The Basics Examples

Solve this system of equations using the substitution method:

-2m + 6n = 8
5m – 6n = -2

Solve the first equation for m (by subtracting 6n from each side, then dividing each side by -2).

m = 3n – 4

Now, take this and substitute the 3n – 4 into either equation, we'll choose the 2nd equation.

5(3n – 4) – 6n = -2
15n – 20 – 6n = -2
9n = 18
n = 2

Now, we just need to substitute this value of n back into the m equation (or any of them for that matter), but this is easiest.

m = 3(2) – 4 = 2

Solution: (2, 2)

Solve this system of equations using the elimination method:

-2m + 6n = 8
5m – 6n = -2

When using the elimination method, you should notice if there's anything you need to multiply either or both equations by before you add them together. And we got lucky on this one. If we add the two equations together, we can knock out the n's without doing any math at all (except adding, of course). "Knock Knock." "Who's there?" "Not the n's". (Boo get off the stage.)

3m = 6
m = 2

Now, subbing back into any equation, we can get the n value.

2(2) + 6n = 8
-4 + 6n = 8
6n = 12
n = 2

Solution: (2, 2)

Solve this system of linear equations using the elimination method:

4x – 6y = – 12
10x – 5y = 10

Since we will start dealing with matrix operations in the next few sections, let's call the first equation Row 1 and the second equation Row 2. If we multiply Row 1 by 5 and Row 2 by -6, that will give us a -30y and a +30y which will cancel out when we add the two equations together. Watch the process:

5(4x – 6y = -12)
-6(10x – 5y = 10)

20x – 30y = -60
-60x + 30y = -60

+

-40x = -120
x = 3
4(3) – 6y = -12
12 – 6y = -12
y = 4

Solution: (3,4)

Solve this non-linear system of equations.

y = x³
y = x

We will do this algebraically by substituting x³ into the y value of y = x.

x³ = x
x³ – x = 0
x(x² – 1) = 0
x(x + 1)(x – 1) = 0
x = 0
x = -1
x = 1

Problem solved. It was nice that the problem was factorable which allowed us to use the zero product property to find all the x values. Now we just need to plug in those x values to one of the equations to find the y values. Let's be lazy and use the y = x equation. That will give us y = 0, y = -1 and y = 1. So our non-linear system of equations intersects in 3 spots: (0, 0), (-1, -1), (1, 1). Can we solve a non-linear system of equations? Heck yeah.

Use a graphing calculator to find a solution for this system of equations:

y = x⁴ + 0.1
y = -5x³

Use your favorite graphing calculator or web graphing calculator to find the intersection of these two equations. Don't know which to use? Check out our suggestions for online graphing calculators on the Best of the Web section for this chapter. The intersection of these two equations is at (-0.276, 0.106). Look below at the graphs.