Question

# Find k for which the system 2x + 3y − 5 = 0 and 4x + ky − 10 = 0 has an infinite number of solutions.

Open in App
Solution

## The given system is $2x+3y-5=0.....\left(i\right)\phantom{\rule{0ex}{0ex}}4x+ky-10=0.....\left(ii\right)$ Here, ${a}_{1}=2,{b}_{1}=3,{c}_{1}=-5.{a}_{2}=4,{b}_{2}=k\mathrm{and}{c}_{2}=-10$. For the system, to have an infinite number of solutions, we must have $\frac{{a}_{1}}{{a}_{2}}=\frac{{b}_{1}}{{b}_{2}}=\frac{{c}_{1}}{{c}_{2}}\phantom{\rule{0ex}{0ex}}⇒\frac{2}{4}=\frac{3}{k}=\frac{-5}{-10}\phantom{\rule{0ex}{0ex}}⇒\frac{1}{2}=\frac{3}{k}=\frac{1}{2}\phantom{\rule{0ex}{0ex}}⇒k=6$ Hence, k = 6.

Suggest Corrections
0
Related Videos
Graphical Solution
MATHEMATICS
Watch in App