Main content

## 8th grade (Eureka Math/EngageNY)

### 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

© 2023 Khan AcademyTerms of usePrivacy PolicyCookie Notice

# Systems of equations with elimination: 4x-2y=5 & 2x-y=2.5

Sal solves the system of equations 4x - 2y = 5 and 2x - y = 2.5 using elimination. Created by Sal Khan and Monterey Institute for Technology and Education.

## Want to join the conversation?

- What is the point of standard form? It seems to me that all it is good for is solving systems of equations like this. But can't you just use slope-int form?(18 votes)
- Plus when you get to matrices, standard form is the best/only way to fromulate them(5 votes)

- I cannot figure this out at all:

-5x +2y = 22

9x + 8y= 30

please help(5 votes)- The answer above was not correct... you take -5x+2y=22 and multiply the whole equation by four to get: -20x+8y=88, and then you can subtract the other original equation: 9x+8y=30 from it... and you get -29x=58, so you divide -29 on both sides, which gives you x=-2... then to find y you simply plug this into the equation: 9(-2)+8y=30 and you get y=6... hope that helps sorry it's 9 months late though!;)(5 votes)

- what if the problem is a ratio(3 votes)
- At2:10, instead of changing 4x - 2y = 5 to slope-intercept form in order to graph it, how (if you even can) would you graph the equation while keeping it in standard form?(2 votes)
- It is possible to do that but it is harder. So just put it into slope intercept form.(1 vote)

- if -3x-6y=-63, how do you know if the "6y" is negative?(1 vote)
- You always take the sign that is before the number, for example you can separate the expression into -3x and -6y.(2 votes)

- if the answer is o=o is there any solutions?(1 vote)
- If you end up with 0 = 0, you will have infinite solutions.(2 votes)

- What is the solution of 4x+y=57 and 7x-2y=21?(1 vote)
- First at all,we need to simplify both equations. The first one we will isolated the y, and we get y=-4x+57 and the second one y=3.5x -10.5.And we need to multiply -1 to any one of the equations.I choose to do with the second one, then -y=-3.5x+10.5,and add this to the first one,we get 0=-7.5x+67.5,and you can solve for x. x=9 Then substitute x to any of this equation.You will get y=21.You can check it out.

Remember this method!(2 votes)

- How do i solve:

8x-7y=21

2x=3y+4

using addition/elimination? can u help me with the steps?(2 votes)- 8x-7y=21 2x=3y+4

What i am going to start with is removing the Ys.

So. with the first equation i'll multiply it by 3, the second, i'll multiply it by 7, that should cancel the Ys out when i add the two equations.

8x-7y=21 = 24x-21y=63 2x=3y+4 = 14x=21y+28 Now i'll add the equations together.

24x-21y=63 +

28+21y=14x =

24x+28=14x+63 Now i'll subtract 14x from both sides, then 28 from both sides.

24x+28=14x+63

-14x -14x =

10x+28=63

-28 -28

10x=35 i'll divide by 10 and then if i haven't messed up we'll have your x value.

10x=35*____*

10 =

x=3.5 So now i'll substitute 3.5 for x to solve for y.

7=3y+4 That's your second equation, with 3.5 for x, therefore 7 for 2x. i'll subtract 4.

-4 -4 =

3=3y Divide by 3 and...*__*

3 =

1=y Now i'll see if that checks out with you first equation (x=3.5, and y=1)

28-7=21 Add seven.

+7 +7

28=28 Yes it does, there's your answer and i hope that helps make things a bit clearer.

Good fortune problem solving!(1 vote)

- what are the rules to follow when calculating simultaneous equations(1 vote)
- Can you always use both adding and subtraction to do elimination?(1 vote)
- Yes, in the end you get the same answer either way.(1 vote)

## Video transcript

We're told to solve and graph
the solution for the system of equations right here. And the first thing that jumps
out at me, is that we might be able to eliminate one
of the variables. And if we just focus on the x,
we have a 4x here and we have a 2x right here. If we were to just add them
right now, we would get a 6x. So that wouldn't eliminate it. But if we can multiply this 2x
by negative 2, it'll become a negative 4x, and then when you
add it, they would cancel out. So let's multiply this
equation, this second equation, by negative 2. So I'm going to multiply both
sides of this equation by negative 2. And the whole motivation is
so that this 2x becomes a negative 4x. And, of course, I can't just
multiply only the 2x. Anything I do to the left-hand
side of the equation I have to do to every term, and
I have to do to both sides of the equation. So the second equation becomes
negative 4x-- that's negative 2 times 2x-- plus-- we have
negative 2 times negative y-- which is plus 2y is equal to 2.5
times negative 2, is equal to negative 5. I just rewrote the second
equation, multiplying both sides by negative 2. Now, this top equation-- I'll
write it on the bottom now-- we have 4x minus 2y is
equal to positive 5. And now we can eliminate it. We can say, hey, look, the
negative 4x and the positive 4x should cancel out, or
they will cancel out. So let's add these
two equations. Let's add the left side to the
left side, the right side to the right side, and we can
do that because these two things are equal. We're doing the same thing to
both sides of the equation. So what do we get? If we take our negative 4x
plus our 4x, well, those cancel out. So you're left with nothing. Maybe I could write a 0 there. 0x if you want. And then you have your plus
2y and your negative 2y. Those also cancel out. So you're also left with 0y. And then that equals negative
5 plus 5 is equal to 0. So this just simplifies to 0
equals 0, which is true, but it's kind of bizarre. We had all these x's and y's. Everything canceled out. So let's explore this
a little bit more. Let's graph it and see what this
0 equals 0 is telling us when we try to solve this
system of equations. So let me graph this top guy. I'll do it in blue. So right now it's in
standard form. Let's put it in slope-intercept
form. So we have 4x minus
2y is equal to 5. Let's subtract 4x
from both sides. I want the x terms on
the right-hand side. So then I'm left with negative
2y is equal to negative 4x plus 5. Now we can divide both
sides by negative 2. And we are left with y is equal
to positive 2x, right, that's positive 2x, minus 2.5. So let's graph that. The y-intercept is
negative 2.5. So negative 2.5 right there, and
then it has a slope of 2. So if we move up 1, if we move
up in the x-direction, if we move to the right 1 in the
positive x-direction, we will move up 2. So 1, 2. Right there. And if we were to do it again,
we move up 1, 2. Just like that. So the line's going to look
something like this. I'll try my best to draw
a straight line. This is the hardest part
about a lot of these problems. There you go. So that's the top equation. Now, let me draw the
bottom equation. Let me draw and I'll do it
in this green color. So this bottom equation was 2x
minus y is equal to 2.5. And we can subtract 2x
from both sides. The left-hand side becomes
negative y is equal to 2x plus-- or is equal to negative
2x plus 2.5. Now let's multiply or divide
both sides by negative 1. And you get y is equal to
positive 2x minus 2.5. And let's try to graph this, and
you already might notice something interesting about
these two equations. You try to graph this, the
y-intercept is at negative 2.5, right there. The slope is 2. So it's going to be this
exact same line. And you saw that
algebraically. I didn't have to graph it. These two lines have the exact
same equation when you put them in slope-intercept form. That's the first equation. That's the second equation. So what this 0 equals 0 is
telling us is actually that these are the same line. That these actually have an
infinite number of solutions. Any point on this line, which
is both of those lines, will satisfy both of these
equations. You give me an arbitrary y,
solve for x in the top equation, that x
and y will also satisfy the bottom equation. So this actually has an infinite
number of solutions. These are the same line.