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Call the one integer x.
call the other one y
x = 4y - 8
xy = 60
Solve this.

can you help solve it?

Substitute the x=4y-8 into the second equation

you will get (4y -8)y = 60
solve for y.
then solve for x

i got y=-15 and x=-68?

im not positive that thats correct

no =( I don't think so.

Try again.

i dont know what i did wrong :/

Show me your work, okay?

ok 60=(4y-8)y
60=4y-8y
60/-4 = -4y/-4
y=-15

The problem lies in line 2

x= 4(-15)-8
x= -60-8
x=-68

what the problem? :o

(4y-8)y = 4y^2 -8y

distribute, sweet thang.

Yes =) Take out a common factor of 4 first though.

2 and 2

It makes it a lot easier.

Nooo
show me your factoring, please.

4? isnt is 2 and 2 or 4 and 1?

noope. show me your factoring.

from 4y^2 -8y-60 = 0

4y*y ?

ok

I only say that because it's a whole different issue. We can work on it another time =)

Re-read the problem and tell me if we have answered the question yet.

i did but i don't know won't that only find x and not y?

Right. Well actually, it finds y. Right?

i think so?

yes it seems so confusing
i got fractions for both answers
-1/12 and 1/20
is that correct?

=/ I don't know what mistake you made.

i dont know either :(

\[\frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-(-8) \pm \sqrt{(-8)^2-4(4)(-60)}}{2(4)}\]

You probably just messed up on a sign somewhere. Did you plug in like that?

oh i see what i did im sorry i was doing 2ac instead of 2a

Ah =)

x= 5 and x= -3?

Good. Except we were solving for y. It's important because of the next step.

which variable do i plug into and do i use the 5 or -3?

lol

sorry and yes we know 2 answer to y so then which one would you substitute for x?

ok i chose 5 and plugged it into x=4y-8 and i got 12

Good =)
So one possible solution is:
x=12 y =5
Understand? Find the other possible solution.

the other possible solution is 55?

Woah there. Where'd 55 come from? You makin' guesses or somethin'?

whoops is it 63?

wait ! lol sorry im using the wrong equation

I have no idea what you're doing. Lol.

x= -20 ?

That's not wrong... but it's not a complete answer.

but isnt it x= 12, -20

but theres 2 x and y values

Ooooh, goodness. How to explain this...

lol i dont know what should i do? should i just leave 2 values for both variables?

Kind of.

Alright, stop worrying so much about the correct answer and let's just consider the question, okay?

ok

Two solution sets. (5,12) and (-3, 20) That's all you need to know

And we're just trying to figure out two numbers that those things are true for.

Right?

right

Well, it turns out, there's more than one possible pair of numbers.

yup so i just leave those 2 answers?

Right. And what are those two answers?

y= 5, -3
x= 12, -20

mmmm, I don't like how you wrote that.

why?

Is x=5 y = -20 a solution?

Try plugging it in if you need to check.

no i put y=5

and x= -20

Sorry, my mistake.
Is y = 5 x=-20 a solution?

That's what I meant to ask.

wait no it isnt i just tried it out

Oh my. =(

Okay so... is y=5 still good?

It's not good when x=-20, we decided.

yes

Alright, why is it still good?

because when we plug it in -20=4(5)-8
it gives us 12 which is not equal to -20

Okay, that tells me why
y=5 x=-20 is not a solution.
Is y =5 bad in general then?
Is x=-20 bad?

yes its bad the -20

wait no the 5 is bad

if we plug in -3 it will work

Ahhhhhh
my point isn't that they are bad COMPLETELY.
My point is just that they don't work TOGETHER.

oh sorry lol so i can just leave both numbers?

-3 and -20

Great. That's one answer.

As a SECOND. SEPARATE answer

you could do 5 and 12

Goodnight. =)

goodnight and thanks so much once again! you have been an amazing help!

My pleasure =)