Question

find the value of k for which the system of equation has no unique solution 3x-4y-7=0 , kx +3y-5=0

Answer 1

Answer:

No Solution!

Value of k = $\frac{-9}{4}$

Step-by-step explanation:

Given Problem:

Find the value of k for which the system of equation has no unique solution 3x-4y-7=0 , kx +3y-5=0

To Find:

Value of k

No Solution:

$=>\frac{a^1}{a^2} = \frac{b^1}{b^2} \neq \frac{c^1}{c^2}$

$=> a^1 = 3^x ,b^1 = 4y ,c^1 = 7$

$=>3^x - 4y + 7 = 0$

$=>a^2 = k^x , b^2 = 3y , c^2 = 5$

$=>k^x - 3^y - 5 = 0$

$=>\frac{3}{k} =\frac{-4}{3}$

$=>9 = -4k$

$=>k = \frac{-9}{4}$

Answer 2

Answer:

$\large k=\dfrac{-9}{4}$

Step-by-step explanation:

Given :

3 x - 4 y - 7 = 0  .....( i )

k x + 3 y - 5 = 0 ....( ii )

Now know for no solution formula

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

Comparing from equation ( i ) we get

$\large a_1=3 ,\ b_1=-4 \ and \ c_1=-7$

Again from equation ( ii ) we get

$\large a_2=k ,\ b_2=3 \ and \ c_2=-5$

putting values in formula

$\large \dfrac{3}{k}=\dfrac{-4}{3}\neq \dfrac{-7}{-5} \\\\\\\large \dfrac{3}{k} =\dfrac{-4}{3} \\\\\\\large k=\dfrac{-9}{4}$

Thus we get $\large k=\dfrac{-9}{4}$