The value of k for which the system of equations x + 2y − 3 = 0
Question:

The value of $k$ for which the system of equations $x+2 y-3=0$ and $5 x+k y+7=0$ has no solution, is

(a) 10

(b) 6

(c) 3

(d) 1

Solution:

The given system of equations are

$x+2 y-3=0$

 

$5 x+k y+7=0$

For the equations to have no solutions,

$\frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}} \neq \frac{c_{1}}{c_{2}}$

$\frac{1}{5}=\frac{2}{k}=\frac{-3}{7}$

If we take

$\frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}}$

$\frac{1}{5}=\frac{2}{k}$

$k=10$

Therefore the value of k is10.

Hence, correct choice is $a$.

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