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### Course: 8th grade (Eureka Math/EngageNY) > Unit 4

Lesson 4: Topic D: Systems of linear equations and their solutions- Systems of equations: trolls, tolls (1 of 2)
- Systems of equations: trolls, tolls (2 of 2)
- Testing a solution to a system of equations
- Solutions of systems of equations
- Systems of equations with graphing
- Systems of equations with graphing
- Systems of equations with graphing: 5x+3y=7 & 3x-2y=8
- Systems of equations with graphing: y=7/5x-5 & y=3/5x-1
- Systems of equations with graphing: chores
- Systems of equations with graphing
- Systems of equations with elimination: 3t+4g=6 & -6t+g=6
- Systems of equations with elimination
- Systems of equations with elimination: x+2y=6 & 4x-2y=14
- Systems of equations with elimination: -3y+4x=11 & y+2x=13
- Systems of equations with elimination: 2x-y=14 & -6x+3y=-42
- Systems of equations with elimination: 4x-2y=5 & 2x-y=2.5
- Systems of equations with elimination: x-4y=-18 & -x+3y=11
- Systems of equations with elimination
- Systems of equations with elimination: 6x-6y=-24 & -5x-5y=-60
- Systems of equations with elimination challenge
- 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
- Systems of equations with elimination: apples and oranges
- Systems of equations with elimination: TV & DVD
- Systems of equations with elimination: King's cupcakes
- Systems of equations with elimination: Sum/difference of numbers
- Systems of equations with elimination: potato chips
- Systems of equations with elimination: coffee and croissants
- Systems of equations with substitution: coins
- Systems of equations with substitution: potato chips
- Systems of equations with substitution: shelves
- Systems of equations word problems
- Age word problem: Imran
- Age word problem: Ben & William
- Age word problem: Arman & Diya
- Age word problems
- Solutions to systems of equations: consistent vs. inconsistent
- Systems of equations number of solutions: fruit prices (1 of 2)
- Systems of equations number of solutions: fruit prices (2 of 2)
- Systems of equations number of solutions: y=3x+1 & 2y+4=6x
- Solutions to systems of equations: dependent vs. independent
- Number of solutions to a system of equations graphically
- Number of solutions to a system of equations graphically
- Forming systems of equations with different numbers of solutions
- Number of solutions to a system of equations algebraically
- Comparing Celsius and Fahrenheit temperature scales
- Converting Fahrenheit to Celsius

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# Systems of equations with substitution: y=-1/4x+100 & y=-1/4x+120

Learn to solve the system of equations y = -1/4x + 100 and y = -1/4x + 120 using substitution. Created by Sal Khan and Monterey Institute for Technology and Education.

## Want to join the conversation?

- What about this problem

Y=3x+7 and...........Y=3x+7(10 votes)- X can be any real number and y will be whatever that real number is times three and plus seven.(31 votes)

- Wht about a problem like this

-5x+y=-3

3x-8y=24(9 votes)- Let -5x+y=-3 be the first equation, and 3x-8y=24 be the second. Let's manipulate the first equation by adding 5x to both side; resulting in y=-3+5x, or 5x-3. Now let's plug in 5x-3 for wherever y has been in the second formula. 3x-8(5x-3)=24. This is now pretty straightforward to solve for x. We distribute the 8 into (5x-3) and get 3x-40x-24=24. -48=27x. x=-48/27, x=-16/9.

Now let's plug in the x for either one of the equations to get y. 3(-16/9)+8y=24. -16/3+8y=24. 8y=24+16/3. 24y=72+16. 24y=88. y=88/24, y=11/3.

Hope it helped!(5 votes)

- How do you know what you're supposed to multiply the equations by?(4 votes)
- give an example and i will probably be able to help(6 votes)

- solve for y in terms of x

5x +380y= -2280(5 votes)- Regina,

Solve for y in terms of x just means "isolate y on one side of the equation".

You do the same thing to both sides of the equation to keep it equal while eliminating everything but the y on the left.

If you first subtract 5x from both sides, the left will have 5x-5x which becomes 0 and disappears leaving you with only 380y on the left.

If you then multiply each side by the reciprocal of 380 (which is 1/380, the 380/380 on the left becomes 1 and disappears leaving you with only y on the left.

Make sure as you do this, you do the same thing to the right side.

When the y becomes isolated on the left, you will have solved for y in terms of x.

