Question

for what value of x will the following pair of linear equation have infinatily many solution kx + 3y -(k-3) + 0 , 12x + ky - k = 0

Answer 1

Answer:

If equation has infinitely many solution then it will satisfy the condition

$\frac{a1}{a2} = \frac{b1}{b2} = \frac{c1}{c2} \\ then \: \: in \: the \: given \: equation \\ a1 = k \: \: b1 = 3 \: \: c1 = - (k - 3) \\ a2 = 12 \: \: b2 = k \: \: c2 = - k \\ so \frac{k}{12} = \frac{3}{k} = \frac{ - (k - 3)}{ - k} \\ as \: all \: are \: equal \: above \: so \: we \: can \: put \\ \frac{k}{12} = \frac{3}{k} \: \: \: and \: \: \: \frac{3}{k} = \frac{k - 3}{k} \\ cross multiply \: both \: and \: in \: second \: case \: base \: is \: same \: as \: (k) \: so \: eliminate \: k\\ {k}^{2} = 12 \times 3 \: \: \: \: \: \: 3 = k - 3 \\ {k}^{2} = 36 \: \: \: \: \: \: \: \: k = 3 + 3 \\ k = \sqrt{36} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: k = 6 \\ there fore \: k = + 6 \: and \: - 6$