Find the error in the following computation. Then correct the error by rewriting the statement.

Question

If y varies directly with x², and y = 2 when x = 4, then y = 3 when x = 9.

Please describe each step so I can understand. Thank you!

Answer 1

y varies directly with x² means:

 

y = kx² 

 

where k is an unknown constant.  We can find the value of k by plugging in the given point y = 2 when x = 4:

 

y = kx²

2 = k4²

2 = k*16

2/16 = k

1/8 = k

 

So the full equation of the direct variation is:

 

y = (1/8)x²

 

Now, is y = 3 when x = 9?  Let's check by plugging in x = 9 and see if we get y=3:

 

y = (1/8)x²

y = (1/8)9²

y = 81/8 = 10 ¹/₈

 

10 ¹/₈ does not equal 3.  So the correct statement would be "then y = 10 ¹/₈ when x = 9".

Answer 2

since, y varies directly with x^2, we can write the equation as -

  y= x²........................(1)

where, k is a constant.

 

Note that, in equation (1) when x increases y will also increase. 

now, if x=2, then for y we will get y=2² => y=4

same way, when x=3, we get y = 3² => y=9

Answer 3

The Y term must be squared to get the X term, thus when X is 2 then X-squared would be 4. Likewise, when Y is 3, then x-squared is 9. So, the statement should be: If Y varies directly with X-squared, and Y = 4 when x = 2, then Y = 9 when X = 3.