Question

The solution of the system of linear equations $\frac{2}{x}+\frac{5}{y}=10and\frac{3}{x}+\frac{2}{y}=10$ is

• A. x = 11, y = 10
• B. $x=\frac{11}{5},y=\frac{11}{12}$
• C. $x=\frac{11}{5},y=12$
• D. $x=\frac{11}{5},y=\frac{11}{20}$

Answer 1

The correct option is D $x=\frac{11}{5},y=\frac{11}{20}$

Given: $\frac{2}{x}+\frac{5}{y}=10and\frac{3}{x}+\frac{2}{y}=10$

Let $\frac{1}{x}=pand\frac{1}{y}=q$

Therefore, the pair of linear equations become
2p + 5q = 10 ..…(i)
and, 3p + 2q = 5 …..(ii)
Multiplying (i) by 2 and (ii) by 5, we get
4p + 10q = 20 ..…(iii)
15p + 10q = 25 ..…(iv)
Subtracting (iv) from (iii), we get
(4p + 10q) – (15p + 10q) = 20 – 25
$\Rightarrow$ 4p + 10q – 15p – 10q = 20 – 25
$\Rightarrow$ –11p = –5
$p=\frac{5}{11}$
Now, putting $p=\frac{5}{11}$ in (i), we get

$2\times \frac{5}{11}+5q=10\frac{10}{11}+5q=105q=10-\frac{10}{11}5q=10-\frac{100}{11}q=\frac{20}{11}$

Thus, $p=\frac{5}{11}andq=\frac{20}{11}Therefore,x=\frac{1}{p}=\frac{11}{5}andy=\frac{1}{q}=\frac{11}{20}.$
Hence, the correct answer is option (4).