Algebra: Linear Equations 4

Welcome to the presentation on level four linear equations.

So, let's start doing some problems.

So.

Let's say I had the situation-- let me give me a couple of

problems-- if I said three over x is equal to, let's just say five.

So, what we want to do -- this problem's a little unusual from

everything we've ever seen.

Because here, instead of having x in the numerator, we actually

have x in the denominator.

So, I personally don't like having x's in my denominators,

so we want to get it outside of the denominator into a

numerator or at least not in the denominator as

soon as possible.

So, one way to get a number out of the denominator is, if we

were to multiply both sides of this equation by x, you see

that on the left-hand side of the equation these two

x's will cancel out.

And in the right side, you'll just get five times x.

So this equals -- the two x's cancel out.

And you get three is equal to fivex.

Now, we could also write that as fivex is equal to three.

And then we can think about this two ways.

We either just multiply both sides by one / five, or you could just

do that as dividing by five.

If you multiply both sides by one / five.

The left-hand side becomes x.

And the right-hand side, three times one / five, is equal to three / five.

So what did we do here?

This is just like, this actually turned into a level

two problem, or actually a level one problem,

very quickly.

All we had to do is multiply both sides of this

equation by x.

And we got the x's out of the denominator.

Let's do another problem.

Let's have -- let me say, x plus two over x plus one is

equal to, let's say, seven.

So, here, instead of having just an x in the denominator,

we have a whole x plus one in the denominator.

But we're going to do it the same way.

To get that x plus one out of the denominator, we multiply both

sides of this equation times x plus one over one times this side.

Since we did it on the left-hand side we also have

to do it on the right-hand side, and this is just seven / one,

times x plus one over one.

On the left-hand side, the x plus one's cancel out.

And you're just left with x plus two.

It's over one, but we can just ignore the one.

And that equals seven times x plus one.

And that's the same thing as x plus two.

And, remember, it's seven times the whole thing, x plus one.

So we actually have to use the distributive property.

And that equals sevenx plus seven.

So now it's turned into a, I think this is a level

three linear equation.

And now all we do is, we say well let's get all the x's on

one side of the equation.

And let's get all the constant terms, like the two and the seven, on

the other side of the equation.

So I'm going to choose to get the x's on the left.

So let's bring that sevenx onto the left.

And we can do that by subtracting sevenx from both sides.

Minus sevenx, plus, it's a minus sevenx.

The right-hand side, these two sevenx's will cancel out.

And on the left-hand side we have minus sevenx plus x.

Well, that's minus six plus two is equal to, and on the

right all we have left is seven.

Now we just have to get rid of this two.

And we can just do that by subtracting two from both sides.

And we're left with minus six x is equal to six.

Now it's a level one problem.

We just have to multiply both sides times the reciprocal

of the coefficient on the left-hand side.

And the coefficient's negative six.

So we multiply both sides of the equation by negative one / six.

Negative one / six.

The left-hand side, negative one over six times negative six.

Well that just equals one.

So we just get x is equal to five times negative one / six.

Well, that's negative five / six.

And we're done.

And if you wanted to check it, you could just take that x

equals negative five / six and put it back in the original question

to confirm that it worked.

Let's do another one.

I'm making these up on the fly, so I apologize.

Let me think.

three times x plus five is equal to eight times x plus two.

Well, we do the same thing here.

Although now we have two expressions we want to get

out of the denominators.

We want to get x plus five out and we want to get

this x plus two out.

So let's do the x plus five first.

Well, just like we did before, we multiply both sides of

this equation by x plus five.

You can say x plus five over one.

Times x plus five over one.

On the left-hand side, they get canceled out.

So we're left with three is equal to eight times x plus five.

All of that over x plus two.

Now, on the top, just to simplify, we once again

just multiply the eight times the whole expression.

So it's eightx plus forty over x plus two.

Now, we want to get rid of this x plus two.

So we can do it the same way.

We can multiply both sides of this equation by

x plus two over one.

x plus two.

We could just say we're multiplying both

sides by x plus two.

The one is little unnecessary.

So the left-hand side becomes threex plus six.

Remember, always distribute three times, because you're

multiplying it times the whole expression.

x plus two.

And on the right-hand side.

Well, this x plus two and this x plus two will cancel out.

And we're left with eightx plus forty.

And this is now a level three problem.

Well, if we subtract eightx from both sides, minus eightx, plus-- I

think I'm running out of space.

Minus eightx.

Well, on the right-hand side the eightx's cancel out.

On the left-hand side we have minus fivex plus six is equal

to, on the right-hand side all we have left is forty.

Now we can subtract six from both sides of this equation.

Let me just write out here.

Minus six plus minus six.

Now I'm going to, hope I don't lose you guys by

trying to go up here.

But if we subtract minus six from both sides, on the left-hand

side we're just left with minus fivex equals, and on the

right-hand side we have thirty-four.

Now it's a level one problem.

We just multiply both sides times negative one / five.

Negative one / five.

On the left-hand side we have x.

And on the right-hand side we have negative thirty-four / five.

Unless I made some careless mistakes, I think that's right.

And I think if you understood what we just did here, you're

ready to tackle some level four linear equations.

Have fun.