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### Course: 8th grade > Unit 4

Lesson 3: Solving systems with substitution- Systems of equations with substitution: 2y=x+7 & x=y-4
- Systems of equations with substitution
- Systems of equations with substitution: y=4x-17.5 & y+2x=6.5
- Systems of equations with substitution: -3x-4y=-2 & y=2x-5
- Systems of equations with substitution: 9x+3y=15 & y-x=5
- Systems of equations with substitution
- Systems of equations with substitution: y=-5x+8 & 10x+2y=-2
- Systems of equations with substitution: y=-1/4x+100 & y=-1/4x+120
- Substitution method review (systems of equations)

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# Systems of equations with substitution: 2y=x+7 & x=y-4

When solving a system of equations using substitution, you can isolate one variable and substitute it with an expression from another equation. This will allow you to solve for one variable, which you can then use to solve for the other. Created by Sal Khan and Monterey Institute for Technology and Education.

## Want to join the conversation?

- this impossible help me(8 votes)
`Ok, let's take the same problem and break it down, very carefully.`

**2y = x + 7**

&

x = y - 4

so the second equation above denotes that*x_ correlates to _y - 4*. So, let us substitute*y - 4*on the top of the equation replacing the position of the value x. Making it*2y = (y-4) + 7*. Let us find the value of y.**2y = y + 3**<-- Here, we have combined like terms first. (*negative 4 plus 7*).

*-y + 2y = y - y + 3* <-- Now, we have subtracted y from one side to the other (*variable*) side of the equation.**y = 3**<-- Now, we found the value of the variable,*y_. Which is _3*.

Let us now substitute the value of y, for the second term first -**x = y - 4****x =***3_ - 4**<-- Replaced the value of _y*with the constant value*3_.**!!**x = - 1**<-- We found the value of *_x

Now,*x is equal to negative 1*and*y is equal to 3*.

To prove that these answers as valid, try to substitute into both equations.

By substituting these*x_ and _y*values, we should be able to view both sides of the equation, as true.

Thanks :)(32 votes)

- Trying to solve substitution problem.

x=3y-4

4y-2x=13

Do I need to use a negative fraction with X?(16 votes)- Just substitute the value of x in the first equation into the other equation and solve for y.

4y - 2x = 13

4y - 2(3y - 4) = 13

Once you have solved that, substitute the value of y back into the first equation and solve for x.(14 votes)

- How do I solve an equation like this? X=y+8. And 3x-y=16 ? I am trying to solve for x and y(10 votes)
- x=y+8. Substitute that into the second equation. 3(y+8)-y=16. Distribute the 3.

3y+24-y=16. add like terms.

2y+24=16. subtract 24 from both sides.

2y=-8. divide by 2

y=-4. Now that you know y=-4, plug that in the first equation.

x=-4+8.

x=4(12 votes)

- Can't we just put it into Desmos caculator and solve it that way?(8 votes)
- Why does substitution work?

The two equations are separate equations. Why does this work? The graph seems easy to get, but this is a bit harder.

I am having a hard time visualizing substitution or just understanding why it works. thanks!(5 votes) - how do I use substitution if the both equations have the y variable by itself?(4 votes)
- you don't! you already have your answer there.(2 votes)

- But how do I graph a system of equations when my x or y is a decimal?(2 votes)
- change it to a mixed number :DD(3 votes)

- at4:35that was fascinating(3 votes)
- How do i solve 5y(2y-3)+(2y-3)?(2 votes)
- 5y(2y-3) + (2y-3) = (2y-3) (5y+1) = 10y^2 -13y - 3

Here, I took the (2y-3) as common factor using the additive distributive identity we had studied earlier. Hope you understood.

Another method will be 5y(2y-3)+(2y-3) = 10y^2-15y+2y-3 = 10y^2-13y-3(2 votes)

- -6x+6y=-24

-3x-12y=-12(2 votes)- y=0

x=4

is one way to solve that problem(2 votes)

## Video transcript

Use substitution to
solve for x and y. And we have a system
of equations here. The first equation is
2y is equal to x plus 7. And the second equation here
is x is equal to y minus 4. So what we want to do,
when they say substitution, what we want to do
is substitute one of the variables
with an expression so that we have an equation
and only one variable. And then we can solve for it. Let me show you what
I'm talking about. So let me rewrite
this first equation. 2y is equal to x plus 7. And we have the second
equation over here, that x is equal to y minus 4. So if we're looking
for an x and a y that satisfies both constraints,
well we could say, well look, at the x and y have to
satisfy both constraints, both of these constraints
have to be true. So x must be equal to y minus 4. So anywhere in this top
equation where we see an x, anywhere we see an
x, we say well look, that x by the second constraint
has to be equal to y minus 4. So everywhere we see an
x, we can substitute it with a y minus 4. So let's do that. So if we substitute y minus
4 for x in this top equation, the top equation becomes 2y
is equal to instead of an x, the second constraint
tells us that x needs to be equal to y minus 4. So instead of an x,
we'll write a y minus 4, and then we have a plus 7. All I did here is I
substituted y minus 4 for x. The second constraint tells
us that we need to do it. y minus 4 needs to be
equal to x or x needs to be equal to y minus 4. The value here is now
we have an equation, one equation with one variable. We can just solve for y. So we get 2y is equal to y, and
then we have minus 4 plus 7. So y plus 3. We can subtract y from both
sides of this equation. The left hand side,
2y minus y is just y. y is equal to-- these
cancel out. y is equal to 3. And then we could go
back and substitute into either of these
equations to solve for x. This is easier right over
here, so let's substitute right over here. x needs
to be equal to y minus 4. So we could say that x
is equal to 3 minus 4 which is equal to negative 1. So the solution to this system
is x is equal to negative 1 and y is equal to 3. And you can verify that it
works in this top equation right over here. 2 times 3 is 6 which is indeed
equal to negative 1 plus 7. Now I want to show you that
over here we substituted-- we had an expression that,
or we had an equation, that explicitly solved for x. So we were able to
substitute the x's. What I want to show
you is we could have done it the
other way around. We could have solved for y and
then substituted for the y's. So let's do that. And we could have substituted
from one constraint into the other
constraint or vice versa. Either way, we would have
gotten the same exact answer. So instead of saying x
is equal to y minus 4, in that second
equation, if we add 4 to both sides of this equation,
we get x plus 4 is equal to y. This and this is the
exact same constraint. I just added 4 to
both sides of this to get this
constraint over here. And now since we've solved
this equation explicitly for y, we can use the first
constraint, the first equation. And everywhere
where we see a y, we can substitute it with x plus 4. So it's 2 times--
instead of 2 times y, we can write 2 times x plus 4. 2 times x plus 4 is
equal to x plus 7. We can distribute this 2. So we get 2x plus 8
is equal to x plus 7. We can subtract x from both
sides of this equation. And then we can subtract 8 from
both sides of this equation, subtract 8. The left hand side,
that cancels out. We're just left with an x. On the right hand
side, that cancels out, and we are left
with a negative 1. And then we can
substitute back over here we have y is equal to x plus 4,
or so y is equal to negative 1 plus 4 which is equal to 3. So once again, we
got the same answer even though this time we
substituted for y instead of substituting for x. Hopefully you found
that interesting.