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What value of (k) makes (kx+y=3) and (4x+2y=8) have no solution?

Answer and explanation

Correct answer: 2

Two linear equations have no solution when \(\frac{a_1}{a_2}=\frac{b_1}{b_2}\neq\frac{c_1}{c_2}\). Here, \(\frac{k}{4}=\frac{1}{2}\), giving \(k=2\). The first equation then becomes \(2x+y=3\); multiplying it by 2 gives \(4x+2y=6\), whereas the second equation is \(4x+2y=8\). Thus, the two lines have the same slope but different intercepts, so they do not intersect and have no solution. Exam tip: equate the ratios of the coefficients of \(x\) and \(y\), then verify that the constants have a different ratio.

Related tags

Linear EquationsSolvability ConditionsNo SolutionParameterParallel LinesConditions For SolvabilityPair Of Linear Equations In Two VariablesMathematics

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