The Power of Elimination
How would you solve the system of linear equations below?
x – y = 5
2x + y = 13
There are two primary approaches for solving systems of linear equations:
1) Substitution Method
2) Elimination Method
The Substitution Method
With this method, we take one of the equations and solve for a certain variable. For example, we might take x – y = 5 and add y to both sides to get x = y + 5.
Then we take the second equation (2x + y = 13) and replace x with y + 5 to get: 2(y + 5)+ y = 13
From here, we have an equation we can solve for y: y = 1
Now that we know the value of y, we can take one of the equations and replace y with 1 to find the value of x: x = 6
So, the solution is x = 6 and y = 1.
The Elimination Method
With this method, we notice that, if we add the two original equations (x – y = 5 and 2x + y = 13), the y’s cancel out (i.e., they are eliminated), leaving us with: 3x = 18.
From here, when we divide both sides by 3, we get: x = 6, and from here we can find the value of y: y = 1.
Okay, so that’s how the two methods work. What’s my point?
The point I want to make is that, although both methods get the job done, the Elimination method is superior to the Substitution method. And by “superior,” I mean “faster.”
First, the Elimination method can often help us avoid using fractions. Consider this system:
5x – 2y = 7
3x + 2y = 17
To use the Substitution method here, we’d have to deal with messy fractions. For example, if we take the equation 5x – 2y = 7 and solve for x, we get x = (2/5)y + 7/5. Then when we take the second equation (3x + 2y = 17) and replace x with (2/5)y + 7/5, we get: 3[(2/5)y + 7/5] + 2y = 17. Yikes!!
Alternatively, we can use the Elimination method and add the two original equations (5x – 2y = 7 and 3x + 2y = 17). When we do this, the y’s cancel, leaving us with: 8x = 24, which means x = 3. No messy fractions.
It has been my experience that many students rely solely on the Substitution method to solve systems of equations, and this can potentially eat up a lot of time on test day. So, be sure to learn the Elimination method soon. In fact, if I were you, I’d drop the Substitution method from my repertoire; it isn’t very useful.
If you still believe that the Substitution method is just as good as the Elimination method, try solving this question using the Substitution method.
If 5x – 8y = 11, and 4x – 9y = 4, what is the value of x + y?
To view the solution to the above question, see my post, The Reasonable Test-Maker.