Pre-Algebra: Equations with Variables on 2 Sides

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SOLVING EQUATIONS WITH VARIABLES ON BOTH SIDES

All right, let’s try one where we see lots and lots and lots of unknowns throughout. 5c plus 1 minus 6c equals negative 7 plus 2c plus 2. Wow. Well, we should try to combine like terms to get this thing going. The first thing I am going to do is to put all the c’s together on each side. Here I’ve got 5 c’s; then I’m taking away 6 c’s. That leaves me with negative 1 c, or negative c, and I’ve got a plus 1, so I combine these like terms, and the 1 just hangs out there. That is going to equal. Here I have a negative 7 plus 2; negative 7 plus 2 is negative 5 plus 2c. Now, I want to bring the c’s together. I am going to reunite the c’s. So, I am going to add c to both sides. Negative c plus c is 0. I’m just left with the 1 equals negative 5. I’ve got 2c; someone else gives me another c, and I’ve got a total of 3 c’s. What would I do now? Now I can add 5 to both sides, because I want to have just the c’s alone. There are no c’s here, so I see 6 is equal to 3c, divide both sides by 3, and I see that c equals 6 divided by 3, which is 2. So the answer is c equals 2. Cool.

Let’s try another one. All right, look at this one. Lots of denominators here, which might bother some people, including me. So, I like to try and clean off the denominators. I could multiply through by a common denominator. In fact, I could look at the least common denominator. If you are looking at 4, 5, 8, and 2, 8 certainly has within it factors of 4 and 2, but I need a 5. So, it looks like 40 is the least common denominator. Let’s multiply everything through by 40. That means this side by 40 and this side by 40. To multiply this side by 40, I’ve got to distribute that 40, and it has got to hit every single term; 40 times this, 40 times this, 40 times this, 40 times this, 40 times that, 40 times everything. Do you understand? All right. So, when I distribute I see 40 times this, and I see I can simplify this, and this just becomes 10 times 3 times w, which is 30w. Minus, when I multiply this by 40, you can see I get the cancellation of the 40 with the 5, and move an extra factor of 8 on top with this 8, which leads me to a 64w. When I multiply this by 40; what do I see? I see I am going to get a 5, because the 40 and the 8 can simplify a little bit. And that is going to equal: when I multiply this by 40, I see negative 40w, when I multiply this by 40, I see 40 divided by 2, which is just 20.

Now, it is an equivalent equation, but now there is no denominator, so I am happy. The numbers have been inflated a little bit. Of course, there is no such thing as a free math lunch. If you don’t want denominators, it comes with the price of having slightly larger numerators, but that’s okay. I don’t mind. They are all whole numbers. So, who cares? All right. Let’s see, I’ve got a 30w minus 64w, which leaves me with a—what? A negative 34w plus 5 equals negative 40w plus 20. How about we add 40w to both sides to bring those w’s together; to reunite them? Let’s reunite those w’s. Let’s hear it for the w’s. Let’s reunite them. Come on, right here on our stage. So here I have negative 34w plus 40w is going to be 6w plus 5 equals, and negative 4w plus 4w is 0, and I get 20. Right now I subtract 5—there are no w’s there. There are no w’s in 5. Why would we possibly want to keep those 5s there? I don’t know. 6w, that adds to give 0, equals 15. So, if I divide both sides by 6, I see that w equals fifteen-sixths, which if I simplify that, leaves me with just 5 over 2.

So the answer is w equals five-halves. Pretty cool.