Question

Find the value of k so that the graph of the following system of equations has no solution. 3x-2y-12=0 ; kx+6y-10=0

Answer 1

k = -9

Explanation:

The equations are of the form

ax+by+c=0

Two equations have no solution when

`(a_1)/(a_2)=(b_1)/(b_2)\\eq (c_1)/(c_2)`

3x-2y-12=0 ; kx+6y-10=0

`(3)/(k)=(-2)/(6)\\\Rightarrow k=3* (6)/(-2)\\\Rightarrow k=-9`

`(c_1)/(c_2)=(-12)/(-10)=1.2`

So,

`(3)/(-9)=(-2)/(6)\\eq 1.2`

Hence, k = -9