Question

Solve:

$3(2u+v)=7uv$ and
$3(u+3v)=11uv$.

Answer 1

Clearly, the given equations are not linear equations in the variables u and v but can be reduced to linear equations by appropriate substitution.

If we put $u=0$ in either of the two equations, we get $v=0$.
So, $u=0,v=0$ form a solution of the given system of equations.

To find the other solutions, we assume that $u\ne 0,v\ne 0$.
Now, $u\ne 0,v\ne 0\Rightarrow uv\ne 0$.

On dividing each one of the given equations by uv, we get

$\frac{6}{v}+\frac{3}{u}=7$ (i)

$\frac{3}{v}+\frac{9}{u}=11$ (ii)

Taking $\frac{1}{u}=x$ and $\frac{1}{v}=y$, the given equations become

$3x+6y=7$ ..(iii)

$9x+3y=11$ .(iv)

Multiplying equation (iv) by $2$, the given system of equations becomes

$3x+6y=7$ .(v)

$18x+6y=22$ .(vi)

Substracting equation (vi) from equation (v), we get

$-15x=-15\Rightarrow x=1$

Putting $x=1$ in equation (iii), we get

$3+6y=7\Rightarrow y=\frac{4}{6}=\frac{2}{3}$

Now, $x=1\Rightarrow \frac{1}{u}=1\Rightarrow u=1$

and, $y=\frac{2}{3}\Rightarrow \frac{1}{v}=\frac{2}{3}\Rightarrow v=\frac{3}{2}$.

Hence, the given system of equations has two solutions given by

(i) $u=0,v=0$

(ii) $u=1,v=3/2$.