Question

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

Answer 1

Finding the value of $\text{k}$:

First, we compare the given equations with the general equations .

The general equations are ${\text{a}}_1{\text{x+b}}_1{\text{y+c}}_1\text{=0}$ and ${\text{a}}_2{\text{x+b}}_2{\text{y+c}}_2\text{=0}$

After comparing the equations we get ${\text{a}}_1{\text{=1, b}}_1{\text{=2, c}}_1\text{=-3}$ and ${\text{a}}_2{\text{=5, b}}_2{\text{=k, c}}_2\text{=7}$

It is given that the system has no solution it means the system is inconsistent have a set of parallel lines, So $\frac{{\text{a}}_1}{{\text{a}}_2}=\frac{{\text{b}}_1}{{\text{b}}_2}\ne \frac{{\text{c}}_1}{{\text{c}}_2}$

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

$$\begin{gathered} \Rightarrow \frac{1}{5}=\frac{2}{\text{k}}, \\ \Rightarrow \text{k=5×2=10} \end{gathered}$$

Hence, the value of k is $10$.