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Lesson 110: Solving Equations with a Variable on Both Sides
In this lesson, we'll learn how to solve algebraic equations that have a variable on both sides of the equals sign. It just involves a few extra steps.
First, be sure to review all of the
previous lessons on algebra. Make sure that you fully
understand them. In this lesson learn how to solve equations
that look like this:
5x - 7 = 3x + 9
Let's do that example. What's different is that we have a variable on both sides of the equals sign. The general rule for problems like this is that we want to all of the variable terms on one side of the equals sign, and all of the constant (non-variable) terms on the other side. Then the problem will look just like what we've already learned how to solve.
Let's start with the constants. First, I must decide if I want to try to get them all onto the left, or onto the right side of the equals sign. Students often are not comfortable with having to make the decision, even though there is no wrong answer. Here is how I'll look at the problem. To get rid of the 9 on the right side, I'll have to subtract 9 from both sides of the equation, like we learned how to do. On the left, we would then have -7 - 9, and that will give us a negative number. That is no big deal, and would work just fine, but given the choice, I'd rather avoid the negative number.
Instead, what I'll do is see that I have a -7 on the left, and so I'll add 7 to each side. On the left the 7's cancel. We now have 5x = 3x + 9 + 7. All the constants are now on one side. We'll combine them, and now we have 5x = 3x + 16. Now that the constants are all on the right, so our variables must go on the left. We'll need to subtract 3x from each side of the equation, to get rid of it on the right. We have 5x - 3x = 16. Combine to get 2x = 16. Divide each side by 2 to get x = 8. As always, be sure to check your answer by substituting the answer into the original equation, and seeing if it works.
Note that if you were solving this problem on paper, you would be expected to line up each step on top of each other, and solve the problem step-by-step as shown in previous lessons. In some cases, you will lose points on a test if you do not clearly show each step, or if you skip steps and do them in your head.
It's very important to understand that we could have chosen to do
get our variables and constants onto the opposite sides of what we
did in this problem. We will get the exact same answer, but
we'll have to work with some negatives. Here is the problem
again:
5x - 7 = 3x + 9
Let's try getting our constants onto the left. We'll subtract 9 from each side. We now have 5x - 7 - 9 = 3x. Combining, we have 5x - 16 = 3x. Be careful with your subtraction. Now we'll need to get our variables onto the right. We'll subtract 5x from each side, and are left with -16 = -2x. Again, be very careful when subtracting. Now we'll have to divide each side by -2. Remember that a negative divided by a negative equals a positive, so we get x = 8, just like above.
Let's now do the example that we put aside in Lesson 107, when we learned how to combine like terms. After combining, we ended up with 2x + 6 = 3x + 2. We can follow the procedure above. I'll choose to get my x terms onto the right, and my constants onto the left, but we could have done the exact opposite. First we'll subtract 2x from each side, therefore eliminating the x term on the left. We now have 6 = x + 2. Now I'll subtract 2 from each side, eliminating the constant on the right. We have x = 4. Always check your answer in the original equation to see if it works, and it does.
You'll have more practice with this in later lessons. Just study the way that we solved this equation, and make sure that you understand the steps, and the overall idea. Make sure that you understand that it doesn't matter which side you get the variables onto, and which side the constants, as long as you get them onto opposite sides.
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.