Question

For what value of k will the following pair of linear equation will have no solutions: Kx+3y+(2-k)=0 12x+ky-k=0
please answer fast

Answer 1

Answer:

$k_{12}$ = ± 6

Step-by-step explanation:

Two linear equations

y = ($m_{1}$)x + $b_{1}$ and

y = ($m_{2}$)x + $b_{2}$

are parallel if

$m_{1}$ = $m_{2}$ and $b_{1}$ ≠ $b_{2}$

Rewrite standard form of equation in slope-intercept form

y = (- $\frac{k}{3}$)x + $\frac{k-2}{3}$

y = (- $\frac{12}{k}$)x + $\frac{k}{k}$

$\frac{k}{3}$ = $\frac{12}{k}$ after cross multiplication k² = 36 ===> $k_{12}$ = ± 6

and $\frac{k-2}{3}$ ≠ 1 ====> k ≠ 5