Question

Find the values of k for which the following system of equations have infinitely many solutions.
(i) 2x + 3y = 7
(ii) (k + 2)x + (2k + 1)y = 3(2k - 1)

$\left[3\right]$

Answer 1

System of equations:
$a_{1}x+b_{1}y+c_1=0$
$a_{2}x+b_{2}y+c_2=0$
have infinite solutions if:
$\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}$
$\left[1\right]$

The given system of equations will have infinitely many solutions if:
$\frac{2}{k+2}=\frac{3}{(2k+1)}=\frac{-7}{-3(2k-1)}\Rightarrow \frac{2}{k+2}=\frac{3}{2k+1}and\frac{3}{(2k+1)}=\frac{-7}{-3(2k-1)}$
$\left[1\right]$

$\Rightarrow$ 4k + 2 = 3k + 6 and 18k - 9 = 14k + 7
$\Rightarrow$ 4k - 3k = 6 - 2
$\Rightarrow$ k = 4
Also, 18k - 9 = 14k + 7
$\Rightarrow$ 18k - 14k = 7 + 9
$\Rightarrow$ 4k = 16
$\therefore$ Value of k = 4
$\left[1\right]$