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# Linear equations with unknown coefficients

CCSS Math: HSA.REI.B.3

## Video transcript

- [Voiceover] So we have an equation. It says, a-x plus three-x
is equal to b-x plus five. And what I want to do
together is to solve for x, and if we solve for x it's going to be in terms of a, b, and other numbers. So pause the video and
see if you can do that. All right now, let's do this together, and what I'm going to do, is I'm gonna try to
group all of the x-terms, let's group all the x-terms
on the left-hand side. So, I already have a-x and
three-x on the left-hand side. Let's get b-x onto the left-hand
side as well, and I can do that by subtracting b-x from both sides. And if I subtract b-x from both sides, I'm going to get on the right-hand side, I'm going to have a, or
on the left-hand side, a-x plus three-x minus b-x, so I can do
that in that color for fun, minus b-x, and that's
going to be equal to... Well, b-x minus b-x is just zero, and I have five. It is equal to five. And now what I can do is, I can factor an x out
of this left-hand side of this equation, out of all of the terms. So, I can rewrite this as x times... Well, a-x divided by x is a. Three-x divided by x is three, and then negative b-x divided by x is just going to be negative b. I could keep writing
it in that pink color. And that's all going to be equal to five. And now, to solve for x I can just divide both sides by, the thing that
x is being multiplied by, by a plus three minus b. So, I can divide both sides by a plus three minus b. A plus three minus b. On this side, they cancel out. And, I have x is equal to five over a plus three minus b, and we are done. Let's do one more of these. So, another equation here. We have a... Here we have a times the
quantity, five minus x, is equal to b-x minus eight. So, once again, pause the video and see if you can solve for x. Well, the way I like to approach these is let's just expand everything all out. So, let me just distribute this a, and then I'm gonna collect
all the x-terms on one side, and all of the non-x-terms
on the other side, and essentially do what I
just did in the last example. So, let's first distribute this a. So the left-hand side becomes five-a, I could say a
times five or five-a, minus a-x, a-x, that is going to be equal to b-x minus eight. Now we can subtract b-x from both sides. So, we're gonna subtract
b-x from the left-hand side, b-x from the right-hand side. And, once again, the whole
reason I'm doing that, I want all the x-terms on the left, and all the non-x-terms on the right. And, actually, since I want all
the non-x-terms on the right, I can actually subtract five-a
from both sides, as well. So I'm kind of doing
two steps at once, here, but hopefully it makes sense. I'm trying to get rid of the b-x here, and I'm trying to get
rid of the five-a here. So, I subtract five-a there, and I'll subtract five-a there, and then let's see what this give us. So, the five-a's cancel out. And, on the left-hand side, I have negative a-x, negative a-x, minus b-x, minus, you know, in that same green color, minus b-x. And on the right-hand side, I have... This is going to be equal to, the b-x's cancel out, and I have negative eight minus five-a. Negative eight minus, in that same magenta color, minus five-a. And let's see, I have all my x's on one side, all my
non-x's on the other side. And here I can factor out an x, and if I factor out an x, what do I have? And, actually, one thing
that might be nice. Let me just multiply both
sides by negative one. If I multiple both sides by negative one, I get a-x plus b-x, plus b-x is equal to eight plus five-a. That just gets rid of all
of those negative signs. And now I can factor out an x here. So let me factor out an x, and I get x times a plus b. A plus b is going to be equal to eight plus five-a. Eight plus five-a. And we're in the home stretch now. We can just divide both sides by a plus b. So we could divide both sides by a plus b. A plus b. And we're going to be left with, x is equal to eight plus five-a, plus five-a, over... Actually, I'll write a and
b in our original colors. A plus, so this was that a, that a, plus b. So this is the b, that's a and b, a plus b, and we're done. We have now solved for x in terms of a's and b's and other things. And we are all done.