How do I solve: 9y^4/x^-2 + (x^-1/y^2)^-2?

Question

I'm still confused with this one: every solution seems to combine x^2 + x^2 as x^2; I don't see why it wouldn't be 2x^2.

Answer 1

remember x^-2 = 1/ x^2 and x6-1 = 1/x^1 =1/x

lets simplify the terms separately

9y^4/ x^-2 = (9y^4)/ (1/x^2)
= (9y^4) X (x^2/1)
= 9 y^4 x^2
=9 x^2 y^4

(x^-1/y^2)^-2 = (x^-1)^-2 / (y^2)^-2
= ( x^2) / (y^-4) {y^-4 = 1/y^4}
= x^2 X (1/y^4)
= x^2 X y^4

Now combine the terms
9 x^2 X y^4 + x^2 X y^4
9 x^2y^4 + x^2y^4 {these two are like terms}
10 x^2y^4

remember in this question we have to follow the BEDMAS rule so we cant add as you are saying. multiplications are done first.

e.g 3x2+2x4 = 6+8 and NOT 3x4x4

Answer 2

You can't solve this because there isn't an "=" in it. You can simplify:

9y^4 * x^2 + x^2 * y^4

= x^2(9y^4 + y^4)

=10y^4 * x^2

Answer 3

you can't answer it without equal mark I will give you the steps and you will continua
9y^4*x^2 + x^2 * y^4=(the value of sum)
x^2 [9y^4+y^4] =
x^2= then x=
or 9y^4+y^4=
10y^4=
then y=

Answer 4

9y^4/x^-2 + (x^-1/y^2)^-2
=9y^4x^2 + 1/(xy^2)^-2
=9y^4x^2 + x^2y^4
=10 x^2y^4

Answer 5

9y^4/x^-2 + (x^-1/y^2)^-2

=9y^4x^2 + (1/xy^2)^-2
=9y^4x^2 + (1/xy^2)^-2
=9y^4x^2 + (1/x-2y^-4)
=9y^4x^2 + (x^2y^4)
=10x^2y^4

Answer 6

9y^4/x^-2+(x^-1/y^2)^-2=
=9y^4x^2+(1/xy^2)^-2=
=9(xy^2)^2+(xy^2)^2=
=10(xy^2)^2=
=10x^2y^4

Answer 7

Why would you want yo?