Question

Find the value of k for which the following pair of linear equations has infinitely many solutions.
2x-3y=7, (k+1)x+(1-2k)y = (5k -4)​

Answer 1

Answer

The given equations are,

$\sf 2x - 3y - 7 =0$

$\sf (k + 1)x + (1-2ky + (4-5k) = 0$

The equations are of the form

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

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

where $\sf {a}_{1} = 2, {b}_{1} = -3, {c}_{1} = -7$

and $\sf {a}_{2} = (k + 1), {b}_{2} = (1 - 2k), {c}_{2} = (4 - 5k)$

The system equation is infinity having many solution.

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

$\dashrightarrow \sf \dfrac{2}{(k + 1)} = \dfrac{-3}{(1 - 2k)} = \dfrac{-7}{(4 - 5k)}$

$\dashrightarrow \sf \dfrac{2}{(k + 1)} = \dfrac{3}{(3k - 1)} = \dfrac{7}{(5k - 4)}$

$\dashrightarrow \sf \dfrac{2}{(k + 1)} = \dfrac{3}{(2k - 1)} \: \: and \: \: \dfrac{3}{(2k - 1)} = \dfrac{7}{(5k - 4)}$

$\sf \dashrightarrow 4k - 2 = 3k + 3 \: \: and \: \: 15k - 12 = 14k - 7$

$\sf \dashrightarrow k = 5 \: \: and \: \: k = 5$

Hence, value of k is 5.