 # Systems of equations with elimination (and manipulation) | High School Math | Khan Academy

January 13, 2020

Let’s solve a few more systems
of equations using elimination, but in these it
won’t be kind of a one-step elimination. We’re going to have to massage
the equations a little bit in order to prepare them
for elimination. So let’s say that we have an
equation, 5x minus 10y is equal to 15. And we have another equation,
3x minus 2y is equal to 3. And I said we want to do
this using elimination. Once again, we could use
substitution, we could graph both of these lines and figure
out where they intersect. But we’re going to
use elimination. But the first thing you might
say, hey, Sal, you know, with elimination, you were
subtracting the left-hand side of one equation from another,
sides. And I could do that, because it
was essentially adding the same thing to both sides
of the equation. But here, it’s not
obvious that that would be of any help. If we added these two left-hand
sides, you would get 8x minus 12y. That wouldn’t eliminate
any variables. And on the right-hand
side, you would just be left with a number. And if you subtracted,
that wouldn’t eliminate any variables. So how is elimination
going to help here? And the answer is, we can
multiply both of these equations in such a way that
maybe we can get one of these terms to cancel out with
one of the others. And you could really
pick which term you want to cancel out. Let’s say we want to cancel out
the y terms. So I’ll just rewrite this 5x minus
10y here. 5x minus 10y is equal to 15. Now, is there anything that
I can multiply this green equation by so that this
negative 2y term becomes a term that will cancel out
with the negative 10y? So I essentially want to make
this negative 2y into a positive 10y. Right? Because if this is a positive
10y, it’ll cancel out when I add the left-hand sides
of this equation. So what can I multiply
this equation by? Well, if I multiply it by
negative 5, negative 5 times negative 2 right here would
be positive 10. So let’s do that. Let’s multiply this equation
times negative 5. So you multiply the left-hand
side by negative 5, and multiply the right-hand
side by negative 5. And what do you get? Remember, we’re not
fundamentally changing the equation. We’re not changing the
information in the equation. We’re doing the same thing
to both sides of it. So the left-hand side of the
equation becomes negative 5 times 3x is negative 15x. And then negative 5 times
negative 2y is plus 10y, is equal to 3 times negative
5 is negative 15. And now, we’re ready to
do our elimination. If we add this to the left-hand
side of the yellow equation, and we add the
negative 15 to the right-hand side of the yellow equation, we
are adding the same thing to both sides of the equation. Because this is equal to that. So let’s do that. So 5x minus 15y– we have this
little negative sign there, we don’t want to lose that–
that’s negative 10x. The y’s cancel out. Negative 10y plus
10y, that’s 0y. That was the whole point behind
multiplying this by negative 5. Is going to be equal to–
15 minus 15 is 0. So negative 10x is equal to 0. Divide both sides by negative
10, and you get x is equal to 0. And now we can substitute back
into either of these equations to figure out what y
must be equal to. Let’s substitute into
the top equation. So we get 5 times 0, minus
10y, is equal to 15. Or negative 10y is
equal to 15. Let me write that. Negative 10y is equal to 15. Divide both sides
by negative 10. And we are left with
y is equal to 15/10, is negative 3/2. So if you were to graph it,
the point of intersection would be the point
0, negative 3/2. And you can verify that it also
satisfies this equation. The original equation over
here was 3x minus 2y is equal to 3. 3 times 0, which is 0, minus 2
times negative 3/2 is, this is 0, this is positive 3. Right? These cancel out, these
become positive. Plus positive 3 is equal to 3. So this does indeed satisfy
both equations. Let’s do another one of these
where we have to multiply, and to massage the equations, and
then we can eliminate one of the variables. Let’s do another one. Let’s say we have 5x plus
7y is equal to 15. And we have 7– let me do
another color– 7x minus 3y is equal to 5. Now once again, if you just
added or subtracted both the left-hand sides, you’re
not going to eliminate any variables. These aren’t in any way kind of
have the same coefficient or the negative of their
