Question

Find the numbers $u\&v$ such that satisfies $2uv+3v=u$ and $3u+v=2uv.$

• A. $u=\frac{-5}{2}v=\frac{15}{12}$
• B. $u=5v=\frac{-15}{11}$
• C. $u=-15v=\frac{5}{11}$
• D. $u=15v=\frac{5}{11}$

Answer 1

The correct option is A $u=\frac{-5}{2}v=\frac{15}{12}$

Given, $2uv+3v=u---\left(i\right)3u+v=2uv---\left(ii\right)$

Lets convert it to linear equations.
Divide $\left(i\right)\&\left(ii\right)$ with $uv$
$\frac{2uv}{uv}+\frac{3v}{uv}=\frac{u}{uv}\Rightarrow 2+\frac{3}{u}=\frac{1}{v}\frac{3}{u}-\frac{1}{v}=-2---\left(iii\right)\frac{3u}{uv}+\frac{v}{uv}=\frac{2uv}{uv}\Rightarrow \frac{1}{u}+\frac{3}{v}=2---\left(iv\right)$

Lets consider $x=\frac{1}{u}\&y=\frac{1}{v}$
So, $\left(iii\right)\&\left(iv\right)$ becomes
$3x-y=-2---\left(v\right)x+3y=2---\left(vi\right)$

Multiply $\left(v\right)$ by 3
$9x-3y=-6---\left(vii\right)$

Add $\left(vi\right)\&\left(vii\right)$
$9x-3y=-6\underset{–}{x+3y=2}10x=-4\Rightarrow x=\frac{-2}{5}$
Substitute $x$ in $\left(vi\right)$
$\frac{-2}{5}+3y=2\Rightarrow y=\frac{12}{15}$

So numbers are
$u=\frac{1}{x}=\frac{-5}{2}v=\frac{1}{y}=\frac{15}{12}$