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## 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 elimination: 2x-y=14 & -6x+3y=-42

Sal solves the system of equations 2x - y = 14 and -6x + 3y = -42 using elimination. Created by Sal Khan and Monterey Institute for Technology and Education.

## Want to join the conversation?

- can there be a subtraction elimination?(20 votes)
- Adding a negative and subtracting a positive are the same thing. Technically, with the existence of negative numbers, adding and subtracting are the same thing. You could,if you so desired, call this method "subtraction elimination" and just add a negative of the second equation; it would be the same process.(18 votes)

- Can there be a subtraction elimination besides the ones there are?(1 vote)
- He should not say that when x=1 or y=2 that they intersect in exactly one place. Just a little error. I may have missed something.(0 votes)
- He is referring to the solutions, because your looking for the point of intersection.(3 votes)

- i think i got the answer on accident

it plugged them in and it seemed to work too

y=-7

x=3.5(1 vote) - What would the answer be if you had a problem such as 9x + y = -17 and -4x + y = -5? How would I continue the problem of eliminations if there is 5x + 2y = -22?(1 vote)
- Instead of adding the two equations, just subtract one equation from another to eliminate y >>> (9x+y)-(-4x+y) = -17-(-5) >>> 9x+y+4x-y = -12 >>> 13x = -12 >>> x = -12/13. Substitue the value of x in one of the equation to solve for the value of y (which when solved, y = -113/13) Hope this help :)(1 vote)

- How would I go about using one of these 3 methods for a problem like this:

Joe spent 201 dollars on shirts and pants. Shirts cost 27 each, and pants cost 22 each. Eight were bought in total. How many were bought of each kind?(1 vote) - Sal, can there be a elimination where you subtract?(1 vote)
- Yes, you eliminate the one variable by subtracting it off.(1 vote)

- why in the first exercise has to multiply x 3??.(1 vote)
- At2:00, is a dependent system means both equation are the same?(2 votes)
- Partially. A
**dependent system**is a system where both lines have the same slope.

For example:`x + y = 10`

2x + 2y = 20

These are different equations, but the slopes are the same. This is because if you divide both sides by 2 in the second equation, you'll get`x + y = 10`

.(0 votes)

- I am faced with this problem...

3x+5y =30

5x+8y=49

and I don't know how to start the problem(1 vote)- If the two equations are part of a system you could solve that by elimination. One way would be to multiple the first equation by five and the second by negative three, so you'll have 15x and -15x cancelling each other out when you add the two equations. From there you'll just have to solve for y. Once you have the value of y, you can substitute it back to one of the original equations to solve for x.(1 vote)

## Video transcript

Solve for x and y. And we have the system
of equations right here. We have 2x minus
y is equal to 14, and negative 6x plus 3y
is equal to negative 42. So we could try to solve
this by elimination, and let's see if we
can eliminate our y variables first. We have a 3y here, and we
have a negative y up here. And they won't eliminate. If you just add negative y
plus 3y, that won't eliminate. But if we could turn this
negative y into a negative 3y, then it would cancel
out with the 3y. And the best way to turn a
negative y into a negative 3y is to multiply this
entire top equation by 3. So let's do that. Let me get some space
over to the left Let's multiply this
entire top equation by 3. So I'm going to
multiply it by 3. So I multiply 2x by 3. I get 6x. I multiply negative y by 3. I get negative 3y. And then I multiply 14 by 3. 3 times 14 is 42, right? 3 times 10 is 30,
plus 12, it's 42. And then we can add
both of these equations. Something interesting
should already maybe be showing up on your radar. But let's add both
of these equations. Let's add the left-hand side. Negative 6x plus 6x? Well, those cancel out. We get 0. Then we have 3y minus 3y. Those cancel out. We get another 0. And then finally, you
get negative 42 plus 42. Well, that's a 0. So we end up with just
0 is equal to 0, which is clearly true, but it's
not putting any constraints on the x or y. And that's because whenever
you have a situation like this, where you just get something
that's obviously true-- 0 equals 0, or 1
equals 1, or 5 equals 5-- what we're dealing
in this situation is where both of
our constraints, both of our equations, are
actually the same equation. So this right here is
a dependent system. And you see it right over here. If you take that first
equation, you multiply it by 3, you got 6x minus
3y is equal to 42. If we then multiplied
it by negative 1, you would get the
exact same equation as the second equation. You would get negative 6x plus
3y is equal to negative 42. Or another way to
think about it, if you want to go from
the first equation to the second equation,
you just multiply both sides of the
equation times negative 3. So both of these constraints are
actually the same constraints. They're just kind of a scaled-up
multiple of each other. So if you were to
graph them-- and I might as well graph
them for you right here. This first equation right
here, let me do it over here. It's 2x minus y is equal to 14. You could subtract
2x from both sides, and you would get-- let
me just subtract it. Subtract 2x, subtract 2x. On the left-hand side, you're
left with just negative y. On the right-hand side, you
have negative 2x plus 14. Multiply both sides
by negative 1, and you get y is
equal to 2x minus 14. So this first
equation over here, if I were to draw my axes--
so that is my y-axis, and then that is my x-axis. This graph right over here,
its y-intercept is negative 14. So this is 0, negative 14. And it has a slope of 2. So it's going to look
something like this. It's going to look
something like that. And then this second
equation, if you were to graph it, because
this is a dependent system, it's the exact same line. If you were to put this in
slope-intercept form and graph it, you would get
the exact same thing. It would go right on top of it. So there's actually an
infinite number of solutions. These two lines
are the same line. So they intersect everywhere on
each of the respective lines. They're the same line. And when you get
something like this, 0 equals 0 or 1 equals 1, that's
the telltale sign that you're dealing with a dependent system. If you got something
like 0 equals 1, then you would have no solution. So this one right here would
be an inconsistent system. And this would be the situation. So let me make it very clear. This is the situation. We're dealing with
the same lines. The lines are the same lines. This is a situation where
you have parallel lines. And so they never intersect. And then, obviously, you
have the easiest situation. I think this is one most of
us are familiar with, where you have something like x is
equal to 1, or y is equal to 2, or anything like this. It doesn't have to be 1 or 2. And then this is
clearly a situation where you have an
independent system, where you have two
different lines that intersect in exactly one place. Anyway, this-- there's an
infinite number of solutions. Any x and y that satisfied
the first equation will also satisfy
the second equation, because they're fundamentally
the same constraint.