Question

Find the value of k for which the given system of equations has infinitely many solutions:
x+(k+1)y=5 ---(1)
(k+1)x+9y+(1-8k)=0

Answer 1

Step-by-step explanation:

Given:

$x+(k+1)y=5...eq1\\(k+1)x+9y+(1-8k)=0...eq2$

To find: Find the value of k for which the given system of equations has infinitely many solutions.

Solution:

Tip: If two lines have infinitely many solutions then ratios of coefficients of x,y and constant term are same.

if lines are

$a_1x + b_1y + c_1 = 0 \\ a_2x + b_2y + c_2 = 0$

then

$\boxed{\bold{\blue{ \frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}}}}$

Step 1: Write coefficients

$a_1 =1 \\ a_2 = k + 1 \\ b_1 = k + 1 \\ b_2 = 9 \\ c_1 = - 5 \\ c_2 = 1 - 8k$

Step 2:Put the coefficients into the condition

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

Step 3: Take first two terms

$\frac{1}{k + 1} = \frac{k + 1}{9} \\ \\ ( {k + 1)}^{2} = 9 \\ \\ k + 1 = ±3 \\ \\ k = 2 \\ \\ or \\ \\ k = - 4$

Step 4: Take last two ratios

$\frac{k + 1}{9} = \frac{ - 5}{1 - 8k} \\ \\ (k + 1)(1 - 8k) = - 45 \\ \\ k - 8 {k}^{2} + 1 - 8k + 45 = 0 \\ \\ - 8 {k}^{2} - 7k + 46 = 0 \\ \\ or \\ \\ 8 {k}^{2} + 7k - 46 = 0$

solution of this quadratic equation gives k=2 and k=-23/8

Step 5: Take first and last term

$\frac{1}{k + 1} = \frac{ - 5}{1 - 8k} \\ \\ 1 - 8k = - 5k - 5 \\ \\ - 8k + 5k = - 5 - 1 \\ \\ - 3k = - 6 \\ \\ k = 2$

Step 6: Find value of k

(k=2 or k=-4) and (k=2 or k=-23/8) and (k=2)

Thus,for k=2 lines have infinite many solutions

Final answer:

For lines have infinite many solutions,value of k must be

$\bold{\red{k = 2}}$

Hope it helps you.

Remark: One can verify by putting the value of k in condition of coefficients.

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