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?
A. $$ - \frac{7}{3}$$
B. $$\frac{7}{3}$$
C. $$\frac{3}{7}$$
D. $$ - \frac{3}{7}$$
Answer: Option B
Solution(By Examveda Team)
$$\eqalign{ & {\left( {x + y} \right)^2} = {x^2} + {y^2} + 2xy \cr & \Rightarrow {\left( {x + y} \right)^2} = 29 + 2 \times 10 \cr & \Rightarrow {\left( {x + y} \right)^2} = 49 \cr & \therefore x + y = 7 \cr & {\left( {x - y} \right)^2} = {x^2} + {y^2} - 2xy \cr & \Rightarrow {\left( {x - y} \right)^2} = 29 - 2 \times 10 \cr & \Rightarrow {\left( {x - y} \right)^2} = 9 \cr & \therefore \left( {x - y} \right) = 3 \cr & \therefore \frac{{x + y}}{{x - y}} = \frac{7}{3} \cr} $$Related Questions on Algebra
If $$p \times q = p + q + \frac{p}{q}{\text{,}}$$ then the value of 8 × 2 is?
A. 6
B. 10
C. 14
D. 16
A. $$1 + \frac{1}{{x + 4}}$$
B. x + 4
C. $$\frac{1}{x}$$
D. $$\frac{{x + 4}}{x}$$
A. $$\frac{{20}}{{27}}$$
B. $$\frac{{27}}{{20}}$$
C. $$\frac{6}{8}$$
D. $$\frac{8}{6}$$
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