For what value of k , the system of equations kx+4y-k+4=0 and 16x+ky=…

The correct answer is Option (3) → 8

Two equations \(a_1x+b_1y=c_1\) and \(a_2x+b_2y=c_2\) have infinitely many solutions when the two lines coincide, that is, when \(\displaystyle\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}\).

The two equations (written in this form):

• \(kx+4y=k-4\) and \(16x+ky=k\)

From the \(x\) and \(y\) coefficients (equal ratios):

• \(\displaystyle\frac{k}{16}=\frac{4}{k}\) gives \(k^2=64\)

• So \(k=8\) or \(k=-8\)        (1)

From the \(y\) coefficients and the constants (equal ratios):

• \(\displaystyle\frac{4}{k}=\frac{k-4}{k}\) gives \(4=k-4\)

• So \(k=8\)        (2)

The value of \(k\):

• Only \(k=8\) satisfies both (1) and (2)

• Then every ratio is \(\frac{1}{2}\)