If (3y)/x = 4 and x^2 = 36, what are the values for x and y? pls , it's the last question and i'm really confused. i added a picture of the problem just in case you needed to see it or something

x=±6 and y=±8.

Step-by-step explanation:

We have to find value of x and y.For this ,two equations are given

3y/x=4

x²=36

from above equation,we can find value of x by taking square root.

x²=36

taking square root to both sides of above equation,we get

√x²=√36

as 36 is square of ±6,so

x=±6

putting  x=6 in first equation to find the value of y,we get

3y/(6)=4

multiplying by 6 to both sides of above equation,we get

6(3y/6)=6(4)

3y=24

multiplying by 1/3 to both sides of above equation,we get

1/3(3y)=1/3(24)

y=8

Now,putting  x=-6 in first equation to find the value of y,we get

3y/(-6)=4

multiplying by 6 to both sides of above equation,we get

-6(3y/-6)=-6(4)

3y=-24

multiplying by 1/3 to both sides of above equation,we get

1/3(3y)=1/3(-24)

y=-8

so, x=±6 and y=±8.

asdsaddsa

step-by-step explanation: