Condition for no solution in system of linear equations

Question

For some constant real numbers $p$, $k$ and $a$, consider the following system of linear equations in $x$ and $y$: $$px - 4y = 2$$ $$3x + ky = a$$ A necessary condition for the system to have no solution for $(x, y)$, is

Accepted Answer

A. For a system of two linear equations $a_1x + b_1y = c_1$ and $a_2x + b_2y = c_2$ to have no solution, the lines must be parallel and distinct: $$\frac{a_1}{a_2} = \frac{b_1}{b_2} 
eq \frac{c_1}{c_2}$$ Substituting the given coefficients: $$\frac{p}{3} = \frac{-4}{k} 
eq \frac{2}{a}$$ From $\frac{-4}{k} 
eq \frac{2}{a}$, cross-multiplying gives $-4a 
eq 2k \implies 2a + k 
eq 0$. Thus, $2a + k 
eq 0$ is a necessary condition for the system to have no solution.

Answer

$2a + k 
eq 0$

Answer

$ap - 6 = 0$

Answer

$ap + 6 = 0$

Answer

$kp + 12 
eq 0$

Condition for no solution in system of linear equations

Explanation

Related Algebra questions

Practise this under exam conditions