Question

Find the value of k for which each of the following systems of linear equations has an infinite number of solutions:

$2x+3y=7,(k-1)x+(k+2)y=3k.$

Answer 1

The given system may be written as

2x+3y-7=0
(k−1)x+(k+2)y-3k=0

The given system of equation is of the form

$a_{1}x+b_{1}y+c_1$ = 0

$a_{2}x+b_{2}y+c_2$ = 0

Where, $a_1$=2,$b_1$=3,$c_1$=−7

$a_2$=k,$b_2$=k+2,$c_2$=3k

For unique solution,we have

$\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}$
$\frac{2}{k-1}=\frac{3}{k+2}=\frac{-7}{-3k}$

$\frac{2}{k-1}=\frac{3}{k+2}$ and $\frac{3}{k+2}=\frac{-7}{-3k}$

⇒2k+4=3k−3 and 9k=7k+14
⇒k=7and k=7

Therefore, the given system of equations will have infinitely many solutions, if k=7.