Which relation is correct for the equations (14x+21y=98) and (2x+3y=17)?

Correct answer: \(\frac{14}{2}=\frac{21}{3} \ne \frac{98}{17}\)

Here, \(a_1=14, b_1=21, c_1=98\) and \(a_2=2, b_2=3, c_2=17\). We get \(\frac{a_1}{a_2}=\frac{14}{2}=7\) and \(\frac{b_1}{b_2}=\frac{21}{3}=7\), but \(\frac{c_1}{c_2}=\frac{98}{17}\) is not 7. Hence, \(\frac{a_1}{a_2}=\frac{b_1}{b_2}\ne\frac{c_1}{c_2}\), so the two lines are parallel and the pair has no solution. Option B is incorrect because \(\frac{98}{17}\ne7\). Exam tip: compare all three ratios; equality of only the first two ratios means no solution.