Question

23. Determine the value of k for which the given system of equations has no solution 3x – 4y + 7 = 0, kx + 3y – 5 = 0

Answer 1

Hey There,

Let us first understand the answer with General Form. 

Let the two equations be:

 $a_1x+b_1y+c_1 \, and \\ \\ a_2x+b_2y+c_2$


For the system of equations to have no solution, the condition is  


$\frac{a_1}{a_2}=\frac{b_1}{b_2}\neq\frac{c_1}{c_2}$ 

Here, your equations are: 

3x - 4y + 7 = 0   and
kx + 3y - 5 = 0 
Here we can easily see that

$\frac{b_1}{b_2}\neq\frac{c_1}{c_2}$



So, using $\frac{a_1}{a_2}=\frac{b_1}{b_2}$ we have 


$\frac{3}{k}=\frac{-4}{3} \\ \\ \implies k=\frac{3\times 3}{-4} \\ \\ \\ \implies \boxed{k=\frac{-9}{4}}$


Hope it helps
Purva
Brainly Community

Answer 2

$\sf \: a_1 x + b_1y+c_1= 0$


$\sf \: a_2 \: a=x+b_2 \: y+c_1=0$


$\color{blue} \sf \: No solution = > \frac{a_1}{a_2} = \frac{b_1}{b_2} \: \cancel{ = } \: \frac{c_1}{c_2}$


$\sf \: 3x - 4y + 7 = 0 \: \: ... \boxed1$


$\sf \: k_x + 3y - 5 = 0 \: \: ... \boxed2$


$\sf \: a_1 = 3$

$\sf \: b_1 = 4$

$\sf \: c_1 = + \: 7$


$\sf \: a_2 = k$


$\sf \: b_2 = 3$


$\sf \: c_2 = - 5$

$\sf \: \frac{3}{k} = \frac{ - 4}{3} \: \cancel{ = } \: \frac{7}{ - 5}$


$\sf \: \frac{3}{k} = \frac{ - 4}{3}$

$\sf \: \frac{k}{3} = \frac{ - 3}{4} = \boxed{ \sf \: k = \frac{ - 9}{4} }$