Question

Solve: $3(2x+y)=7xy$, $x\ne 0,y\ne 0$
$3(x+3y)=11xy$

• A. $\frac{2}{3},\frac{1}{2}$
• B. $\frac{1}{3},\frac{3}{2}$
• C. $1,\frac{3}{2}$
• D. $\frac{1}{2},\frac{3}{2}$

Answer 1

The correct option is C $1,\frac{3}{2}$

The given equations are to be reduced to linear equations by dividing it by $xy$
$\frac{6}{y}=\frac{3}{x}=7$ $\Rightarrow$ $\frac{3}{x}+\frac{6}{y}=\mathrm{7.......}\left(1\right)$
$\frac{3}{y}+\frac{9}{x}=11$ $s\Rightarrow$ $\frac{9}{x}+\frac{3}{y}.....\left(2\right)$
On observation, it is found that the coefficients of $x$ or $y$ are easy
multiplies. So, using the method of elimination
$\frac{3}{x}+\frac{6}{y}=7$,
Multiplying $\left(2\right)$ by $2$
$\frac{18}{x}+\frac{6}{y}=22$
--------------------------------------------
$\frac{15}{x}=15,x=1$
Put $x=1$ in $\left(1\right)$
We get $y=\frac{3}{2}$