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### Course: Algebra (all content) > Unit 5

Lesson 4: Solving systems of equations with substitution# Solving linear systems by substitution (old)

An old video where Sal introduces the substitution method for systems of linear equations. Created by Sal Khan.

## Want to join the conversation?

- How would u solve this problem:

{Y=-5x+3

{10x+2y=0(40 votes)- you have to isolate the y before you substitute(3 votes)

- I can't figure out how to use substitution to solve an equation with 2 variables.

For example:

y= -2x + 10

y= x - 1

I don't get it please help. :-)(4 votes)- Good evening.

One way would be to substitute the y in Your first equation with the entire right-hand side of Your second equation (which, as You can see, is equal to y):

`y = -2x + 10 :*now substitute y for x - 1*.

x - 1 = -2x + 10 :*add 2x + 1 to both sides*.

3x = 11 :*divide both sides by 3*.**x = 11/3**

Use that value to solve for y using one of Your two initial equations, for example:

y = x - 1 :*substitute x for 11/3*.

y = 11/3 - 1

y = 11/3 - 3/3**y = 8/3**

And voilà! You're done!(4 votes)

- This isn't a question but this helped a lot soooooo thanks(3 votes)
- Great! Glad it helped. But you should put it in the tips/thanks next time.(3 votes)

- when writing the equation don't the variables go first?(2 votes)
- It could be place at the end or the front of the equation, but it is recommended to put the variable in the front.(2 votes)

- What would one do if the equation came out X+(-10-8X)=8

I got y=-5-4x though in the equation it 2y did i do it right?(2 votes) - How would I use substitution method here?

3x + 2y = 15

2x + 6y = 11(2 votes)- Solve one of the equations for x or y, and then substitute that into the other equation.(2 votes)

- how would you solve y+2+4x

Y+Y=Y(2 votes) - Ok so far we did questions like y=4x-2 and such but i was wondering how do make the linear line if the equation is y=x²+5 and so...(2 votes)
- What if it was like this ?

x= 2y + 9

x = 5y + 20(2 votes)- This question isn't exactly the ideal question to use substitution but here:

First step: You have the "x" variables isolated in both equations but you only need it to be isolated in one equation. Therefore, you can move the -9 in the first equation to the other side.

x - 9 = 2y

Second step: You'd want to isolate the y in the first equation so you'd divide by 2 on both sides.

(x - 9) / 2 = y

Third step: In order to make the equation only have one variable, you substitute the second equation in for "x".

((5y + 20) - 9) / 2 = y

Fourth step: Simplify!

(5y + 11) / 2 = y

Fifth step: Keep simplifying!

y = (5/2)y + 11/2

Sixth step: Subtract (5/2)y from both sides to isolate "y".

(-3/2)y = 11/2

Seventh step: Divide both sides by (-3/2).

y = -22/6 = -11/3.

Eighth step: Plug -11/3 in as "y" for the first equation.

x = 2(-11/3) + 9

Ninth step: Simplify!

x = -22/3 + 9

Tenth step: You need common denominators to add fractions so change "9" to "27/3".

x = -22/3 + 27/3

Eleventh step: Add!

x = 5/3.

Your answer would be (5/3, -11/3).

Hope this helped :)(1 vote)

- How would you solve this problem?

{X+y=1000

{0.05x+0.06y=57(2 votes)

