Question

find K, If the equation 2x+3y-2=0 2x+4y-k=0 and x-2y+3k=0 are consistent

Answer 1

16

Answer 2

We have three lines:

$$\begin{aligned} L_1:~& 2x+3y-2=0 \quad\Rightarrow\quad 2x+3y = 2 \quad(1) \\ L_2:~& 2x+4y-k=0 \quad\Rightarrow\quad 2x+4y = k \quad(2) \\ L_3:~& x-2y+3k=0 \quad\Rightarrow\quad x-2y = -3k \quad(3) \end{aligned}$$

For consistency, the three lines must have a common point.

1. Solve (1) and (2):

Subtract (1) from (2):

$$(2x+4y) - (2x+3y) = k - 2 \quad\Rightarrow\quad y = k - 2.$$

Substitute $y = k-2$ in (1):

$$2x + 3(k-2) = 2 \quad\Rightarrow\quad 2x = 2 - 3k +6 = 8 - 3k,$$

thus,

$$x = \frac{8-3k}{2}.$$

2. Substitute into (3):

$$\frac{8-3k}{2} - 2(k-2) + 3k = 0.$$

Multiply through by 2:

$$8-3k - 4(k-2) + 6k = 0.$$

Expand and simplify:

$$8-3k - 4k +8 +6k = 0 \quad\Rightarrow\quad 16 - k = 0.$$

Hence,

$$k = 16.$$