Solving Systems of Equations
(without graphs) 2

Alright everyone.  You now know the substitution method.  Another method is called the elimination method or the addition or subtraction method.  In this way of solving systems of equations, one variable is eliminated by adding or subtracting the equations (understand their names now? =P).

4x + 5y = 14
-4x - 3y = -10

Adding them would give 2y = 4

4x + 5y = 14
+ -4x - 3y = -10
2y = 4

As you can see, the 4x and -4x cancelled out, therefore eliminating the variable x, leaving an equation with only one variable (y), able to be solved.

2y = 4
y = 2

Now that you have a value for y, you must find one for x.  To do this, just substitute the value for y into either original equation, and solve it for x

4x + 5(2) = 14
4x + 10 = 14
4x = 4
x = 1

Your solution for these two equations is (1, 2).

Notice that only because 4x and -4x, when added, produce 0 (cancel out), the equation can be solved.  Their coefficients are opposites of each other.  That is why it worked.

Try another one.

6x - 2y = 18
6x - 7y = 3

-1(6x - 7y = 3)
-6x + 7y = -3

6x - 2y = 18
+ -6x + 7y = -3
5y = 15

y = 3
6x - 2(3) = 18
6x - 6 = 18
6x = 24
x = 4

The solution is (4, 3).

Try another one, just for practice.

5x - 7y = 17
-9x - 7y = -11

-1(-9x - 7y = -11)
9x + 7y = 11

5x - 7y = 17
+ 9x + 7y = 11
14x      = 28

x = 2
5(2) - 7y = 17
10 - 7y = 17
-7y = 7
y = -1

The solution is (2, -1).

But what if you want to solve a system using this method where no coefficients are the same or opposites of each other?