I hope that helps make it click for you.(4 votes)

- What is x-y=12(3 votes)
- You can't figure this out with the provided information. It could have multiple answers, which include: (12,0), (100, 88), and infinitely many others.(4 votes)

- I am still a little bit confused about why the substitution method actually works. Can someone please explain it to me more?(3 votes)
- How do you solve with x/3-y/2=1 and y=2x-2(3 votes)
- Gabby,

x/3-y/2=1

y=2x-2

From the second equation you know that y and 2x-2 are the same so you can substitute (2x-2) for the y in the first equation

x/3-y/2=1 Substiture (2x-2) for the y

x/3 - (2x-2)/2 = 1

Now distribute the -1/2

x/3 - 2x/2 + 2/2 = 1 The 2/2s both become 1

x/3 - 1x +1 = 1 Subtract 1 from both sides

x/3 - 1x = 0

Now, you might just see that x=0 is the solution for x,

Or to combine the x terms, you must find a common denominator.

x/3 - (3/3)x = 0 now 1/3 - 3/3 = -2/3 so

(-2/3)x = 0 Now multiply both sides by the reciprocal of -2/3 which is -3/2

x=0

Now find y by substituting 0 for x in either equation.

y= 2(0) -2

y= 0-2

y= -2

Now double check by substituting 0 for x and -2 for y in the other equation

0/3 - (-2/2) = 1

0-(-1) = 1

1=1

This is true so the answers double check.

x=0 y=-2 or (0,-2)

I hope that helps make it click for you.(4 votes)

- So, what if the X was solved already for us and the Y would be the thing we would have to solve. Would this apply the same way it does for solving X? Solving Y?(3 votes)
- Yes, solving Y is the same thing as solving X. Lets say that you have x already solved lets say x=3. Now you have to solve for Y. It is the exact same thing. Lets plug what we know into an equation. Lets say that 3+4y=15. Do the same thing you would to find X. If you do your same steps, you will get that Y=4 even doing the same thing you use for X.(0 votes)

- How do i solve this problem?

6x - 6y = -12

-4x + y = 11(2 votes)- Solve one of the equations for one of the variables, then substitute into the second equation and solve for the second variable, then complete the problem by solving for the first variable.

y = 11 + 4x

6x - 6(11 + 4x) = -12

6x - 66 - 24x = -12

-18x = 54

x = -3

y = 11 + 4x-3

y = 11 -12

y = -1

Check your anser...

6(-3) - 6(-1) = -12

-18 + 6 = -12

-12 = -12

-4(-3) - 1 = 11

12 - 1 = 11

11 = 11(2 votes)

- So i have a question about an Equation that im doing for math homework. y= 2x+1 and y = x-4, I really do not get how to do this could you help me this is very confusing(2 votes)
- You do just like the video. Since Y is equal to both (2x+1) and (x-4) you can set these equal to each other like this 2x+1 = x-4 If you solve this equation for x you will come out with x= -5 If you replace both your equations at the top with (-5) you will find that both answers equal -9. So the answer is -5(1 vote)

## Video transcript

We're given a system of
equations here, and we're told to solve for x and y. Now, the easiest thing to do
here, since in both equations they're explicitly solved for y,
is say, well, if y is equal to that, and y also has to equal
this second equation, then why don't we just set
them equal to each other? Or another way to think about
it is, if y is equal to this whole thing right over here--
that's what that first equation is telling us-- and if
we have to find an x and a y that satisfy both of these
equations, if y is equal to that, why can't I just
substitute that right here for y? And if we do that, the left-hand
side of this bottom equation becomes negative
1/4x plus 100. And then that is going to be
equal to this right-hand side-- and I'll do it in the
same color-- is equal to negative 1/4x plus 120. Now, the first thing we might
want to do is maybe get all of our x terms onto the
left- or the right-hand side of the equation. And if we wanted to get rid
of these x terms from the right-hand side, get them on the
left-hand side, the best thing to do is to add 1/4x to
both sides of this equation. So let me do that. So we're going to add 1/4x here,
add 1/4x here, and you might already be sensing that
something shady is going on. So let's do it. So negative 1/4x plus 1/4x. They cancel out. You get 0x. So the left side of the
equation is just 100. And then the right side of
the equation, same thing. Negative 1/4x plus 1/4x. They cancel out. No x's. And you're just left with
is equal to 120. Which we know is definitely
not the case. 100 is not equal to 120. We got this nonsensical
equation here, that 100 equals 120. So this type of system
has no solution. You know it has no solution
because in order for it to have any solution, these two
numbers would have to be equal to each other, and they are
not equal to each other. And if you look at the original
equations, it might jump out at you why they
have no solutions. Both of these lines, or both
of these equations, if you view them as lines, have
the exact same slope. But they have different
y-intercepts. So if I just were to do a
really quick graph here. That's my y-axis, that is my
x-axis, so it's y and x. This first graph over here,
its y-intercept is 100. Let me do it a little
bit lower. Its y-intercept-- let's say
that that is 100, so it intersects right there. And there's a slope
of negative 1/4. So maybe it looks something
like this. That's that first line. This second line-- I'll do it
in pink right here-- y is equal to negative 1/4x plus 120,
its y-intercept might be right here at 120. But it has the same slope,
negative 1/4, so its slope, the line would look something
like this. So you see that there are no x
and y points that satisfy both of these equations. Another way to think about it. If y-- you take an x. This first equation says, OK,
you take your x, multiply it by negative 1/4, and
add 100, and that's going to give you y. Now, here we say, well, you
take that same x, and you multiply it by negative 1/4 and
add 120, and that has to be equal to y. Well, the only way that that
would ever be true is if 100 and 120 were the same
number, and they're not the same number. So you're never going to have
a solution of this system. These two lines are never going
to intersect, and that's because they have the
exact same slope.