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## Equations with variables on both sides

Current time:0:00Total duration:4:50

## Video transcript

We have the equation 3/4x plus
2 is equal to 3/8x minus 4. Now, we could just, right from
the get go, solve this the way we solved everything else, group
the x terms, maybe on the left-hand side, group the
constant terms on the right-hand side. But adding and subtracting
fractions are messy. So what I'm going to do, right
from the start of this video, is to multiply both sides of
this equation by some number so I can get rid of
the fractions. And the best number to do it
by-- what number is the smallest number that if I
multiply both of these fractions by it, they won't be
fractions anymore, they'll be whole numbers? That smallest number
is going to be 8. I'm going to multiply 8 times
both sides of this equation. You say, hey, Sal, how
did you get 8? And I got 8 because I said,
well, what's the least common multiple of 4 and 8? Well, the smallest number that
is divisible by 4 and 8 is 8. So when you multiply
by 8, it's going to get rid of the fractions. And so let's see what happens. So 8 times 3/4, that's
the same thing as 8 times 3 over 4. Let me do it on the
side over here. That's the same thing as 8 times
3 over 4, which is equal to 8 divided by 4 is just 2. So it's 2 times 3, which is 6. So the left-hand side becomes
8 times 3/4x is 6x. And then 8 times 2 is 16. You have to remember, when you
multiply both sides, or a side, of an equation by a
number, you multiply every term by that number. So you have to distribute
the 8. So the left-hand side is 6x plus
16 is going to be equal to-- 8 times 3/8, that's pretty
easy, the 8's cancel out and you're just
left with 3x. And then 8 times negative
4 is negative 32. And now we've cleaned up the
equation a good bit. Now the next thing, let's try to
get all the x terms on the left-hand side, and all the
constant terms on the right. So let's get rid of this
3x from the right. Let's subtract 3x from
both sides to do it. That's the best way I can think
of of getting rid of the 3x from the right. The left-hand side of this
equation, 6x minus 3x is 3x. 6 minus 3 is 3. And then you have a plus 16 is
equal to-- 3x minus 3x, that's the whole point of subtracting
3x, is so they cancel out. So those guys cancel out, and
we're just left with a negative 32. Now, let's get rid of the 16
from the left-hand side. So to get rid of it, we're going
to subtract 16 from both sides of this equation. Subtract 16 from both sides. The left-hand side of the
equation just becomes-- you have this 3x here; these 16's
cancel out, you don't have to write anything-- is equal to
negative 32 minus 16 is negative 48. So we have 3x is equal
to negative 48. To isolate the x, we can just
divide both sides of this equation by 3. So let's divide both sides
of that equation by 3. The left-hand side of the
equation, 3x divided by 3 is just an x. That was the whole point
behind dividing both sides by 3. And the right-hand side,
negative 48 divided by 3 is negative 16. And we are done. x equals negative 16
is our solution. So let's make sure that this
actually works by substituting to the original equation
up here. And the original equation
didn't have those 8's out front. So let's substitute in the
original equation. We get 3/4-- 3 over 4-- times
negative 16 plus 2 needs to be equal to 3/8 times negative
16 minus 4. So 3/4 of 16 is 12. And you can think
of it this way. What's 16 divided by 4? It is 4. And then multiply that
by 3, it's 12, just multiplying fractions. So this is going to
be a negative 12. So we get negative 12 plus
2 on the left-hand side, negative 12 plus 2
is negative 10. So the left-hand side
is a negative 10. Let's see what the right-hand
side is. You have 3/8 times
negative 16. If you divide negative 16 by 8,
you get negative 2, times 3 is a negative 6. So it's a negative 6 minus 4. Negative 6 minus 4
is negative 10. So when x is equal to negative
16, it does satisfy the original equation. Both sides of the equation
become negative 10. And we are done.