Solve a System of Three Linear Equations Using Substitution

By Mary Jane Sterling

In a finite mathematics course, you can expect to run into a lot of problems that involve systems of linear equations. When you’re working with a system of three or more linear equations, you’ll find that using substitution to solve the system involves one variable in terms of another in terms of another, and so on.

The main thing to remember is your target variable. You want to substitute equivalences of one of the variables into all the others—pick a variable and stick with it.

To solve the following system of equations, which variable would you choose?

FNTMATH_0701

None of them looks particularly inviting at first, but the x variable does have a coefficient of 1 in the second equation, so take that as your choice.

First, solving for x in the second equation, you subtract 3y from each side and add 5z to each side, giving you x = 14 – 3y + 5z. Then substitute that expression into the first and third equations.

Substituting into the first equation: 3(14 – 3y + 5z) – 2y + 4z = 1. This simplifies first to –11y – 19z = –41, which becomes 11y – 19z = 41.

In the third equation, 4(14 – 3y + 5z) – 4y + 3z = 1 simplifies to 16y – 23z = 55.

So the new system of equations, in just two variables, is

FNTMATH_0702

The choices of variable to solve for aren’t great, but the smallest number is 11, so the first equation is the easiest choice. Solving for y in the first equation, you get

FNTMATH_0703

Put that into the second equation and solve for z following these steps:

  1. Substitute in for y:
    FNTMATH_0704
  2. Multiply each term by 11 to get rid of the fraction:
    16(19z + 41) – 253z = 605
  3. Simplify:
    304z + 656 – 253z = 605, which becomes 51z = –51
  4. Divide each side by 51:
    z = –1

Now solve for y by putting z = –1 into

FNTMATH_0705

You get

FNTMATH_0706

And now, armed with y = 2 and z = –1, you can put those values in one of the original equations to solve for x. A good choice would be the second original equation, x + 3y – 5z = 14, because the coefficient of x is 1. Using that equation, you get x + 3(2) – 5(–1) = 14, which simplifies to x = 3. The solution of the system as the coordinates of a point is (3, 2, –1).