Question

determine the value of k for which following pair of linear equation has unique solution(k+1)x+5y=4 3x+5y=7

Answer 1

$$\begin{gathered} \mathrm{The}\mathrm{given}\mathrm{linear}\mathrm{equations}\mathrm{are}: \\ \left(\mathrm{k}+1\right)\mathrm{x}+5\mathrm{y}-4=0 \\ 3\mathrm{x}+5\mathrm{y}-7=0 \\ \mathrm{Comparing}\mathrm{with}\mathrm{standard}\mathrm{equations},\mathrm{that}\mathrm{are} \\ {\mathrm{a}}_1\mathrm{x}+{\mathrm{b}}_1\mathrm{y}+{\mathrm{c}}_1=0\mathrm{and}{\mathrm{a}}_2\mathrm{x}+{\mathrm{b}}_2\mathrm{y}+{\mathrm{c}}_2=0,\mathrm{we}\mathrm{get} \\ {\mathrm{a}}_1=\left(\mathrm{k}+1\right);{\mathrm{b}}_1=5;{\mathrm{c}}_1=-4 \\ {\mathrm{a}}_2=3;{\mathrm{b}}_2=5;{\mathrm{c}}_2=-7 \\ \mathrm{For}\mathrm{unique}\mathrm{solution},\mathrm{we}\mathrm{have} \\ \frac{{\mathrm{a}}_1}{{\mathrm{a}}_2}\ne \frac{{\mathrm{b}}_1}{{\mathrm{b}}_2} \\ \Rightarrow \frac{\left(\mathrm{k}+1\right)}{3}\ne \frac{5}{5} \\ \Rightarrow \frac{\mathrm{k}+1}{3}\ne 1 \\ \Rightarrow \mathrm{k}+1\ne 3 \\ \Rightarrow \mathrm{k}\ne 2 \\ \mathrm{So},\mathrm{the}\mathrm{given}\mathrm{system}\mathrm{of}\mathrm{equation}\mathrm{has}\mathrm{unique}\mathrm{solution}\mathrm{for}\mathrm{k}\ne 2. \end{gathered}$$