Question

If x = `("a"^2 + 3"a" - 4)/(3"a"^2 - 3)` and y = `("a"^2 + 2"a" - 8)/(2"a"^2 - 2"a" - 4)` find the value of x²y^–2  

Answer 1

x = `("a"^2 + 3"a" - 4)/(3"a"^2 - 3)` 

= `(("a" + 4)("a" - 1))/(3("a"^2 - 1))`

= `(("a" + 4)("a" - 1))/(3("a" + 1)("a" - 1))`

= `(("a" + 4))/(3("a" + 1))`

y = `("a"^2 + 2"a" - 8)/(2"a"^2 - 2"a" - 4)`

a² + 2a – 8 = (a + 4) (a – 2)

2a² – 2a – 4 = 2(a² –a – 2)

= 2(a – 2)(a + 1)

y = `(("a" + 4)("a" - 2))/(2("a" - 2)("a" + 1))`

= `("a" + 4)/(2("a" + 1))`

The value of x²y^–2  = `(x^2)/(y^2) = (x/y)^2`

= `[("a" + 4)/(3("a" + 1)) ÷ ("a" + 4)/(2("a" + 1))]`

= `[(("a" + 4))/(3("a" + 1)) xx (2("a" + 1))/(("a" + 4))]^2` 

= `(2/3)^2`

= `4/9`