor of two. The missing 2 belongs in the denominator: $h = v_{0}^{2} sin^{2} θ / (2 g)$ . One substitution of a convenient angle found a structural error that would otherwise have been carried through an entire question.

A third flavour of the same idea catches the average-speed trap. Someone drives 60 miles at 30 mph, then 60 miles at 60 mph, and reports the average as 45 mph. Send the second leg's speed to infinity: the averaging formula runs off to infinity with it, but the whole trip can never average more than 60 mph, because the first leg alone already consumed two hours of a 120-mile journey. Go back to the definition instead. Total time is $2 + 1 = 3$ hours for 120 miles, giving 40 mph — lower than the naive average, and lower for a reason you can state: you spend more time at the slow speed than at the fast one.

5. Exploit symmetry

Many answers are obliged to be arranged symmetrically, and the arrangement is far easier to verify than the values. The roots of a quadratic sit at equal distances either side of the vertex line. The integral of an odd function over an interval centred on zero is zero. A system unchanged by swapping two variables must have solutions that come in swapped pairs.

Worked example

2 x^{2} - 12 x + 10 = 0

1
The submitted working divides by 2 rather than by 2a. With a = 2 the denominator should be 4.
$x = \frac{12 \pm 144 - 80}{2} = \frac{12 \pm 8}{2}$
2
That produces two perfectly plausible-looking integers.
$x = 10 or x = 2$
3
But the roots of any quadratic are symmetric about the vertex line, and that line is fixed by the coefficients alone.
$x = - \frac{b}{2 a} = \frac{12}{4} = 3$
4
The midpoint of 10 and 2 is 6, not 3. No quadratic has roots arranged like that, so at least one value is wrong — and you knew it before checking either one by substitution.
$\frac{10 + 2}{2} = 6 \neq = 3$
5
Vieta's relations make the same point arithmetically: the roots must sum to -b/a and multiply to c/a. Ten and two sum to 12, not 6.
$r_{1} + r_{2} = \frac{12}{2} = 6, r_{1} r_{2} = \frac{10}{2} = 5$
6
Fixing the denominator gives roots that pass both tests: they sum to 6, multiply to 5, and straddle 3 at a distance of 2.
$x = \frac{12 \pm 8}{4} = 5 or 1$

Answer

x = 5 or x = 1

In calculus, the symmetry check is often the entire calculation. Every term of $x^{5} - 3 x^{3} + x$ is odd, and the interval $[- 1, 1]$ is centred on zero, so

\int_{- 1}^{1} (x^{5} - 3 x^{3} + x) d x = 0

without computing an antiderivative. If your evaluation returns anything other than zero, you made a sign error substituting the negative limit — most often by treating an even power of a negative number as negative.

Symmetry also audits systems. The pair $x + y = 10$ , $x y = 21$ is unchanged if you swap the names $x$ and $y$ , so its solutions must come in swapped pairs: here $(3, 7)$ and $(7, 3)$ . If your working produced only one ordered pair, you dropped a root — the two solution methods are compared side by side in our guide to substitution, elimination, and graphing.

Trigonometry supplies the same leverage through parity. Cosine is even, so $cos (- θ) = cos θ$ ; sine and tangent are odd, so they pick up a minus sign. If you simplify an expression and the result changes parity, the manipulation lost a sign somewhere. Which identities are worth holding in memory for this kind of move is the subject of the trig identities worth memorising.

6. Reverse the operation

The most complete check and the slowest: undo what you did and see whether you land back where you started. Differentiate an integral. Multiply out a factorisation. Multiply the quotient by the divisor and add the remainder. Unlike the other five, this one verifies the answer exactly rather than testing a property of it.

Worked example

\int x e^{2 x} d x

1
A submitted answer from integration by parts, taking u to be x and dv to be the exponential part. Both constants look reasonable at a glance.
$\frac{x e ^{2 x}}{2} - \frac{e ^{2 x}}{2} + C$
2
Differentiate it. The first term needs the product rule, and the inner derivative of the exponential contributes a factor of 2.
$\frac{d}{d x} [\frac{x e ^{2 x}}{2}] = \frac{e ^{2 x}}{2} + x e^{2 x}$
3
The second term is straightforward, and its factor of 2 cancels the halving.
$\frac{d}{d x} [- \frac{e ^{2 x}}{2}] = - e^{2 x}$
4
Adding them leaves a term that was never in the integrand, so the answer is wrong.
$x e^{2 x} - \frac{e ^{2 x}}{2} \neq = x e^{2 x}$
5
The size of the leftover locates the error. It is exactly half of the exponential, which points at the constant on the second term: because v is half the exponential, the remaining integral contributes a quarter of it, not a half.
$\int \frac{e ^{2 x}}{2} d x = \frac{e ^{2 x}}{4}$
6
Differentiating the corrected answer now collapses to the integrand exactly.
$\frac{e ^{2 x}}{2} + x e^{2 x} - \frac{e ^{2 x}}{2} = x e^{2 x}$

Answer

\frac{x e ^{2 x}}{2} - \frac{e ^{2 x}}{4} + C

Reversal is nearly free on factorisations, and almost nobody does it. A student factors $6 x^{2} - 5 x - 6$ as $(3 x - 2) (2 x + 3)$ . Expanding gives $6 x^{2} + 9 x - 4 x - 6 = 6 x^{2} + 5 x - 6$ : right coefficients, wrong sign on the middle term. Swapping which bracket carries the minus fixes it, since $(3 x + 2) (2 x - 3) = 6 x^{2} - 9 x + 4 x - 6 = 6 x^{2} - 5 x - 6$ . Ten seconds of expansion, and the sign error never reaches the marker.

