If \( \Large \frac{x}{y}=\frac{3}{4} \), the value of \( \Large \frac{6}{7}+\frac{y-x}{y+x} \) is:
Correct Answer: Description for Correct answer:
\( \Large \frac{x}{y}=\frac{3}{4}, \frac{6}{7}+\frac{y-x}{y+x}=? \)
= \( \Large \frac{6}{7}+\frac{y \left(1-\frac{x}{y}\right) }{y \left(1+\frac{x}{y}\right) } \)
= \( \Large \frac{6}{7}+\frac{y \left(1-\frac{3}{4}\right) }{y \left(1+\frac{3}{4}\right) } \)
= \( \Large \frac{6}{7}+\left[ \frac{4-3}{4} \times \frac{4}{ \left(4+3\right) } \right] \)
= \( \Large \frac{6}{7}+\frac{1}{7} = \frac{7}{7} = 1 \)
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