# Stumped

#### krisolaw

##### New member
If y varies inversely with the square of x, and y is 45 when x is 4, find x when y is 5.

#### stapel

##### Super Moderator
Staff member
Variation problems translate as follows:

. . . . ."y varies directly with x": y = kx
. . . . ."y varies inversely with x": y = k/x
. . . . ."y varies jointly with x and z": y = kxz

Translate the given relationship. Plug "45" in for "y" and "4" in for "x". Solve for the value of "k".

Rewrite the variation equation, this time using the value you found for "k". Plug "5" in for "y". Solve for x.

Eliz.

#### krisolaw

##### New member
Thank you so much. I have a better understaning now. #### krisolaw

##### New member
IF:
. . . . ."y varies directly with x": y = kx
. . . . ."y varies inversely with x": y = k/x
. . . . ."y varies jointly with x and z": y = kxz

Then what about the problem:

If y varies inversely with the square of x, and y is 45 when x is 4, find x when y is 5.

Instead of it being y = k/x wouldn't it be y = k/x^2? Thanks!

#### stapel

##### Super Moderator
Staff member
krisolaw said:
If y varies inversely with the square of x,...wouldn't it be y = k/x^2?
Yes. As stated, "if y varies inversely with" whatever, then the equation is "y = k/(whatever)".

It is common mathematical practice to use variables for unknowns, which is what the "x" stood for. Sorry for the confusion.

Eliz.