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

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,

or adding the two, and then adding the two right-hand

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

to negative 25. Now, we can start with this top

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.

DAMN 'IT YOUTUBE an error occurred please try again later

@JonasM108 I did the same thing. It just so happens that in this problem, 5/4 is the answer for both x and y.

Wow! This helped me so much!! Thanks for posting this~!

… The pretty colors helped, too. ^-^

OMG! THANK YOU SOOOO MUCH

All the other tutorials were for adding and subracting and i taught thos were the only two ways you could do it and my exam is tommrow

THANK YOUUUUUU!!!!

I understand

i get it now!

thank you so much for this homework help, i think i finally get it!!!

THANKS

omg i forgot i had this app and i was really struggling!!

Thank you so muchhhh

I love this man i would not be passing algebra with out him

I thank you, my teacher sucks ass. Can't teach for shut.

Shit*

thank you for posting this

Neither can mine. They need to get rid of MOST of the teachers and just use Youtube for all the classes!

very helpful! you explained it better than my teacher ever could

<3

your voice is like butter being rubbed against my ears.

thats a good point. not spam

refresh

Super helpful

This is so helpful! My math teacher is a complete idiot!

What is the software that you use? I want to be able to do this on my own computer.

you saved my grades

Thank You know online school doesn't mean staying up till midnight doing HW thank you I also proved my dad wrong 2! he thinks he knows it all… he only doesn't know one thing… he doesn't know all the crap in da world!!!!!

THANK YOU! πΊπ

This was really helpful. Thankyou!

I'm here against my will but… …………you cool pie butt poo

I dont get it..

THANK YOU!

who invented this crap

Wouldn't subtracting -10y by +10 give you a -20

Ya if you add it you are right but he used subtraction that is why 5x – 15x gave him -10x but if you use addition it would give you a whole different answer so he made a mistake. I guess !!

Its three am I have a math test today and I am ready to die

what do I do with equations that start with a singular y for example y=2x-3 and the other one would be y=x+4

thank you

memes

I am more confused than ever. The fractions threw me off.

It would probably be easier to find either x or y for either equation and then just plug into the other equation.

I am in 8th grade and we learned this this is going to help me a lot in my finals thanks so much your the best

Is there any way to predict for which variable to solve first? Sometimes I need to solve complex circuits and I end up with three or four equations. However, it can sometimes be a bit difficult to choose the fastest path and trying out different paths takes some extra time, which isn't a very good thing.

You saved my life

Sal is amazing

How would you solve this with y=4x+1 and x+y=15?

It been 7 years since he uploaded this…

Yes! You saved the day again Khan Academy! I dunno why I can't understand my Math teacher lol.

none of these videos help its not just this one its everyone one them im not saying there bad but my teachers just wants us to suffer

what if you get a decimal in a equation

You just saved me from a 0 on my homework

Thank you Khan

Thank the lord for kan Academy!!ππ

I stg if there r fractions on my test tomorrow somebody is gonna get hurt

how you do x + 3x = y 3x + 4y = 7

I have a D+ in math π«π«π« all because of a group quiz

mAsSaGethe equationthank you for the helpful vids