SYSTEMS OF LINEAR EQUATIONS: SOLVING BY SUBSTITUTION

The method of solving "by substitution" works by solving one of the equations (you choose which one) for one of the variables (you choose which one), and then plugging this back into the other equation, "substituting" for the chosen variable and solving for the other. Then you back-solve for the first variable. Here is how it works. (I'll use the same systems as were in a previous page.) Solve the following system by substitution.

The idea here is to solve one of the equations for one of the variables, and plug this into the other equation. It does not matter which equation or which variable you pick. There is no right or wrong choice; the answer will be the same, regardless. But — some choices may be better than others.


For instance, in this case, can you see that it would probably be simplest to solve the second equation for "y =", since there is already a y floating around loose in the middle there? I could solve the first equation for either variable, but I'd get fractions, and solving the second equation forx would also give me fractions. It wouldn't be "wrong" to make a different choice, but it would probably be more difficult. Being lazy, I'll solve the second equation for y:





Now I'll plug this in ("substitute it") for "y" in the first equation, and solve for x:


Now I can plug this x-value back into either equation, and solve for y. But since I already have an expression for "y =", it will be simplest to just plug into this:


Then the solution is (x, y) = (5, 4).



Warning: If I had substituted my "–4x + 24" expression into the same equation as I'd used to solve for "y =", I would have gotten a true, but useless, statement:



Twenty-four does equal twenty-four, but who cares? So when using substitution, make sure you substitute into the other equation, or you'll just be wasting your time.