## The Power of Elimination

How would you solve the system of linear equations below?

** x – y = 5**

** 2 x + 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 (2*x* + *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 2*x* + *y* = 13), the *y*’s cancel out (i.e., they are eliminated), leaving us with: 3*x* = 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:

** 5 x – 2y = 7**

** 3 x + 2y = 17**

To use the Substitution method here, we’d have to deal with messy fractions. For example, if we take the equation 5*x* – 2*y* = 7 and solve for *x*, we get *x* = (2/5)*y* + 7/5. Then when we take the second equation (3*x* + 2*y* = 17) and replace *x* with (2/5)*y* + 7/5, we get: 3[(2/5)*y* + 7/5] + 2*y* = 17. Yikes!!

Alternatively, we can use the Elimination method and add the two original equations (5*x* – 2*y* = 7 and 3*x* + 2*y* = 17). When we do this, the *y*’s cancel, leaving us with: 8*x* = 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.

What would happen if both lines are parallel ?

how would we solve the unknowns ?

Substitution and elimination should work both for equations of parallel lines and equations of lines that are not parallel. Hope this helps. 🙂

In the first example of the Elimination Method you ADDED the two linear equations but on the second example displayed on the hyperlinked page you used SUBTRACTION. Is there a rule for this or are you simply combining the equations with either addition or subtraction exclusively as best fits the particular problem?

Hi Myles,

You can use add or subtract the equations according to the method that best fits each question. As long as you do the same thing across the entire equation (you can’t add one part of the equation and subtract another part), it is perfectly fine!

Both are straight line eq’s-a even faster way would be just to graph both mentally and get an intercept.

Hi George,

Yes, mentally graphing the lines and finding their point of intersection will, indeed, work. I’m just not sure how many people are able to do this. It would certainly be difficult to do so with this system:

5x – 8y = 11

4x – 9y = 4

Cheers,

Brent

In example 1; the value of x is 6.

as the calculated value of y is 1, x-1=5

which makes x=6.

You’re absolutely right, sanket.

We’ll fix that right away.

Cheers,

Brent

thanks