Linear equations with unknown coefficients | Mathematics I | High School Math | Khan Academy

So we have an equation. It says ( ax + 3x = bx + 5 ).

And what I want to do together is to solve for ( x ). 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. What I'm going to do is I'm going to try to group all of the ( x ) terms. Let's group all the ( x ) terms on the left-hand side. I already have ( ax ) and ( 3x ) on the left-hand side. Let's get ( bx ) onto the left-hand side as well. I can do that by subtracting ( bx ) from both sides.

If I subtract ( bx ) from both sides, I'm going to get, on the right-hand side, I'm going to have ( 5 ). On the left-hand side, I have ( ax + 3x - bx ). I could do that in that color for fun: ( - bx ). And that's going to be equal to, well, ( bx - bx ) is just ( 0 ), and I have ( 5 ). It is equal to ( 5 ).

Now what I can do is I can factor an ( x ) out of the left-hand side of this equation, out of all of the terms. So I can rewrite this as ( x \times ( \frac{a}{x} + 3 - \frac{b}{x} ) ), where ( ax \div x = a ), ( 3x \div x = 3 ), and ( -bx \div x = -b ). That’s all going to be equal to ( 5 ).

Now to solve for ( x ), I can just divide both sides by the thing that ( x ) is being multiplied by, ( a + 3 - b ). So I can divide both sides by ( a + 3 - b ). On this side, they cancel out, and I have ( x = \frac{5}{a + 3 - b} ).

And we are done!

Let's do one more of these. So another equation here, we have ( a ) here, we have ( a \times (5 - x) = bx - 8 ). 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 out. So let me just distribute this ( a ), and then I'm going to 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 ). The left-hand side becomes ( 5a - ax ). That is going to be equal to ( bx - 8 ).

Now we can subtract ( bx ) from both sides. So we're going to subtract ( bx ) from the left-hand side and from the right-hand side. Once again, the whole reason I'm doing that is I want all the ( x ) terms on the left and all the non-( x ) terms on the right.

Actually, since I want all the non-( x ) terms on the right, I can also subtract ( 5a ) from both sides. I'm kind of doing two steps at once here, but hopefully, it makes sense. I'm trying to get rid of ( bx ) here and get rid of ( 5a ) here, so I subtract ( 5a ) there and I'll subtract ( 5a ) there.

Then let’s see what this gives us. The ( 5a )s cancel out, and on the left-hand side, I have ( -ax - bx ). I’m doing that same green color: ( - bx ). On the right-hand side, this is going to be equal to ( -8 - 5a ) (let’s say magenta color).

Now I’ve separated all my ( x )s on one side and all my non-( x )s on the other side. Here I can factor out an ( x ). If I multiply both sides by ( -1 ), I get ( ax + bx = 8 + 5a ).

That just gets rid of all those negative signs. Now I can factor out an ( x ). I get ( x \times (a + b) = 8 + 5a ).

Now we can just divide both sides by ( a + b ). So we divide both sides by ( a + b ), and we're going to be left with ( x = \frac{8 + 5a}{a + b} ).

And we are done! We have now solved for ( x ) in terms of ( a )s and ( b )s and other things, and we are all done.