Question

the value of k for which the system of equation x+2y-3=0 and 5x+ky+7=0 has no solution is

Answer 1

The condition for no solution is...
a1/a2 = b1/b2 not equals to c1/c2,
so,
a1 = 1 , a2= 5 , b1= 2 , b2 = k
1/5 = 2/k,
now,
k = 10 (cross multiplication)


Hope it helps...

Answer 2

Answer:

The value of k is $k=10$ and $k\neq-\frac{14}{3}$  

Step-by-step explanation:

Given : Equation $x+2y-3=0,\ 5x+ky+7=0$

To find : The value of k for which the 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  

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

Comparing with equations,

a=1,b=2,c=-3,d=5,e=k,f=7

Substituting the values,

$\frac{1}{5}=\frac{2}{k}\neq \frac{-3}{7}$

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

$k=2\times 5$

$k=10$

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

$2\times 7\neq-3\times k$

$14\neq-3k$

$k\neq-\frac{14}{3}$

Therefore, The value of k is $k=10$ and $k\neq-\frac{14}{3}$