Question

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

Answer 1

Answer:

The value of k is $k=-\frac{9}{4}$ and $k\neq-\frac{15}{7}$

Step-by-step explanation:

Given : Equation $3x-4y+7=0,kx+3y-5=0$

To find : The value of k for which the following system of equation has no solution.

Solution :

When the system of equation is in form $ax+by+c=0, dx+ey+f=0$ then the condition for no solutions is

$\frac{a}{d}=\frac{b}{e}\neq \frac{c}{f}$

Compare and substituting the values,

$\frac{3}{k}=\frac{-4}{3}\neq \frac{7}{-5}$

$\frac{3}{k}=\frac{-4}{3}\neq \frac{7}{-5}$

Taking 1,  $\frac{3}{k}=\frac{-4}{3}$

$3\times 3=-4\times k$

$9=-4k$

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

Taking 2,  $\frac{3}{k}\neq \frac{7}{-5}$

$3\times -5\neq7\times k$

$-15\neq7k$

$k\neq-\frac{15}{7}$

Therefore, The value of k is $k=-\frac{9}{4}$ and $k\neq-\frac{15}{7}$

Answer 2

Answer:

$value \: of \: k = \frac{-9}{4}\:and \: k ≠\frac{-15}{7}$

Step-by-step explanation:

$Compare\: given\: two\\equations \: 3x-4y+7=0,\:kx+3y-5=0\\with\:a_{1}x+b_{1}y+c_{1}=0,\\a_{2}x+b_{2}y+c_{2}=0,we\:get$

$a_{1}=3,\:b_{1}=-4,\:c_{1}=7;$

$a_{2}=k,\:b_{2}=3,\:c_{2}=-5;$

$\frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}}≠\frac{c_{1}}{c_{2}}$

( Given no solution )

$\frac{3}{k}=\frac{-4}{3}≠\frac{7}{-5}$

$i)\implies \frac{3}{k}=\frac{-4}{3}$

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

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

$ii))\frac{3}{k}≠\frac{7}{-5}$

$\implies \frac{-5}{7}\times 3 ≠ k$

$\implies \frac{-15}{7}≠k$

Therefore,

$value \: of \: k = \frac{-9}{4} \: and \: k ≠\frac{-15}{7}$

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