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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 elimination: apples and oranges

Sal solves a word problem about the price of apples and oranges by creating a system of equations and solving it. Created by Sal Khan.

## Want to join the conversation?

- Why was 11 turned into -11 for the bottom problem? I understand the chages to the 5x & 5y but dont understand the change is it because 10 is less than 11?(8 votes)
- The 11 was "turned into" -11, by an operation which prepares the whole equation for the elemination process used later on in order to solve this problem. Essentialy, the whole equation 5x+ 5y = 11 is multiplied by -1 and becomes -5x -5y = -11. This "new equation" still has the same meaning it had before you multiplied it by -1, because you did the same thing to both sides. If you had just multiplied the left side (5x + 5y) by -1, you would have changed the meaning of this equation and the results for x any y you got after the elemination process would have been wrong.(13 votes)

- At2:45, I could not undeerstand why Sal multiplied the bottom equation -1.(4 votes)
- Sal multiplies the bottom equation by -1 to combine the equations and CANCEL out the x values. When he multiples the entire bottom equation by -1 it does NOT change the value of the equation, so he can do this and then add together the equations.

Because he now has a -5x value and a 5x value these two variables cancel out when the equations are added together leaving only the y variable. With one variable left, we can now solve for y.

When you have solved for y using this method, you can then plug that value of y back into an equation and solve for x.

The key here is to narrow the equation down to only one variable somehow. We cannot solve the equation if there is two unknowns.(4 votes)

- three times the larger of two numbers is equal to four times the smaller. the sum of the numbers is 21. find the numbers(3 votes)
- Robert,

You need to convert the words to equations.

The first sentence is

"three times the larger of two numbers is equal to four times the smaller."

First convery the number to number symbols.

"3 times the larger of two numbers is equal to 4 times the smaller."

The words times can be changed to "*"

"3 * the larger of two numbers is equal to 4 * the smaller."

The words "is equal to" can be replaced with an "=" sign.

"3 * the larger of two numbers = 4 * the smaller."

Your unknowns are "the larger of two numbers" and "the smaller"

Let's use the variable L for "the larger " and S for the smaller

3 * L = 4 * S

The second sentence is

"The sum of the numbers is 21"

Sum means add and "the numbers are the same unknowns in the first sentence L and S so

L+S is 21 The is means equals so

L+S = 21

So your two equations are

3L = 4S

L+S = 21

I hope that helps make it click for you.(5 votes)

- so how do I graph this?(2 votes)
- i would just go to desmos and type the equations in. desmos is an online graphing calculator.(3 votes)

- I still don't get how to find the other answer do we simpfly?(3 votes)
- You plug in what you got for the first answer into the other variable. Say you are solving 3x+y=10 and you know that y=3. All you have to do is this- 3x+3=10 and solve from there. Simplifying is only necessary at the end when you are dividing.(1 vote)

- the sum of the two digits in a two digit number equal 7 when reversed the new number is 45 more than the original number what is the new number(2 votes)
- It isn't straightforward to express that in terms of a linear system of equations. This is a different class of problem. There are only 6 numbers to try though that have two digits with digits adding to 7. Just start at 16 and work your way up from there, testing 16, 25, 34, 43, 52, 61. Of course I immediately see that 16 works, so you only have to try one number.(2 votes)

- Is 5x=6?

5x+4y=10

5x+5y=11

so 5x could be 6 and y could be 1(1 vote)- Yes, You can subtract the first problem from the second problem. You get 5x+5y-5x-4y= 11-10. You are left with 0x-1y=1 or simply y=1. Substitute 1 for y in either equation- I will use the 1st for an example- 5x+(4*1)=10 or 5x+4=10. Subtract 4 from either side and yes you are left with 5x=6 as you asked above. . Solving for x you divide each side by 5 and get x=6/5.(3 votes)

- How much does it cost to run an 800 watt microwave for 17 hours if the power company charges $.11 per Kilowatt-hour?(3 votes)
- Total kilowatt hour = 800*17/1000 = 13.6 kw-hr (I divided by 1000 to convert w-hr to kw-hr)

Total charge = 13.6*0.11 = $1.496.

