Question

Solve for u and v,
where u, v$\ne 0$
$2(3u-v)=5uv2(u+3v)=5uv$

• A.
  $2and1$
• B.
  $2and3$
• C.
  $4and0$
• D.
  $7and4$

Answer 1

The correct option is A

$2and1$
$2(3u-v)=5uv\frac{6u-2v}{uv}=\frac{5uv}{uv}\frac{6u}{uv}-\frac{2v}{uv}=\frac{5uv}{uv}\frac{6}{v}-\frac{2}{u}=\frac{5uv}{uv}\frac{6}{v}-\frac{2}{u}=\mathrm{5...}\left(1\right)2(u+3v)=5uv\Rightarrow \frac{2u+6v}{uv}=\frac{5uv}{uv}\Rightarrow \frac{2u}{uv}+\frac{6v}{uv}=\frac{5uv}{uv}\Rightarrow \frac{2}{v}+\frac{6}{u}=\mathrm{5....}\left(2\right)Let\frac{1}{v}=a,\frac{1}{u}=b6a-2b=5...\left(3\right)2a+6b=\mathrm{5...}\left(4\right)ByCrossMuliplicationMethod$

$\frac{a}{-10-30}=\frac{6}{10-30}=\frac{-1}{36-(-4)}\Rightarrow \frac{a}{-40}=\frac{b}{-20}=\frac{-1}{40}\Rightarrow \frac{a}{-40}=\frac{-1}{40}\Rightarrow a=\frac{40}{40}\Rightarrow a=1\Rightarrow \frac{6}{-20}=\frac{-1}{40}\Rightarrow b=\frac{20}{40}\Rightarrow b=\frac{1}{2}\therefore a=1\Rightarrow \frac{1}{v}=1\therefore v=1b=\frac{1}{2}\Rightarrow \frac{1}{u}=\frac{1}{2}\therefore u=2$

So, option a is correct.