Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

If I am asked to solve a systems of equation, how would I know which method (substitution, or elimination) to use? What set of conditions should I be looking for, or is it that either method should in theory work?

Then say that I am given the following system of equation, but neither substituion nor elimination produces the correct result (based on what I have did thus far)

$xy = 4$

$2x - y - 7 = 0$

How would I go on about solving this type of one?

share|cite|improve this question
Any method will give you the correct result, the choice is based on the one that seems easier in the specific case – Alessandro May 31 '14 at 11:57
up vote 2 down vote accepted

Solve for one variable using the second equation. Let's solve for $y$: $$2x - y - 7 = 0 \iff y = 2x - 7$$

Now plug $y = 2x - 7$ into $xy = 4$ to get a quadratic in $x$:

$$xy = 4 \iff x(2x - 7) = 4 \iff 2x^2 - 7x - 4 = 0 \iff (2x + 1)(x - 4) = 0$$

If the factoring isn't immediately apparent, you can use the quadratic formula to obtain two solutions: $x = -1/2$ or $x = 4$.

Then you go back to your equation for $y$ to find the corresponding $y$-values for each solution $x$.

share|cite|improve this answer

2x-y-7=0 Xy=4



Take lcm




X=-1/2, x=4

If x=-1/2


If x=4


share|cite|improve this answer

To answer the more theoretical question: Substitution and elimination are each valid techniques, and either will work for any problem. It is just that for larger problems, elimination tends to get less messy, and elimination sets you up for learning Gaussian Elimination (row-reduction) if you ever have to take matrices. Once you take Linear Algebra, you will understand why each of these methods has a place. :)

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.