Question

The pair of linear equation 3x+5y=3,6x+ky=8 do not have solution if k=?

Answer 1

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Answer 2

The value of k is 10 and $k\neq\frac{40}{3}$

Step-by-step explanation:

Given : The pair of linear equation 3x+5y=3, 6x+ky=8.

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  

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

On comparing, a=3,b=5,c=-3,d=6,e=k and f=-8

Substituting the values,

$\frac{3}{6}=\frac{5}{k}\neq \frac{-3}{-8}$

$\frac{1}{2}=\frac{5}{k}\neq \frac{3}{8}$

Taking 1, $\frac{1}{2}=\frac{5}{k}$

$1\times k=5\times 2$

$k=10$

Taking 2, $\frac{5}{k}\neq \frac{3}{8}$

$5\times 8\neq 3\times k$

$40\neq 3k$

$k\neq\frac{40}{3}$

Therefore, the value of k is 10 and $k\neq\frac{40}{3}$

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