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Lesson 135: Solving Systems of Linear Equations Algebraically (Part 1 of 2)
In this very important lesson we'll learn how to solve systems of linear equations using algebra, instead of graphically. This is often a better solution when solving graphically is not practical for any reason.
Before starting this lesson, make sure that you review all of the previous lessons on algebra, and graphing linear equations. Make sure that you fully understand them, or you will have difficulty with this lesson.
Very often we are asked to solve a system of linear equations. If we are not specially told to do so graphically, then we can assume that we should do so algebraically. Recall that what we are trying to do is find a value of x and y that will satisfy both equations at the same time.
Let's take the equations y + 2 = 3x, and -x - y = -2. These are both linear equations, even though they are not in slope-intercept form. If we wanted to graph these equations, we would probably first put them in that form, but for what we are doing, we don't have to.
What we need to do, though, is use algebra to manipulate these
equations so that the variables and constants are in the same place
in both equations. There is more than one way to do this, so
we'll look for what seems the easiest. First I'll rewrite the
equations so they are on top of each other.
y + 2 = 3x
-x - y = -2
I can see that there is a y term on the left side of the both
equations, so I'll leave that there. In the first equation,
the constant is on the left of the equals, and in the second one, it
is on the right. What I'll do is subtract 2 from each side of the
first equation, to move our constant to the right. We now
have:
y = 3x - 2
-x - y = -2
Now we have an x term to the right of the equals in the first
equation, but our x term in the second equation is on the left.
I'll subtract 3x from each side of the first equation, to get the x
term on the left. We now have:
y - 3x = -2
-x - y = -2
I'll rewrite the first equation so that the x term comes first, to
line up with the second equation. We have to be careful with
our signs. We get
-3x + y = -2
-x - y = -2
Now our variables and constants line up. What we have to now do is either add or subtract the equations so that a variable drops out. Let's examine what we have. We have a -x term, and a -3x term, so adding or subtracting those will not cause them to drop out. But, we have a y and a -y. If we add the two equations together, the y term will drop out. It's good that it worked that way. In the next example, you'll see what to do if it does not.
What we do is add the two equations "downward." We'll add -3x
+ -x to get -4x. Our y terms drop out. We then
have -2 + -2 which is -4. We now have:
-4x = -4. Solving for x we get 1. Now we can substitute
that value of x into either original equation to determine the value
of y. Doing so, and using basic algebra, we can see that y =
1. We should verify the the solution of (1, 1) does indeed
work in both equations. In fact, that is the only solution to
the problem.
Let's try a more tricky example:
2x + 3y = 6
4x + y = 12
Here the equations are such that the variables and constants are lined up nicely on top of each other. If they weren't, we would have to use basic algebra like we did above to arrange them like that.
Now, though, a quick glance will show that if we add or subtract these equations, it will not cause any variable to drop out. We have a few choice. The first option is to multiply both sides of the first equation by 2. That will give us a 4x in the first equation to match the one in the second equation. We would then subtract the equations to get the x term to drop out. The second option is to multiply both sides of the second equation by 3. That will give us a 3y in the second equation to match the 3y in the first. Again, we'd subtract the two equations. Either option will work, so just pick one.
I'll pick multiplying the second equation by 3. After doing
so, our equations are:
2x + 3y = 6
12x + 3y = 36
Now we'll subtract the second equation from the first by subtracting "downward". We get -10x = -30. As always, be very careful with your signs and with your basic arithmetic. Solving for x, we get x = 3. By the way, sometimes your answer will be a fraction. Don't assume your answer is wrong, just check it.
Now we need to substitute x into either original equation to determine y. I'll choose the first one. Substituting x, and solving for y using basic algebra, we see that y = 0. Our solution is (3,0). As always, make sure that it works in both original equations.
Note that in some cases, you will need to multiply the first
equation by a particular number, and the second equation by a
different number, in order to set up a variable so that it will drop
out. For example, look at these equations:
4x - 2y = 7
3x + 5y = 9
I can't just multiply one equation to set up a variable to drop out. I have two options. The first option is to get the x term to drop out. I can multiply the top equation by 3, and the bottom by 4. That will give us a 12x term in each equation, which would drop out when subtracting. I chose the number that I needed to use by looking at the coefficients 4 and 3. What I need to find was the LCM of those two numbers, and then multiply each equation such that each would have an x term with that LCM as a coefficient. Stated more simply, seeing 4x and 3x, I knew that 12x would have to be my target for each equation.
Another option would be to get the y terms to drop out. The LCM of 2 and 5 is 10. I could multiply the top equation by 5, and the bottom equation by 2, in order to have a 10y term in both equations. Then I'd proceed as above. As you do more problems like this, they will become easier.
In the next lesson you will learn an alternate method for doing these problems which, in some cases, is an easier option.
Remember that you can ask a math question if you have additional questions about a topic, or you can contact me if you have any comments or suggestions for this site.