What is the value of (t) for infinitely many solutions of (6x-ty=18) and (14x-21y=42)?

Correct answer: (t=9)

The direct answer is option C: \(t=9\). For two linear equations to have infinitely many solutions, both equations must represent the same line. For equations \(a_1x+b_1y=c_1\) and \(a_2x+b_2y=c_2\), the condition is \(\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}\), provided the coefficients are written in the same order. Here the ratios are \(\frac{6}{14}=\frac{3}{7}\) and \(\frac{18}{42}=\frac{3}{7}\). The middle ratio must also be \(\frac{3}{7}\): \(\frac{-t}{-21}=\frac{t}{21}=\frac{3}{7}\). Multiplying by 21 gives \(t=9\). Thus option C works. Option A, \(t=6\), gives \(t/21=2/7\), not \(3/7\), so the lines are not identical. Option B, \(t=8\), gives \(8/21\), again not the required ratio. Option C gives all three equal ratios, so the equations are multiples of one another and have infinitely many common points. Option D, \(t=12\), gives \(4/7\), which also fails. A useful check is to multiply the first equation by \(7/3\): it becomes \(14x-\frac{7t}{3}y=42\). To match the second equation, \(-\frac{7t}{3}=-21\), giving \(t=9\). Memory cue: infinitely many solutions means same line, so all three coefficient-to-constant ratios must match.