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How to Write a Linear Equation Solving Linear Equations Practice Problems Solving Linear Equations With Literal Coefficients Problem Solving Using Linear Equations How do I Use a System of Equations Solving a System of Equations With Two Unknown Solving Problems Involving Systems of Equations What is an Inequality Solving Linear Inequalities Practice Problems Horizontal Asymptotes Definition Rules Linear Inequality Solving Graphing Problems

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GMAT: What is a Linear Equation
Solving a System of Equations With Two Unknown

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.

Solving a System of Equations With Two Unknown

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

Think of this like making a sandwich. You need your ingredients to line up. If your tomato is over too far, it's going to slide right out. So, let's line everything up.

x + y = 5

5y - 1 = 2x

Great! Now we need to get coefficients balanced for one of the variables so they'll cancel out. This is like making sure your amounts of mayonnaise and mustard are in balance. Here, we can multiply the top equation by 2 so we get 2x on top to match the 2x on the bottom.

Now, let's subtract. The 2x's cancel out. We're left with -3y = 9. Divide by -3 and y = -3. Now, plug -3 in for y in the first equation. x - 3 = 5. So, x = 8. That's our sandwich! But wait. Let's check our work before we eat it.

In x + y = 5, we get 8 - 3 = 5. That works out. In 2x + 5y = 1, we get 2(8) + 5(-3) = 1. That's 16 - 15 = 1. Success!

Let's try another:

2x - y = 3

-4x + 3y = 1

Hey, look - our terms are neatly stacked. Thanks, sandwich gods! And, we can multiply the top equation by 2 to get rid of the x terms. So, it becomes 4x - 2y = 6. Here, we want to add. -2y + 3y is just y. And, 6 + 1 is 7. So, y = 7.

Now, plug 7 in for y in one of the equations. 2x - 7 = 3. 2x = 10. x = 5.

Let's check our answers. 2x - y = 3 becomes 2(5) - 7 = 3. That's 10 - 7 = 3, so that works. And, -4x + 3y = 1 becomes -4(5) + 3(7) = 1. That's -20 + 21 = 1, so that works too. We did it!