Because differentiation is mechanical and integration is not, reversal is especially lopsided in calculus — checking is far easier than solving. That asymmetry is worth exploiting every time, and it is why the integral calculator shows the differentiation check alongside the result.

Where to put the checkpoints in a long problem

A five-mark question is usually three problems stacked, each feeding a number to the next. Checking only at the end is the worst placement: a slip in the first phase has already contaminated everything downstream, and the final check tells you that something is wrong without telling you which third to redo. Put a checkpoint at each junction instead, and every error is caught within one step of where it was made.

Worked example

A cylindrical tank, r = 1.2 m, h = 3.5 m, drains at 40 L mi n^{- 1} .

1
Phase one: the volume in cubic metres.
$V = π r^{2} h = π (1.2)^{2} (3.5) = 5.04 π \approx 15.83 m^{3}$
2
Checkpoint one, two ways. Units: metres squared times metres gives cubic metres, which is a volume. Magnitude: pi is about 3 and 1.44 times 3.5 is about 5, so expect roughly 15. Both agree, so the value is safe to carry forward.
3
Phase two: convert to litres, because the drain rate is quoted in litres.
$15.83 m^{3} \times 1000 L m^{- 3} \approx 15, 830 L$
4
Checkpoint two is a direction check, and it catches the most common error in this problem. A litre is smaller than a cubic metre, so the number must get bigger. If your conversion made it smaller you divided when you should have multiplied.
5
Phase three: divide the volume by the rate.
$t = \frac{15 , 830 L}{40 L mi n ^{- 1}} \approx 395.8 min$
6
Checkpoint three. Units: litres divided by litres per minute leaves minutes, which is a time. Magnitude: sixteen thousand over forty is four hundred. Then answer the question actually asked, which was how long — in hours.
$395.8 min \approx 6 h 36 min$

Answer

t \approx 6 hours 36 minutes

No checkpoint above repeated any arithmetic. Each tested a property the answer had to have — a unit, a rough size, a direction of change — and each cost a few seconds. That is the economics of checking: you buy information about correctness at a fraction of the price of the calculation itself.

Choosing a check under time pressure

You will not run six checks on one question. Run the cheapest one that could plausibly catch the kind of error you make, which you can look up in your own marked work rather than guess at.

Check	Costs	Catches	Blind to
Substitute back	Seconds	Rearrangement slips, sign errors, extraneous roots	A wrongly set-up equation
Units or degree	Seconds	Missing factors, wrong formula, multiply vs divide	Numerical slips
Order of magnitude	About 20 seconds	Decimal points, calculator entry, dropped percentages	Errors under a factor of two
Limiting case	A minute, on symbolic answers	Structurally wrong formulas and models	Arithmetic in a correct formula
Symmetry	Seconds, when it applies	Lost roots, sign errors at limits	Errors that preserve symmetry
Reverse the operation	As long as the reverse takes	Essentially everything	Nothing, but it is the slowest

A practical routine: estimate before you start, so the expected size of the answer is already written down; substitute back the moment you have a number; and reverse the operation only on the questions carrying the most marks. If you keep an error log — a running list of the mistakes that actually cost you marks, described in how to study for a math test — the right check for you falls out of it within a fortnight.

Check yourself

A student solves 2/(x-3) + 1 = x/(x-3) and reports x = 3. Which single check disposes of this in under ten seconds, and what does it show?

Frequently asked questions

How long should checking take?

Budget about one minute of checking for every five minutes of solving. That is enough for two independent checks on almost any problem. If a check costs more than that, it is really a second solution, and you should ask whether the problem is worth solving twice.

Does substituting the answer back catch every kind of error?

No, and this is the trap. Substitution verifies that your answer satisfies the equation you wrote down. If you translated the question into the wrong equation, the wrong answer will satisfy that wrong equation perfectly and the check will pass. Setup errors are caught by reading the answer back against the original story, by units, and by order-of-magnitude estimates.

Why did my answer satisfy the equation but still get marked wrong?

Usually one of three reasons. You solved a correctly stated equation but answered a different question than the one asked, for example finding x when the question wanted the perimeter. You lost a second solution, which happens constantly with square roots and absolute values. Or the equation itself was a mistranslation of the word problem.

Is estimating worth the time when I have a calculator?

The calculator is the reason it is worth the time. Calculators do not make arithmetic errors; they faithfully compute whatever you typed. Almost every calculator-related mistake is a data-entry error, and those produce answers that are wrong by a factor of ten, a hundred, or a sign. An estimate is the only check that reliably catches them.

Which two checks should I use if the exam is nearly over?

Substitute the answer back into the original equation, and glance at the units or the plausible range. Together they take under thirty seconds per question and catch the two most common failure modes: an arithmetic slip during rearrangement, and a formula applied with a term dropped.

Sources

Keep going

12 algebra mistakes that quietly cost you marksThe errors these checks are designed to catch, each shown wrong then right.Order of operations beyond PEMDASWhy the acronym misleads on division and subtraction, and how a grader reads an expression.Equation solver with stepsEvery operation applied to both sides, so you can compare against your own working.Turning words into equationsThe setup errors that substitution can never catch, and how to avoid making them.