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Welcome to the presentation on
level four linear equations.

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So, let's start doing
some problems.

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So.

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

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problems-- if I said 3 over x
is equal to, let's just say 5.

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

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everything we've ever seen.

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

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have x in the denominator.

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

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

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numerator or at least not in
the denominator as

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soon as possible.

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

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

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

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x's will cancel out.

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And in the right side,
you'll just get 5 times x.

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

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And you get 3 is equal to 5x.

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Now, we could also write
that as 5x is equal to 3.

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And then we can think
about this two ways.

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We either just multiply both
sides by 1/5, or you could just

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do that as dividing by 5.

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If you multiply
both sides by 1/5.

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The left-hand side becomes x.

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And the right-hand side, 3
times 1/5, is equal to 3/5.

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So what did we do here?

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This is just like, this
actually turned into a level

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two problem, or actually a
level one problem,

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very quickly.

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All we had to do is multiply
both sides of this

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equation by x.

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

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Let's do another problem.

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Let's have -- let me say,
x plus 2 over x plus 1 is

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equal to, let's say, 7.

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

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we have a whole x plus
1 in the denominator.

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

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To get that x plus 1 out of the
denominator, we multiply both

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sides of this equation times x
plus 1 over 1 times this side.

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

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to do it on the right-hand
side, and this is just 7/1,

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times x plus 1 over 1.

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On the left-hand side, the
x plus 1's cancel out.

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And you're just left
with x plus 2.

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It's over 1, but we can
just ignore the 1.

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And that equals 7
times x plus 1.

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And that's the same
thing as x plus 2.

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And, remember, it's 7 times
the whole thing, x plus 1.

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So we actually have to use
the distributive property.

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And that equals 7x plus 7.

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

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three linear equation.

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

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one side of the equation.

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And let's get all the constant
terms, like the 2 and the 7, on

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the other side of the equation.

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

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So let's bring that
7x onto the left.

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And we can do that by
subtracting 7x from both sides.

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Minus 7x, plus,
it's a minus 7x.

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The right-hand side, these
two 7x's will cancel out.

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And on the left-hand side
we have minus 7x plus x.

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Well, that's minus 6x plus
2 is equal to, and on the

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right all we have left is 7.

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Now we just have to
get rid of this 2.

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And we can just do that by
subtracting 2 from both sides.

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And we're left with minus
6x packs is equal to 6.

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Now it's a level one problem.

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We just have to multiply both
sides times the reciprocal

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of the coefficient on
the left-hand side.

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And the coefficient's
negative 6.

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So we multiply both sides of
the equation by negative 1/6.

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Negative 1/6.

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The left-hand side, negative
1 over 6 times negative 6.

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Well that just equals 1.

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So we just get x is equal
to 5 times negative 1/6.

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Well, that's negative 5/6.

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And we're done.

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

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equals negative 5/6 and put it
back in the original question

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to confirm that it worked.

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Let's do another one.

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

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Let me think.

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3 times x plus 5 is equal
to 8 times x plus 2.

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Well, we do the
same thing here.

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Although now we have two
expressions we want to get

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out of the denominators.

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We want to get x plus 5
out and we want to get

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this x plus 2 out.

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So let's do the x plus 5 first.

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

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this equation by x plus 5.

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You can say x plus 5 over 1.

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Times x plus 5 over 1.

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

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So we're left with 3 is equal
to 8 times x plus five.

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All of that over x plus 2.

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

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just multiply the 8 times
the whole expression.

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So it's 8x plus 40
over x plus 2.

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Now, we want to get
rid of this x plus 2.

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So we can do it the same way.

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We can multiply both sides
of this equation by

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x plus 2 over 1.

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x plus 2.

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We could just say we're
multiplying both

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sides by x plus 2.

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The 1 is little unnecessary.

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So the left-hand side
becomes 3x plus 6.

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Remember, always distribute
3 times, because you're

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multiplying it times
the whole expression.

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x plus 2.

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And on the right-hand side.

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Well, this x plus 2 and this
x plus 2 will cancel out.

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And we're left with 8x plus 40.

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And this is now a
level three problem.

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Well, if we subtract 8x from
both sides, minus 8x, plus-- I

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think I'm running out of space.

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Minus 8x.

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Well, on the right-hand
side the 8x's cancel out.

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On the left-hand side we have
minus 5x plus 6 is equal

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to, on the right-hand side
all we have left is 40.

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Now we can subtract 6 from
both sides of this equation.

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Let me just write out here.

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Minus 6 plus minus 6.

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

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trying to go up here.

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But if we subtract minus 6 from
both sides, on the left-hand

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side we're just left with
minus 5x equals, and on the

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right-hand side we have 34.

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Now it's a level one problem.

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We just multiply both
sides times negative 1/5.

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Negative 1/5.

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On the left-hand
side we have x.

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And on the right-hand side
we have negative 34/5.

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

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

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ready to tackle some level
four linear equations.

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Have fun.