## Video transcript

In the last video, we saw what
a system of equations is. And in this video, I'm going
to show you one algebraic technique for solving systems of
equations, where you don't have to graph the two lines and
try to figure out exactly where they intersect. This will give you an exact
algebraic answer. And in future videos,
we'll see more methods of doing this. So let's say you had
two equations. One is x plus 2y is equal to 9,
and the other equation is 3x plus 5y is equal to 20. Now, if we did what we did in
the last video, we could graph each of these. These are lines. You could put them in either
slope-intercept form or point-slope form. They're in standard
form right now. And then you could graph each
of these lines, figure out where they intersect, and that
would be a solution to that. But it's sometimes hard to
find, to just by looking, figure out exactly where
they intersect. So let's figure out a way to
algebraically do this. And what I'm going to do is
the substitution method. I'm going to use one of the
equations to solve for one of the variables, and then I'm
going to substitute back in for that variable over here. So let me show you what
I'm talking about. So let me solve for x using
this top equation. So the top equation says x
plus 2y is equal to 9. I want to solve for x, so let's
subtract 2y from both sides of this equation. So I'm left with x is
equal to 9 minus 2y. This is what this first equation
is telling me. I just rearranged
it a little bit. The first equation
is saying that. So in order to satisfy both of
these equations, x has to satisfy this constraint
right here. So I can substitute this
back in for x. We're saying, this top equation
says, x has to be equal to this. Well, if x has to be equal
to that, let's substitute this in for x. So this second equation
will become 3 times x. And instead of an x, I'll write
this thing, 9 minus 2y. 3 times 9 minus 2y, plus
5y is equal to 20. That's why it's called the
substitution method. I just substituted for x. And the reason why that's
useful is now I have one equation with one unknown,
and I can solve for y. So let's do that 3 times 9 is 27. 3 times negative 2 is negative
6y, plus 5y is equal to 20. Add the negative 6y plus the 5y,
add those two terms. You have 27-- let's see, this will
be-- minus y is equal to 20. Let's subtract 27
from both sides. And you get-- let me
write it out here. So let's subtract 27
from both sides. The left-hand side, the 27's
cancel each other out. And you're left with negative y
is equal to 20 minus 27, is negative 7. And then we can multiply both
sides of this equation by negative 1, and we get
y is equal to 7. So we found the y value of the
point of intersection of these two lines. y is equal to 7. Let me write over here, so I
don't have to keep scrolling down and back up.
y is equal to 7. Well, if we know y, we can
now solve for x. x is equal to 9 minus 2y. So let's do that. x is equal to 9 minus 2
times y, 2 times 7. Or x is equal to 9 minus 14, or
x is equal to negative 5. So we've just, using
substitution, we've been able to find a pair of x
and y points that satisfy these equations. The point x is equal to negative
5, y is equal to 7, satisfy both of these. And you can try it out. Negative 5 plus 2 times 7,
that's negative 5 plus 14, that is indeed 9. You do this equation. 3 times negative 5 is negative
15, plus 5 times y, plus 5 times 7. So negative 15 plus
35 is indeed 20. So this satisfies
both equations. If you were to graph both of
these equations, they would intersect at the point
negative 5 comma 7. Now let's use our newly
found skill to do an actual word problem. Let's say that they tell
us that the sum of two numbers is 70. And they differ-- or maybe we
could say their difference-- they differ by 11. What are the numbers? So let's do this word problem. So let's define some
variables. Let's let x be the larger
number, and let y be the smaller number. I'm just arbitrarily creating
these variables. One of them is larger
than the other. They differ by 11. Now, this first statement, the
sum of the two numbers is 70. That tells us that x plus
y must be equal to 70. That second statement, that
they differ by 11. That means the larger number
minus the smaller number must be 11. That tells us that x minus
y must be equal to 11. So there we have it. We have two equations
and two unknowns. We have a system of
two equations. We can now solve it using
the substitution method. So let's solve for x on this
equation right here. So if you add y to both
sides of this equation, what do you get? On the left-hand side, you just
get an x, because these cancel out. And then on the right-hand side,
you get x is equal to 11 plus y, or y plus 11. So we get x is equal
to 11 plus y using the second equation. And then we can substitute it
back into this top equation. So instead of writing x plus
y is equal to 70, we can substitute this in for x. We've already used the second
equation, the magenta one, now we have to use the
top constraint. So if we substitute this in, we
get y plus 11-- remember, this is what x was, we're
substituting that in for x-- plus y is equal to 70. This is x. And that constraint was given to
us by this second equation, or by this second statement. I just substituted this x with
y plus 11, and I was able to do that because that's the
constraint the second equation gave us. So now let's just solve for y. We get y plus 11, plus
y is equal to 70. That's 2y plus 11
is equal to 70. And then if we subtract 11 from
both sides, we get 2y is equal to-- what is that? 59? You subtract 10 from
70, you get 60, so it's going to be 59. So y is equal to 59 over 2. Or another way to write it,
you could write that as 59 over 2 is the same thing as--
let's see-- 25-- 29.5. y is equal to 29.5. Now, what is x going
to be equal to? Well, we already figured out
x is equal to y plus 11. So x is going to be equal to
29.5-- that's what y is, we just figured that out-- plus 11,
which is equal to-- so you add 10, you get 39.5. You add another 1,
you get 40.5. And we're done. If you wanted to find the
intersection of these two lines, it would intersect at
the point 40.5 comma 29.5. And you could have used this
equation to solve for x and then substituted in this one. You could have used this
equation to solve for y and then substituted in this one. You could use this equation
to solve for y and then substitute into that equation. The important thing is, is
you use both constraints. Now let's just verify that
this actually works out. What's the sum of these
two numbers? 40.5 plus 29.5, that
indeed is 70. And the difference between
the two is indeed 11. They're exactly 11 apart. Anyway, hopefully you
found that useful.