Question

For what value of $k$ , will the system of equations $x+2y=5,3x+ky+15=0$has

a) unique solution

b) no solution

Answer 1

Step 1:Finding the values of ${\mathit{a}}_{\mathbf{1}}\mathbf{,}{\mathit{a}}_{\mathbf{2}}\mathbf{,}{\mathit{a}}_{\mathbf{3}}\mathbf{,}{\mathit{b}}_{\mathbf{1}}\mathbf{,}{\mathit{b}}_{\mathbf{2}\mathbf{}}\mathit{a}\mathit{n}\mathit{d}\mathbf{}{\mathit{b}}_{\mathbf{3}}$from the equations

Given,

$$\begin{gathered} x+2y=5 \\ \Rightarrow x+2y-5=0....................................\left(1\right) \\ \Rightarrow 3x+ky+15=0.................................\left(2\right) \end{gathered}$$

Equation (1) and (2) are in standard form

$$\begin{gathered} here{a}_1=1b_1=2c_1=-5 \\ {a}_2=3b_2=k{c}_2=+15 \end{gathered}$$

Step 2:If the pair of equation has unique solution

$\frac{a_1}{a_2}\#\frac{b_1}{b_2}\Rightarrow \frac{1}{3}\#\frac{2}{k}\Rightarrow k\#6$

Therefore the pair of equations has unique solution for all other values except $k=6$

Step 3:If the pair of equations has no solution

$$\begin{gathered} \frac{a1}{a2}=\frac{b1}{b2}\#\frac{c1}{c2} \\ \Rightarrow \frac{1}{3}=\frac{2}{k}\#\frac{-5}{15} \\ Taking\frac{1}{3}=\frac{2}{k}\Rightarrow k=6 \end{gathered}$$

Therefore when $k=6$ the equations has no solution