Question

If $\frac{1}{2(2x+3y)}+\frac{12}{7(3x-2y)}=\frac{1}{2},\frac{7}{(2x+3y)}+\frac{4}{(3x-2y)}=2$, then the value of $(2x+3y)$ is

• A. $7$
• B. $6$
• C. $\frac{1}{7}$
• D. None of these

Answer 1

The correct option is A $7$

$\frac{1}{2\left(2x+3y\right)}+\frac{12}{7\left(3x-2y\right)}=\frac{1}{2}$

$\frac{7}{2x+3y}+\frac{4}{3x-2y}=2$

Let $\frac{1}{2x+3y}=x$ and $\frac{1}{3x-2y}=y$

Then above become equation

$\frac{1}{2}x+\frac{12}{7}y=\frac{1}{2}$

$\Rightarrow 7x+24y=7\left(1\right)\&7x+4y=2\left(2\right)$

$\left(7x+24y\right)-\left(7x+4y\right)=7-2$ $\Rightarrow 20y=5$

$y=\frac{1}{4}$

Putting $y=\frac{1}{4}$ in (2)

$7x+4\times \frac{1}{4}=2$

$\Rightarrow 7x=1$

$x=\frac{1}{7}$

So,

$x=\frac{1}{7}$

$\Rightarrow \frac{.1}{2x+3y}=\frac{1}{7}$

$\Rightarrow 2x+3y=7$