Question

Find the values of m and n for which the following system of linear equations has infinitely many solutions:
3x + 4y = 12,
(m + n)x+2(m - n)y=(5m - 1).​

Answer 1

Answer

The given system of equations is,

$\sf 3x + 4y - 12 = 0$

$\sf (m + n) x + 2 (m - n) y + (1 - 5m) = 0$

These equations are of the form,

$\sf {a}_{1} x + {b}_{1} y + {c}_{1} = 0$

$\sf {a}_{2} x + {b}_{2} y + {c}_{2} = 0$

where,

$\sf {a}_{1} = 3, \: {b}_{1} = 4, \: {c}_{1} = -12$

$\sf {a}_{2} = (m + n), \: {b}_{2} = 2(m - n), \:{c}_{2} = (1 - 5m)$

So,

$\sf \dfrac{{a}_{1}}{{a}_{2}} = \dfrac{3}{(m + n)}, \: \dfrac{{b}_{1}}{{b}_{2}} = \dfrac{2}{(m - n)}, \: \dfrac{{c}_{1}}{{c}_{2}} = \dfrac{12}{(5m - 1)}$

Let the given system of equations have infinitely many solutions. Then,

$\sf \dashrightarrow \dfrac{{a}_{1}}{{a}_{2}} = \dfrac{{b}_{1}}{{b}_{2}} = \dfrac{{c}_{1}}{{c}_{2}}$

$\sf \dashrightarrow \dfrac{3}{(m + n)} = \dfrac{2}{(m - n)} = \dfrac{12}{(5m - 1)}$

$\sf \dashrightarrow \dfrac{3}{(m + n)} = \dfrac{2}{(m - n)} \: \: and \: \: \dfrac{2}{(m - n)} = \dfrac{12}{(5m - 1)}$

Cross multiply the numbers.

$\sf \dashrightarrow 3m - 3n = 2m + 2n \: \: and \: \: 10m - 2 = 12m - 12n$

$\sf \dashrightarrow m - 5n = 0 \: --- (i)$

$\sf \dashrightarrow m - 6n + 1 = 0 \: --- (ii)$

On solving (i) and (ii), we get

$\sf \dashrightarrow m = 5 \: \: and \: \: n = 1$

Thus, the required values are $\sf m = 5$ and $\sf n = 1$