Question

Find the value of k for which the following equations have no solution. 3x + y = 1 and (2k – 1 )x + (k – 1)y = 2k + 1

Answer 1

Answer:

k = 2, k ≠ -2

Step-by-step explanation:

for the equations to have no solutions, we must make them into the format of  $\frac{a1}{a2}$ = $\frac{b1}{b2}$ ≠ $\frac{c1}{c2}$

$\frac{3}{2k-1}$ = $\frac{1}{k-1}$ ≠ $\frac{1}{2k+1}$

3(k-1) = 1(2k-1)

3k - 3 = 2k - 1

k = 2

now we also need to find what k is not equal to

$\frac{b1}{b2}$ ≠ $\frac{c1}{c2}$

$\frac{1}{k-1}$ ≠ $\frac{1}{2k+1}$

2k+1 ≠ k - 1

2k - k ≠ -1 -1

k ≠ -2