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Mathematics 19 Online
OpenStudy (anonymous):

Which ordered pair in the form (x, y) is a solution of this equation? (x + 3)y = 14 a. (5,2) b. (7,2) c. (11,1) d. (3,2)

OpenStudy (fibonaccichick666):

can you solve for y?

OpenStudy (anonymous):

I am completely clueless :(

OpenStudy (fibonaccichick666):

ok, can you tell me where it starts?

OpenStudy (fibonaccichick666):

like your confusion

OpenStudy (anonymous):

wouldnt you start out like this? (x+3) y = 14 xy + 3y = 14

OpenStudy (fibonaccichick666):

you could, but actually, you don't even have to make it that hard. It's one step(think of how you'd solve for b here: (4+9)b=10)

OpenStudy (anonymous):

Oh i kinda get it! the answer would be c right? you can plug in the numbers and (11+3)1 = 14 ;) i think thats it anyways

OpenStudy (fibonaccichick666):

huh, wait no thta's not what I was getting at, but that is a way... not a good way because then you don't get the concept, though it works

OpenStudy (fibonaccichick666):

if you can solve for x or y, you get a form that makes it really to check as opposed to slighty more difficult(YOUR APPROACH)

OpenStudy (fibonaccichick666):

OOPS SORRY FOR THE CAPS!

OpenStudy (fibonaccichick666):

oops again lol

OpenStudy (anonymous):

Thank You so much! :) and its okay xD

OpenStudy (fibonaccichick666):

ok, so real quick, how i would do it: solve for y real quick just divide both sides by (x+3) \[(x+3)y=14\] \[y=\frac{14}{x+3}\] then do your plug and chugs

OpenStudy (fibonaccichick666):

and np

OpenStudy (anonymous):

OH! Okay! I get it from here i guess i just couldn't set up the problem correctly. Thank you!!

OpenStudy (fibonaccichick666):

np best of luck, just tag me if ya need set up help. I am willing to help anyone who puts the effort in. :)

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