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Three main methods come to mind:

  1. Substitution method
  2. Elimination method
  3. Matrix row-reduction.

The first two methods are taught in early grades in most high schools. Matrix row-reduction is taught in senior years in high school, or more likely, in the early years of a university math or science program.

In the substitution method, a variable is isolated (solved for) and then used to substitute in other equation(s).

e.g.

2x + y - 6 = 0

x + 3y - 13 = 0

Solve for y in the first equation. Move the 2x and -6 to the right side, switch signs as we switch sides.

y = -2x + 6

Now take this definition of y and substitute it in the second equation; any place you see y, make the substitution.

x + 3(-2x + 6) - 13 = 0

x + -6x + 18 - 13 = 0

Collect like terms and simplify.

-5x + 5 = 0

-5x = -5

x = 1

Substitute this value of x into the definition for y above.

y = -2(1) + 6

y = -2 + 6

y = 4

So the two lines given intersect at (1, 4).

In the elimination method, multiples of each equation are added (or subtracted) together, eliminating one variable at a time, when they add to 0. Let's use the same equations as before, but use the elimination method this time.

e.g.

2x + y - 6 = 0

x + 3y - 13 = 0

Multiply each term of the first equation by -3. You could just have easily multiplied the second equation by -2. Some of these numbers may look familiar.

-6x - 3y + 18 = 0

x + 3y - 13 = 0

Add the corresponding values in the equations.

-6x + x - 3y + 3y + 18 - 13 = 0 + 0

Adding the positive 3y and the negative 3y yields 0, thus eliminating the y value from further calculations.

-5x + 5 = 0

-5x = -5

x = 1

Now take that value and put it into one of the original equations and solve for the remaining variable.

2(1) + y - 6 = 0

2 + y - 6 = 0

y = -2 + 6 = 4

If we have done things properly, both methods should yield the same result. Which method is easier will depend on the exact specifics of the question.

These methods can be expanded for more than 2 unknown variables defined by 2 equations, but the amount of work required grows quite quickly. In those situations it is often more efficient to use a matrix and perform row reduction. Matrices are a complex topic, and so I won't cover them here. In many simpler situations, elimination and substitution will do the job quite nicely.

* * * * *

There is also the graphical method. Plot the linear equations and the coordinates of point of intersection of the lines is the solution.

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Q: What are the methods in solving systems of linear equation?
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