Question

Find the value of k for which the following system of equations has no solution:
3x-4y+7=0kx+3y-5=0

Answer 1

k = -9/4

Step-by-step explanation:

Given:  

3x -4y +7 = 0

kx +3y -5 = 0

The system of equations has no solution.

a1 = 3, b1 = -4, c1 = 7

a2 = k, b2 = 3, c2 = -5

 So a1/a2 = b1/b2 but not equal to c1/c2

3/k = -4/3

9 = -4k

-9/4 = k

Therefore k = -9/4

Answer 2

$\Large{\underline{\underline{\bf{Solution :}}}}$

As, it is given equations has no solution.

We know the case of no solution.

$\Large{\star{\boxed{\rm{\frac{a_{1}} {a_{2}} = \frac{b_1}{b_2} ≠ \frac{c_1}{c_2}}}}}$

Where,

a1 = 3, a2 = k

b1 = -4, b2 = 3

c1 = 7, c2 = -5

________________[Put Values]

$\sf{→\frac{3}{k} = \frac{-4}{3} ≠ \frac{7}{-5}} \\ \\ \sf{→\frac{3}{k} = \frac{-4}{3}} \\ \\ \sf{→3 \times 3 = k \times -4} \\ \\ \sf{9 = -4k} \\ \\ \sf{→k = \frac{-9}{4}} \\ \\ \Large{\star{\boxed{\sf{→k = \frac{-9}{4}}}}}$