Question

Solve by substitution method $\frac{1}{x}+\frac{1}{y}=4$ and $\frac{2}{x}+\frac{3}{y}=7,x\ne 0,y\ne 0$

Answer 1

Let $\frac{1}{x}=a$ and $\frac{1}{y}=b$
The given equations become
$a+b=4$ ........ (1)
$2a+3b=7$ ......... (2)
Equation (1) becomes b = 4 - a ......... (3)
Substituting b in (2) we get, $2a+3(4-a)=7$
$\Rightarrow 2a+12-3a=7$
$2a-3a=7-12$
$-a=-5\Rightarrow a=5$
Substituting a = 5 in (3) we get, $b=4-5=-1$
But $\frac{1}{x}=a\Rightarrow x=\frac{1}{a}=\frac{1}{5}$
$\frac{1}{t}=b\Rightarrow y=\frac{1}{b}=\frac{1}{-1}=-1$
$\therefore$ The solution is $x=\frac{1}{5},y=-1$