Question

The value of k, for which the system of equations x + (k + 1)y = 5 and (k + 1)x + 9y = 8k – 1 has infinitely many solutions is

Answer 1

Answer:

For infinite many solution,

$\frac{1}{k + 1} = \frac{k + 1}{9} = \frac{5}{8k - 1}$

solving 3 pair,

$\frac{1}{k + 1} = \frac{k + 1}{9} \: \: i.e \: \: (k + 1) {}^{2} = 9 \: \: so \: \: k + 1 = 3 \: \: or \: - 3 \: \: \: then \: \: \: k = 2 \: \: or \: - 4$

now,

$\frac{k + 1}{9} = \frac{5}{8k - 1} \: \: i.e \: \: 8 {k}^{2} + 7k - 1 = 45 \: \: i.e \: \: 8 {k}^{2} + 7k - 46 = 0 \: \: i.e \: \: \: 8 {k}^{2} + 23k - 16k - 46 = 0 \: \: \: i.e \: \: k(8k + 23) - 2(8k + 23) = 0 \: \: \: \: so \: \: \: (8k + 23)(k - 2) = 0 \: \: \: then \: \: \: \: k = 2 \: \: or \: \: - \frac{23}{8}$

now,

$\frac{1}{k + 1} = \frac{5}{8k - 1} \: \: \: i.e \: \: \: 8k - 1 = 5k + 5 \: \: \: i.e \: \: 3k = 6 \: \: then \: \: \: k = 2$

so common solution is X = 2 Ans.

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Answer 2

Answer:

Step-by-step explanation:

For infinite many solution,

solving 3 pair,

now,

now,

K=2