Question

Find the value of k for which the following system of linear equations has infinite solutions:
$x+(k+1)y=5$
$(k+1)x+9y=8k-1$

Answer 1

This system of equation is of the form
$a_{1}x+b_{1}y=c_1$
$a_{2}x+b_{2}y=c_2$

where $a_1=1,b_1=k+1,c_1=5$

and $a_2=k+1,b_2=9$ and $c_2=8k-1$

For infinitely many solutions, we must have
$\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}$

The given system of equations will have infinite solutions, if

$\frac{1}{k+1}=\frac{k+1}{9}=\frac{5}{8k-1}$

$\Rightarrow \frac{1}{k+1}=\frac{k+1}{9}$ and $\frac{k+1}{9}=\frac{5}{8k-1}$

$\Rightarrow (k+1{)}^2=9$ and $(k+1)(8k-1)=45$

Now, $(k+1{)}^2=9$

$\Rightarrow k+1=\pm 3\Rightarrow k=2,-4$

We observe that $k=2$ satisfies the equation $(k+1)(8k-1)=45$ but, $k=-4$ does not satisfy it.

So, for $k=2$ the given system of the linear equations has infinite solutions.