Question

3x+2y=1 (3k-1) X+(2k-1)y=2k+1

Answer 1

Answer:

The given system of equations have solution for any real values of k

Step-by-step explanation:

Concept used:

If the equations

$a_1x+b_1y=c_1$

$a_2x+b_2y=c_2$ have solution, then $a_1b_2-a_2b_1\neq0$

Given equations are

$3x + 2y=1$

$(3k - 1) X + (2k - 1) y = 2k + 1$

Now,

$a_1b_2-a_2b_1$

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

$=6k-3-6k+2$

$=-1\neq0$

Therefore the given system of equations have solution for any real values of k

Answer 2

Answer:

x = 2k + 3

y = -3k - 4

Step-by-step explanation:

3x+2y=1     - eq 1

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

3kx - x + 2ky - y = 2k + 1

=> k(3x + 2y) -(x + y) = 2k + 1

putting 3x+2y=1     from eq 1

=> k - (x + y) = 2k + 1

=> -(x + y) = k + 1  

multiplying by 2

-2x - 2y = 2k + 2  

Adding with eq 1

x = 1 + 2k + 2

=> x = 2k + 3

putting in eq 1

3(2k + 3) + 2y = 1

=> 6k + 9 + 2y = 1

=> 2y = -6k - 8

=> y = -3k - 4