Question
If $${x^2} + {y^2} = 29$$ and xy = 10 where x > 0, y > 0, x > y, then the value of $$\frac{{x + y}}{{x - y}}$$ is?
Answer: Option B
$$\eqalign{
& {\left( {x + y} \right)^2} = {x^2} + {y^2} + 2xy \cr
& \Rightarrow {\left( {x + y} \right)^2} = 29 + 2 \times 10 \cr
& \Rightarrow {\left( {x + y} \right)^2} = 49 \cr
& \therefore x + y = 7 \cr
& {\left( {x - y} \right)^2} = {x^2} + {y^2} - 2xy \cr
& \Rightarrow {\left( {x - y} \right)^2} = 29 - 2 \times 10 \cr
& \Rightarrow {\left( {x - y} \right)^2} = 9 \cr
& \therefore \left( {x - y} \right) = 3 \cr
& \therefore \frac{{x + y}}{{x - y}} = \frac{7}{3} \cr} $$
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$$\eqalign{
& {\left( {x + y} \right)^2} = {x^2} + {y^2} + 2xy \cr
& \Rightarrow {\left( {x + y} \right)^2} = 29 + 2 \times 10 \cr
& \Rightarrow {\left( {x + y} \right)^2} = 49 \cr
& \therefore x + y = 7 \cr
& {\left( {x - y} \right)^2} = {x^2} + {y^2} - 2xy \cr
& \Rightarrow {\left( {x - y} \right)^2} = 29 - 2 \times 10 \cr
& \Rightarrow {\left( {x - y} \right)^2} = 9 \cr
& \therefore \left( {x - y} \right) = 3 \cr
& \therefore \frac{{x + y}}{{x - y}} = \frac{7}{3} \cr} $$
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