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Solving Equations with a Variable on Both Sides - Problem 4
In any single variable equation, start by simplifying -- distribute and combine any like terms on both sides of the equation. Now you can start solving using inverse operations. The goal is to get the variable on one side of the equal sign and all the numbers on the other side. Since there are variables on both sides, you want to eliminate the variable from one side of the equation. After you have done this, the variable should only be on one side of the equal sign. Next, you want all the constants on the other side of the equal sign. Use inverse operations to eliminate any constants that are on the same side of the equal sign as the variable. Remember that when solving equations, you must work in the reverse order of PEMDAS. In some problems after you bring the variable to one side and all the constants to the other side, the result will be an equation where one side of the equal sign is identical to the other side. For example: 2x + 1 = 2x + 1. In cases like this, the equation has infinite number of solutions, or the variable could be any value. This means that no matter what value is substituted in for the variable, the equation will be true.
So when I'm asked to do this problem, this is the kind of situation where I know it's going to be multiple steps. I know there's multiple steps because I have parentheses everywhere, I have Xs everywhere, oh my gosh. The thing to keep in mind is that you want to simplify each side on it's own before you start solving.
So let's look at this side first. I want to distribute that 2 meaning the 2 is going to get multiplied by both of those terms. I'll have 14x plus 6. Over here I want to do some distributing also 10x plus 10 and then even more distributing 4x take away 4.
Now that I have all my distributing done, I have a lot of combining like terms to do on this right side. This left side I can just leave as it is, there's nothing to combine, no numbers that need to be added together. Whereas here, let's see, I have 10 Xs plus I have 14 Xs total and then here I have 10 take away 4, so that's going to be minus, oh sorry, +6. All I've done so far is simplify both sides of the equation.
Oh-oh look at this you guys. This is weird. We haven't seen this before. I have 14x+6=14x+6 like yeah, I know it equals to each other. What happens in this situation is I have a problem where any x value would work. I could stick any number I want to in here and both sides would be equal. I could stick 8 million and I would still get the same answer on both sides. I could stick -0.3 and I would still get the same answer on both sides. What this means is that x could be any number and that's like the answer you write, x could be any value. It's not a number answer necessarily, it's like a word answer and that's something that you'll start to see more of as you progress through your Maths, Math careers. X could be any value. No matter what number I stick in here, I'm going to get a true statement. It's going to be equal on both sides.
I never would have known that looking at the problem to begin with. There's no give away up here that would tell me that. You have to work through and do your solving, excuse me, simplifying techniques in order to see that this is the case. When you see this situation, don't freak out. This is an answer, sometimes x could be anything.