The solution set of x + 2y - 3 = 0 and 5x + k | vedclass

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(A) For a system of two linear equations $a_1x + b_1y + c_1 = 0$ and $a_2x + b_2y + c_2 = 0$ to have an empty set (no solution),the lines must be para

Vedclass · Accepted Answer

$10$

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(A) For a system of two linear equations $a_1x + b_1y + c_1 = 0$ and $a_2x + b_2y + c_2 = 0$ to have an empty set (no solution),the lines must be parallel.This condition is given by $\frac{a_1}{a_2} = \frac{b_1}{b_2} 
eq \frac{c_1}{c_2}$.Given equations are $x + 2y - 3 = 0$ and $5x + ky + 7 = 0$.Here,$a_1 = 1, b_1 = 2, c_1 = -3$ and $a_2 = 5, b_2 = k, c_2 = 7$.Applying the condition for parallel lines: $\frac{1}{5} = \frac{2}{k}$.Cross-multiplying gives $k = 5 	imes 2 = 10$.Checking the inequality condition: $\frac{c_1}{c_2} = \frac{-3}{7}$. Since $\frac{1}{5} 
eq \frac{-3}{7}$,the condition holds.Therefore,the value of $k$ is $10$.

The solution set of $x + 2y - 3 = 0$ and $5x + ky + 7 = 0$ is an empty set. The value of $k$ is .......

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