Solving a System of Equations with Two Unknowns

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  • 0:07 Life is Complex
  • 0:42 Systems of Equations
  • 1:23 Substitution Practice
  • 3:44 Elimination Practice
  • 5:42 Lesson Summary
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Lesson Transcript
Instructor: Jeff Calareso

Jeff teaches high school English, math and other subjects. He has a master's degree in writing and literature.

When you have two variables and two equations, is solving the problem four times as hard? No! Not when you use the substitution or elimination method to solve systems of equations. We'll practice both in this lesson.

Life Is Complex

If there was always just one unknown thing or one problem at a time, life wouldn't be so tough. Let's say you get invited to a wedding and all you need to figure out is if you want the chicken, fish or vegetarian entrée. That's not so bad, right? But, life isn't that simple.

At one point or another, it's your own wedding. And then, there isn't just chicken or fish, there's this reception hall or that one, this best man or that one, this florist or, geez, when did this town get so many florists? And since when are there so many types of paper and different fonts for invitations?

Systems of Equations

Systems of equations can seem just as overwhelming. A system of equations is a group of two or more equations with the same variables. Multiple equations? Multiple variables? It's enough to make you want to elope. Fortunately, though, solving systems of equations is much more straightforward than it seems.

In this lesson, we're going to practice the two most common methods of solving systems of equations. First, there's the substitution method. The substitution method is when you solve one equation for either variable, then substitute the solution into the other equation.

Then there's the elimination method. The elimination method is when we add or subtract equations together to solve for a variable. Let's try out each method as we solve some equations.

Substitution Practice

Let's start with the substitution method. Here are two equations:

y - 2x = 1 and 5x - 2y = 3

Let's take the first one and solve for y. We add 2x to get y = 1 + 2x. Next, we substitute 1 + 2x for y in the second equation. So, we get 5x - 2(1 + 2x) = 3. Now we solve for x. First, we distribute the -2 and get -2 - 4x. Then, 5x - 4x is just x. We add 2 to both sides to get x = 5.

Now we have our x value. Let's plug that in to either equation and get a y value. Just pick the one you think will be easier. Let's use the first one. y - 2(5) = 1. y - 10 = 1. Add 10 and we get y = 11.

A good check is to put both variables back in both equations. If they don't work, you know you made a mistake somewhere. Let's try that here.

y - 2x = 1 becomes 11 - 2(5) = 1. That's 11 - 10 = 1. And yep, 1 = 1. And, 5x - 2y = 3 becomes 5(5) - 2(11) = 3. That's 25 - 22 = 3. Yep again! 3 = 3. We're good! And, no future in-laws were insulted in the solving of this problem.

Let's try one more. Here are two equations:

y = 3x - 4 and 2y - 5x = 2

In this one, we already have one solved for y, so let's just plug 3x - 4 in for y in the second equation. We get 2(3x - 4) - 5x = 2. 2 * 3x is 6x and 2 * 4 is 8. 6x - 5x is just x. Then, we add 8 to both sides, and we get x = 10.

Now, plug 10 in for x in that first equation. y = 3(10) - 4. y = 30 - 4, or 26.

Okay, let's check our work. We have y = 3x - 4, we get 26 = 3(10) - 4. That's 30 - 4, which is 26! And, 2y - 5x = 2 becomes 2(26) - 5(10) = 2. That's 52 - 50 = 2. Success!

Elimination Practice

Let's try some elimination practice. Here are two equations:

x + y = 5 and 5y - 1 = 2x

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