Question

Solve the following system of equations by substitution method.

$\frac{15}{x}+\frac{2}{y}=17,\frac{1}{x}+\frac{1}{y}=\frac{36}{5},x\ne 0,y\ne 0$

• A. $(5,\frac{1}{7})$
• B. $(2,\frac{1}{7})$
• C. $(5,\frac{2}{7})$
• D. None of these

Answer 1

Answer: (A)

$(5,\frac{1}{7})$

Given equations are

$\frac{15}{x}+\frac{2}{y}=17,$

$\frac{1}{x}+\frac{1}{y}=\frac{36}{5}$

$Let\frac{1}{x}=a,\frac{1}{y}=b$

$\therefore 15a+2b=17$......(i)

and $a+b=\frac{36}{5}$

$\Rightarrow a=\frac{36}{5}-b$....(ii)

Putting equation (ii) in (i) we get,

$15[\frac{36}{5}-b]+2b=17$

$\Rightarrow 108-13b=17$

$\Rightarrow b=7$

$\therefore y=\frac{1}{7}$

Putting value of $b$ in equation (i), we get,

$15a+(2\times 7)=17$

$\Rightarrow a=\frac{1}{5}$

$\therefore x=5$

Hence, $(5,\frac{1}{7})$ is the solution of the given system of equations.