For what value of k will the following pair of linear equations have infinite solutions?
\(k x+3 y-(k-3)=0\)
\(12 x+k y-k=0\)
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 6 only
Two linear equations \(a_1x+b_1y+c_1=0\) and \(a_2x+b_2y+c_2=0\) have infinitely many solutions when they are the same line, that is, when \(\displaystyle \frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}\). Here \(a_1=k,\ b_1=3,\ c_1=-(k-3)\) and \(a_2=12,\ b_2=k,\ c_2=-k\).
From \(\displaystyle \frac{k}{12}=\frac{3}{k}\): \(k^2=36\), so \(k=6\) or \(k=-6\)
From \(\displaystyle \frac{3}{k}=\frac{k-3}{k}\): \(k-3=3\), so \(k=6\)
Only \(k=6\) satisfies both
For \(k=0\) the equations become \(3y+3=0\) and \(12x=0\), which meet at the single point (0, −1), so 0 does not work.