Hope it helps! :)(1 vote)

- the thing is wasn’t it obvious enough that you bought one extra orange so an orange would be 1 dollar, 11-10 =1 and then you could’ve just done 11-5 then divided by 5 and got 1.2… why go through the super long process that Sal did to find the answer(2 votes)
- Sal is just trying to give an example how to solve a system of equations, not necessarily is he giving the easiest method.

Also, see4:19until4:38. His wasn't exactly thinking too deeply.(1 vote)

- How many apples and oranges would you need to have 8 fruits that cost $10 at the same time?(2 votes)
- if we are talking about the video, you need at least 9 fruit to have a total of $10.(1 vote)

## Video transcript

You've gone to a fruit stand
to get some fresh produce. You notice that the
person in front of you gets 5 apples and
4 oranges for $10. You get 5 apples and
5 oranges for $11. Can we solve for the price
of an apple and an orange using this information
in a system of linear equations
in two variables? If yes, what is the solution? If no, what is the
reason we cannot? So we're trying to figure
out the price of an apple and the price of an orange. So I would use a for apple, but
I don't like using o for orange because o looks too
much like a zero. So I'll just say x for apples. Let's let x equal
the price of apples. And let's let y equal
the price of oranges. So let's describe what
happened to the person in line in front of us. They bought 5 apples. So how much did they
spend on apples? Well, they bought 5 apples
times x dollars per apple, so they spent 5x dollars
on their 5 apples. And they bought 4 oranges. They bought 4 oranges
times y dollars per orange. So they spent 4y on oranges. So the total amount that
they spent is 5x plus 4y. And they tell us
that this is $10. This is equal to $10. Now, you get in line,
and you buy 5 apples. So you buy 5 apples, just
like the guy in front of you. And you paid x
dollars per apple. So you're going to pay 5 apples
times the price per apple. This is the amount that
you spend on apples. And then you buy 5 oranges. So you're going to pay 5 oranges
times the price per orange, which is y. So this is how much
you spend on oranges. This is how much you spend on
apples and oranges, the sum. And they tell us
that this is $11. So can we solve
for an x and a y? And it looks like we can. And a big giveaway
right over here is the ratio between the x's and
the y's in these two equations are different. So we're getting some
information here. If the ratios were
exactly the same, if this was 5x plus
4y right over here, and we got a different number,
then we would be in trouble. Because we bought
the same combination, but we got a different price. But the good thing is is that
we have a different combination here. So let's see if we
can work it out. Now, the most obvious
thing that jumps out at me is that I have a 5x here, and
I have a 5x right over here. So if I could subtract
this 5x from that 5x, then I would cancel
out all of the x terms. So what I'm going
to do is I'm going to multiply this bottom
equation by negative 1. So it becomes negative
5x plus negative 5y is equal to negative 11. And then I'm going
to essentially add both of these equations. And I could do that
because I'm doing the same thing to both sides. I already know that this
thing is equal to this thing. So I'm just adding those
things to either side. So on the left hand
side, I have 5x minus 5x. Well, those cancel out. And then I have 4y minus 5y. Well, that's negative y. And that's going to be
equal to 10 minus 11, which is negative 1. And then if we multiply
both sides of this times negative 1, or divide
both sides by negative 1, we're going to get
y is equal to 1. So just like that, we
were able to figure out the price of oranges. It's $1 per orange. So this is equal to 1. Now let's figure out
the price of apples. So we can go back into
either 1 of these equations. I'll go back into this first 1. So 5 times-- so let's go to the
person in line in front of us. They bought 5 apples
at x dollars per apple, plus 4 oranges at $1 per orange,
and they spent a total of $10. So this of course is just 4. Let's subtract 4
from both sides, and we get-- well, 4 times 1
minus 4, that just cancels out. We're just going to be left
with 5x on the left hand side. And on the right hand
side, we have 10 minus 4, which is equal to 6. And we can just divide both
sides by 6 now in order to solve for x. Oh, sorry. We can divide both sides by
5 in order to solve for x. It's late in the day. Brain isn't working. Dividing by 6 wouldn't
have done anything. We would have gotten 5/6x. We just want to get an x here. So dividing both sides by 5, we
get x is equal to 6/5 dollars. Or you could say that
x is equal to 6/5, which is the same
thing as 1 and 1/5, which is the same
thing as $1.20. So it's $1 per orange,
and $1.20 per apple. So we absolutely
could figure out the prices of apples and oranges
using the information given.