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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
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Solution
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`
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