Question
Solve for x and y:
$\frac{5}{\text{x}}+\text{6y}=13,$
$\frac{3}{\text{x}}+\text{4y}=7\ (\text{x}\neq0).$

Answer

Putting $\frac{1}{\text{x}}=\text{u}$ the given equations become 5u + 6y = 13 ...(1) 3u + 4y = 7 ...(2)Multiplying (1) by 4 and (2) by 6, we get
20u + 24y = 52 ...(3)
18u + 24y = 42 ...(4)
Subtracting (4) from (3), we get
2u = 10
⇒ x = 5
Substituting u = 5 in (1), we get
5 × 5 + 6y = 13
⇒ 6y = 13 - 25
⇒ 6y = -12
⇒ y = -2
u = 5
$\Rightarrow\frac{1}{\text{x}}=5$ $\Rightarrow\text{5x}=1$ $\Rightarrow\text{x}=\frac{1}{5}$$\therefore$ The solution is $\text{x}=\frac{1}{5}$ and y = -2

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