Find the value of k for which the following systems of equations have infinitely many solutions:

x + (k + 1)y = 5

(k + 1)x + 9y = (8k - 1)

Option 3 : 2

__Concept:__

Let the two equations be:

a1x + b1y + c1 = 0

a2x + b2y + c2 = 0

Then,

- For
**unique solution**, \(\frac{a_{1}}{a_{2}}≠ \frac{b_{1}}{b_{2}}\) - For
**infinitely many solutions**, \(\frac{a_{1}}{a_{2}}= \frac{b_{1}}{b_{2}}=\frac{c_{1}}{c_{2}}\) -
For

**no solution**, \(\frac{a_{1}}{a_{2}}= \frac{b_{1}}{b_{2}}\neq\frac{c_{1}}{c_{2}}\)

__Formula used:__

For any quadratic equation, ax^{2} + bx + c = 0, the **quadratic formula** is:

\(x = {-b \pm \sqrt{b^2-4ac} \over 2a}\)

__Calculation:__

Given equations are:

x + (k + 1)y = 5

(k + 1)x + 9y = (8k - 1)

On comparing the equations with ax + by + c = 0, we get

a1 = 1, b1 = (k + 1), c1 = -5

a2 = (k + 1), b2 = 9, c2 = -(8k - 1)

So, for infinitely many solutions,

\(\frac{a_{1}}{a_{2}}= \frac{b_{1}}{b_{2}}=\frac{c_{1}}{c_{2}}\)

\(⇒ \frac{1}{(k+1)}= \frac{(k+1)}{9}=\frac{5}{(8k-1)}\)

⇒ (k + 1)^{2} = 9

⇒ k^{2} + 1 + 2k - 9 = 0

⇒ k2 + 2k - 8 = 0

Using the quadratic formula, we get

**⇒ k = 2 or k = -4 **----(1)

Also, (k + 1)(8k - 1) = 45

⇒ 8k^{2} - k + 8k - 1 = 45

⇒ 8k2 + 7k - 1 = 45

Using the quadratic formula, we get

**⇒ k = 2 or k = -23/8 **----(2)

From equation (1) & (2), we get,

**⇒ k = 2**