Question

For what value of k the following system of linear equations has no solution (k+1)x+y=1;3x+(k-1)y=2k+5

Answer 1

Equation 1:

(k+1)x + y = 1.

(k+1)x + y - 1 = 0.

In Equation 1:

$a_{1} = (k + 1) \\ b_{1} = 1 \\ c_{1} = - 1$

Equation 2:

3x + (k - 1)y = 2k + 5.

3x + (k - 1)y - 2k - 5 = 0.

In Equation 2:

$a_{2} = 3 \\ b_{2} = (k - 1) \\ c_{2} = -2k - 5$

For 2 equations to have no common solution:

$\frac{a_{1} }{a_{2}} = \frac{b_{1}}{b_{2}} \ne \frac{c_{1}}{c_{2}}$

Substituting the values in the condition to be satisfied:

$\frac{k + 1}{3} = \frac{1}{k - 1} \\ \\ {k}^{2} - 1 = 3 \\ {k}^{2} - 4 = 0 \\ {k}^{2} = 4 \\ k = \sqrt{4} \\ k = 2$

Therefore, the value of k is 2.

Answer 2

Solution:

Given system of equations are,

(k+1)x+y=1

and, 3x+(k-1)y=2k+5

Here,

a1:k+1, a2:3, b1:1, b2:k-1, c1:1 and c2:2k+5

Now,

According to the given condition,the equations don't have a solution.

Thus,the system of equation is inconsistent and the lines representing these equations are parallel.

For parallel lines (no solution),

a1/a2 = b1/b2 ≠ c1/c2

=> (k + 1)/3 = 1/(k - 1) ≠ 1/(2k+5)

=>(k+1)/3=1/(k-1)

=>k² - 1=3

=>k²=4

=>k=±√4

=>k= ±2

For k=±2,the given equations have no solutions