# Verifying Solutions

A solution to an algebra problem is valid if both sides of the equation are still equal when the problem has been worked out with the chosen solution substituted for the variable(s). Verifying solutions is a good way to double-check the work on any problem. Sometimes it is an essential step to obtaining the correct solution, such as when working with radical equations or completing the square to solve a problem, which may result in an extraneous root.

\(x+7=10\)

This problem can be solved by subtracting 7 from each side.

\(x+7-7=10-7\)

\(x=3\)

Once the problem is solved, the solution can be verified by rewriting the problem with 3 substituted for \(x\).

\(3+7=10\)

\(10=10\)

Both sides are equal, verifying that \(x=3\) is a valid solution.

If a problem is solved incorrectly, the mistake can be caught through this verification process.

\(x+7=10\)A student rushing through her homework might mistakenly write \(x=2\) as the solution to this problem. If she takes a moment to rework the equation with her answer, she will realize the answer is incorrect.

\(x+7=10\)

\(2+7=10\)

\(9=10\)

Since \(9\neq10\), the student knows she needs to go back and find a different solution to the problem.

## Verifying Solutions by Substitution

To verify that certain values are solutions to the given equation, we simply substitute them in and check. This is very similar to trial and error.

\[(2, 3), (3, 5), (4, 4), (6, 3), (10, 0)\]

## How many of the above pairs of integers are solutions to \( 2x + 3y = 20 ?\) \[\]

(A) \(\ \ 1\)

(B) \(\ \ 2\)

(C) \(\ \ 3\)

(D) \(\ \ 4\)

(E) \(\ \ 5\)

Correct Answer: B

Solution:We try each of the pairs of integers:

For \((2, 3)\), we have \( 2 \times 2 + 3 \times 3 = 4 + 9 = 13 \neq 20 \).

For \((3, 5)\), we have \( 2 \times 3 + 3 \times 5 = 6 + 15 = 21 \neq 20 \).

For \((4, 4)\), we have \( 2 \times 4 + 3 \times 4 = 8 + 12 = 20 \). This is a solution.

For \((6, 3)\), we have \( 2 \times 6 + 3 \times 3 = 12 + 9 = 21 \neq 20 \).

For \((10, 0)\), we have \( 2 \times 10 + 3 \times 0 = 20 + 0 = 20 \). This is a solution.Thus, 2 of the pairs are solutions.

Incorrect Choices:

(A),(C),(D), and(E)

See the solution for why these choices are wrong.

## Rearranging Equations

Mathematical equations are flexible: even simple ones can be rewritten a number of ways.

\(3x-4=5\)

This problem could be rewritten to isolate the term with the variable by adding 4 to each side of the equation.

\(3x-4=5\)\(3x-4+4=5+4\)

\(3x=9\)

The equation could also be set equal to zero, moving all terms to one side of the equation.

\(3x-4=5\)\(3x-4-5=5-5\)

\(3x-9=0\)

Any operation applied to one side of the equation must also be carried out on the other side. That way, the sides remain equal, even when they look different. Rearranging an equation is a useful skill for solving problems, and also for verifying the solutions. The value for \(x\) found using any of the above equations can be substituted back into the other equations to verify that the solution is correct.

\(3x=9\) can be solve with division.

\(3x\div3=9\div3\)

\(x=3\)

Then, substitute the answer back into the original form of the equation: \(3x-4=5\)

\(3(3)-4=5\)

\(9-4=5\)

\(5=5\)

The solution is verified.

## Practice Problems

**Cite as:**Verifying Solutions.

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