How to solve Systems of Equations ?

There's a simple method that you can use to solve most of the systems of equations you'll encounter in Calculus. It's called the "substitution method."

This is not the only method - and often, it's not the easiest either. But it's something you can usually fall back on when other things don't work. It's also the easiest to learn, in case you've been having trouble learning other methods.

Step 1. Solve one of the equations for one of the variables.
Step 2. By substitution, eliminate that variable from the other equations.

Now you have fewer equations, and fewer variables . Repeat steps 1 and 2 until you have only one equation and one variable.

Step 3. Solve that equation with one variable. Now you know the value of that one variable.
Step 4. Go back to the other solved equations to find values for the other variables.
Step 5. Check your work!

To demonstrate, let's solve this system of equations:

x² + y = 3 (2a)

15x + 3y = 21 (2b)

Let's solve equation (2a) for y (Step 1).

y = 3 - x²

Since y is equal to 3 - x² , we can substitute 3 - x² for y wherever y occurs. Let's make this substitution in equation (2b) (Step 2):

15x + 3(3 - x² ) = 21.

Notice what we've accomplished: we have an equation with only one variable in it. Now we solve for that variable (Step 3):

15x + 9 - 3x² = 21

-3x² + 15x - 12 = 0

-3(x² - 5x + 4) = 0

-3(x - 4)(x - 1) = 0

x = 4 or x = 1.

Now, Step 4: use these values for x to solve (2a) for the other variable y:

x² + y = 3

(4) 2 + y = 3 or (1) 2 + y = 3

16 + y = 3 or 1 + y = 3

y = -13 or y = 2.

So, the possible solutions are

A. x = 4, y = -13

B. x = 1, y = 2.

Often these are written simply as (4, -13) and (1, 2).

Step 5: Check these solutions with the original equations (2a) and (2b). For example, let's check the possible solution (4, -13) with equation (2a):

x² + y = 3

(4)² + (- 13) = 3

16 - 13 = 3

which is true. You can do the rest of the checks yourself.