MCQ
If $\frac{2}{x}+\frac{3}{y}=13$ and $\frac{5}{x}-\frac{4}{y}=-2$, then $x+y$ equals
- A$\frac{1}{6}$
- B$-\frac{1}{6}$
- ✓$\frac{5}{6}$
- D$-\frac{5}{6}$
✓
Answer
Correct option: C.
$\frac{5}{6}$
(C)$\frac{5}{6}$
Let $\frac{1}{x}=u$ and $\frac{1}{y}=v$. Then, the given system of equations becomes
$
2 u+3 v-13=0
$
and,
$
5 u-4 v+2=0
$
Using cross-multiplication, we obtain
$
\begin{array}{ll}
& \frac{u}{6-52}=\frac{v}{-65-4}=\frac{1}{-8-15} \Rightarrow \frac{u}{-46}=\frac{v}{-69}=\frac{1}{-23} \Rightarrow u=2, v=3 \Rightarrow x=\frac{1}{2}, y=\frac{1}{3} \\
\therefore & x+y=\frac{1}{2}+\frac{1}{3}=\frac{5}{6}
\end{array}
$
Let $\frac{1}{x}=u$ and $\frac{1}{y}=v$. Then, the given system of equations becomes
$
2 u+3 v-13=0
$
and,
$
5 u-4 v+2=0
$
Using cross-multiplication, we obtain
$
\begin{array}{ll}
& \frac{u}{6-52}=\frac{v}{-65-4}=\frac{1}{-8-15} \Rightarrow \frac{u}{-46}=\frac{v}{-69}=\frac{1}{-23} \Rightarrow u=2, v=3 \Rightarrow x=\frac{1}{2}, y=\frac{1}{3} \\
\therefore & x+y=\frac{1}{2}+\frac{1}{3}=\frac{5}{6}
\end{array}
$
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