Question

Solve the following system of equations:
$8v=3u+5uv$
$6v-5u=-2uv$.

Answer 1

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

If we put $u=0$ in either of the two equations, we get $v=0$

Thus, $u=0,v=0$ from one solution of the given system of equations.

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

Since $u\ne 0$, $v\ne 0$. Therefore, $uv\ne 0$

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

$\frac{8}{u}-\frac{3}{v}=5\dots \left(i\right)$

$\frac{6}{u}-\frac{5}{v}=-2\dots \left(ii\right)$

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

$8x-3y=5\dots \left(iii\right)$

$6x-5y=-2\dots \left(iv\right)$

Multiplying equation $\left(i\right)$ by $3$ and equation $\left(ii\right)$ by $4$, we get

$24x-9y=15\dots \left(v\right)$

$24x-20y=-8\dots \left(vi\right)$

Substituting equation $\left(vi\right)$ from equation $\left(v\right)$, we get

$11y=23\Rightarrow y=\frac{23}{11}$

Putting $y=\frac{23}{11}$ in equation (iii), we get

$8x-\frac{69}{11}=5$

$\Rightarrow 8x=\frac{69}{11}+5$

$\Rightarrow 8x=\frac{124}{11}$

$\Rightarrow x=\frac{31}{22}$

Now,

$x=\frac{31}{22}$

$\Rightarrow \frac{1}{u}=\frac{31}{22}$

$\Rightarrow u=\frac{22}{31}$

And,

$y=\frac{23}{11}$

$\Rightarrow \frac{1}{v}=\frac{23}{11}$

$\Rightarrow v=\frac{11}{23}$

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

$\left(i\right)$ $u=0,v=0$

$\left(ii\right)$ $u=\frac{22}{31},v=\frac{11}{23}$.