what is the number of solutions to the system?
x + y = -1
5x - 5y = 7

- anonymous

what is the number of solutions to the system?
x + y = -1
5x - 5y = 7

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- schrodinger

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- anonymous

someone please help me! i dont get this

- M

solve for x then plug it in the second equation

- anonymous

so i would set it up like x + y = -1 = 5x - 5y = 7 ?

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## More answers

- M

x = -1 - y
5(-1 - y) -5y = 7
then you'll know what y is

- M

if you know y then you can plug in y to first equation and find out what x is

- anonymous

wait let me write this down and try to do that

- M

-5 - 5y -5y = 7
-10y = 14
y = -14/10

- M

when you have two unknown (here x and y) then you always want to solve for x or y, then plug it into the other equation so you're working with just one unknown variable

- anonymous

ok i get the -10 y part. buy what about the -5 and where did 14 come from?

- anonymous

ahhh. isee.

- M

you try to get y on left and everything else on right

- anonymous

yeah but you left -5 out. what did you do with it? and where did 14 come from?

- M

you move -5 to right side of = sign

- M

you want to get rid of -5 on left side so you add +5 to both side

- M

so -5 disappears on left and +5 appears on right

- anonymous

Okay. but that equals 12.

- M

whatever you do on left you MUST do same on right to keep equation same

- M

sorry it's 12 :)

- M

12/10

- anonymous

i know that. ahh seee. thaats why i was like huh? lol

- anonymous

u got me confused rii there. its cool tho

- M

so plug in y= -12/10 to find x

- anonymous

isnt it 12/-10 ?

- M

it doesn't matter if it's on top or bottom

- M

it's (-1) (12/10)

- M

^^ that's the correct form

- anonymous

ahhhhh i seee. so the answer is one solution.

- M

there's only one solution for y and one for x

- M

answer is a fraction

- anonymous

i know. but its asking me how many solutions. one two or no solutions

- anonymous

i have this other one. i have to find the ordered pairs. let me see if i can do it.
-3x + 5y = 4
5x +5y = 60

- M

i'm not sure what the question is asking?

- M

there can be infinite combination of x and y and it'll give you those answers

- anonymous

it says give the ordered pairs that satisfies the system of equations

- anonymous

well for the other one that your talking about.... it was asking WHAT IS THE NUMBER OF SOLUTIONS TO THE SYSTEM? and i put ONE SOLUTION

- anonymous

theres another choice that says INFINITELY MANY SOLUTIONS

- M

i think it may be asking for just x and y like (x, y)

- M

so just one solution

- anonymous

yeah it is.

- anonymous

Okay. thanks.

- anonymous

let me see if i can do the other one.

- anonymous

i have a question

- anonymous

i did 5x ( -3x + 5y ) +5y = 60

- M

no you have to solve for y first

- M

y = (1/5)(3x +4)

- M

then you plug it in like you did

- anonymous

so thats how im suppose set it up rii?

- M

i'm not sure what you did

- M

you first either solve for x or y

- M

then plug in the result to the second equation

- anonymous

i want to solve for X

- anonymous

so how do i set it up?

- M

x = (5y - 4)(1/3)

- M

5[(5y - 4)(1/3)] +5y = 60
the bracket [ ] is the x

- M

now you only have one variable y in your equation since we plugged in for x

- anonymous

where did one third come from?

- anonymous

wait let me get this straightt....

- M

i just plugged in for x

- anonymous

ahh okay

- anonymous

so now that we have our equation we have to distribute the 5 to the perenthesis (5y-4) correct?

- M

you can do that then divide everything by 3

- anonymous

i can distribute 5 to 1/3 righT?

- anonymous

i can distribute 5 to 1/3 righT?

- anonymous

i can distribute 5 to 1/3 righT?

- anonymous

i can distribute 5 to 1/3 righT?

- anonymous

i can distribute 5 to 1/3 righT?

- anonymous

i can distribute 5 to 1/3 righT?

- anonymous

i can distribute 5 to 1/3 righT?

- anonymous

i can distribute 5 to 1/3 righT?

- anonymous

i can distribute 5 to 1/3 righT?

- M

you're not distributing since (1/3) is its own. you simply multiply

- anonymous

srry about that

- anonymous

so what i shud have is [ (25y-20) (1/3) ]

- anonymous

[ (25y-20) (1/3) ] + 5y = 60

- M

yeah then solve for y

- M

now you have to distribute (1/3)

- anonymous

im a little confused. so right this is what i have......
[ (25y-20) (1/3) ] + 5y = 60
-5y - 5y
[ (20y- 20) (1/3) ] = 60

- M

no

- M

you don't want to move 5y to other side because you still have 25y

- M

you want to move 20(1/3) to right

- M

then combine 25y(1/3) + 5y

- M

you may want to use calculator

- anonymous

so multiply -20/2 times 1/3

- M

remember we're trying to solve for y

- anonymous

wait ignore that.. typo

- anonymous

so multiply 20/1 times 1/3 then combien 25y times 1/3 + 5y then combine like terms so you'd end up with 30y (1/3)

- M

no

- M

you first distribute (1/3) to 25y and -20

- M

so it's 25y(1/3) + 5y = 60 + 20(1/3)

- M

y(25(1/3) + 5) = 60 + 20(1/3)
y = 60 + 20(1/3) / (25(1/3) + 5)

- M

it looks real messy b/c i didn't simplify anything

- M

\[y = 60 + 20(1/3)\div (25(1/3) + 5)\]

- anonymous

so now im gunna distribute 20 (1/3) and 25 (1/3)

- M

no you just multiply

- M

it's just 20/3 and 25/3

- M

you can multiply by (1/3) or just divide by 3

- anonymous

500\3

- M

where'd that come from

- anonymous

wowww. what a retardd.

- anonymous

sorry im a little bit slow and dumb

- anonymous

math is rlly not my thing.

- M

\[200/3 \div 40/3\]

- M

so it's 200/3

- M

i mean 200/40

- M

= 5

- anonymous

y = 5

- M

yeah

- anonymous

x=7

- M

yeah

- anonymous

Omgg. thank you so much! you have no idea how much you've helped me!

- M

np
just remember to isolate x or y, whichever is easier. then plug that into the other equation

- anonymous

thank youu! but the bad thing is, that on my worksheet, theres different types of problems. they never stick to one thing.

- M

yeah, all you can do is practice. good luck

- anonymous

yeah. thank youu! im sure ill do good on my state exam! thanks alot!

- M

bye :)

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