Question

If (3y)/x = 4 and x^2 = 36, what are the values for x and y?

pls help, 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

Answer 1

First, find the value of x.

x squared is 36. So, take the square root of 36. You end up with 6.
x=6
Next, plug 6 in for x, to find y.
We know that whatever 3y is, it had to be divided by 6 to get 4.
So, working backwards, we can take 6*4=24
3y=24
Then, just divide both sides by 3.
24/3=8
y=8


x=6
y=8

Answer 2

x=±6 and y=±8.

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.