coefficient. So let’s pick a variable
to eliminate. Let’s say we want to eliminate
the x’s this time. And you could literally
pick on one of the variables or another. It doesn’t matter. You can say let’s eliminate the
y’s first. But I’m going to choose to eliminate the x’s
first. And so what I need to do is massage one or both of
these equations in a way that these guys have the same
coefficients, or their coefficients are the negatives
of each other, so that when I add the left-hand sides, they’re
going to eliminate each other. Now, there’s nothing obvious–
I can multiply this by a fraction to make it equal
to negative 5. Or I can multiply this by a
fraction to make it equal to negative 7. But even a more fun thing to do
is I can try to get both of them to be their least
common multiple. I could get both
of these to 35. And the way I can do it is by
multiplying by each other. So I can multiply this
top equation by 7. And I’m picking 7 so that
this becomes a 35. And I can multiply this bottom
equation by negative 5. And the reason why I’m doing
that is so this becomes a negative 35. Remember, my point is I want
to eliminate the x’s. So if I make this a 35, and if
I make this a negative 35, then I’m going to be all set. I can add the left-hand
and the right-hand sides of the equations. So this top equation, when you
multiply it by 7, it becomes– let me scroll up a little bit–
we multiply it by 7, it becomes 35x plus 49y is equal
to– let’s see, this is 70 plus 35 is equal to 105. Right? 15 and 70, plus 35,
is equal to 105. That’s what the top
equation becomes. This bottom equation becomes
negative 5 times 7x, is negative 35x, negative 5 times
negative 3y is plus 15y. The negatives cancel out. And then 5– this isn’t a
minus 5– this is times negative 5. 5 times negative 5 is equal
equation and add the same thing to both sides, where that
same thing is negative 25, which is also equal
to this expression. So let’s add the left-hand
sides and the right-hand sides. Because we’re really adding the
same thing to both sides of the equation. So the left-hand side,
the x’s cancel out. 35x minus 35x. That was the whole point. They cancel out, and on the
y’s, you get 49y plus 15y, that is 64y. 64y is equal to 105 minus
25 is equal to 80. Divide both sides by 64, and you
get y is equal to 80/64. And let’s see, if you divide
the numerator and the denominator by 8– actually
you could probably do 16. 16 would be better. But let’s do 8 first,
just because we know our 8 times tables. So that becomes 10/8, and then
you can divide this by 2, and you get 5/4. If you divided just straight
up by 16, you would’ve gone straight to 5/4. So y is equal to 5/4. Let’s figure out what x is. So we can substitute either into
one of these equations, or into one of the original
equations. Let’s substitute into the
second of the original equations, where we had 7x
minus 3y is equal to 5. That was the original version of
the second equation that we later transformed into this. So we get 7x minus 3 times y,
times 5/4, is equal to 5. Or 7x minus 15/4
is equal to 5. Let’s add 15/4– Oh, sorry,
I didn’t do that right. This would be 7x minus 3 times
4– Oh, sorry, that was right. What am I doing? 3 times is 15/4. Is equal to 5. Let’s add 15/4 to both sides. And what do we get? The left-hand side just
becomes a 7x. These guys cancel out. And that’s going to be
equal to 5, is the same thing as 20/4. 20/4 plus 15/4. Or we get that– let me scroll
down a little bit– 7x is equal to 35/4. We can multiply both sides by
1/7, or we could divide both sides by 7, same thing. Let’s multiply both
sides by 1/7. The same thing as
dividing by 7. So these cancel out and you’re
left with x is equal to– Here, if you divide 35
by 7, you get 5. You divide 7 by 7, you get 1. So x is equal to 5/4 as well. So the point of intersection of
this right here is both x and y are going to
be equal to 5/4. So if you looked at it as a
graph, it’d be 5/4 comma 5/4. And let’s verify that this
satisfies the top equation. And if you take 5 times
5/4, plus 7 times 5/4, what do you get? It should be equal to 15. So this is equal to 25/4,
plus– what is this? This is plus 35/4. Which is equal to 60/4, which
is indeed equal to 15. So it does definitely satisfy
that top equation. And you could check out this
bottom equation for yourself, but it should, because we
actually used this bottom equation to figure out that
x is equal to 5/4.