Question

Find the value of k for which the following system of linear equations has infinite number of solutions:
i) x + ( k + 1)y = 5, ( k +1 )x + 9y = ( 8k – 1 ) ii) 2x + 3y = 7, ( k – 1 )x + ( k + 2 )y = 3k

Answer 1

$$\begin{gathered} \left(1\right)\mathrm{x}+\left(\mathrm{k}+1\right)\mathrm{y}-5=0 \\ \left(\mathrm{k}+1\right)\mathrm{x}+9\mathrm{y}-\left(8\mathrm{k}-1\right)=0 \\ \mathrm{For}\mathrm{infinite}\mathrm{solutions}, \\ \frac{1}{\mathrm{k}+1}=\frac{\mathrm{k}+1}{9}=\frac{5}{8\mathrm{k}-1} \\ \frac{1}{\mathrm{k}+1}=\frac{\mathrm{k}+1}{9}\mathrm{and}\frac{\mathrm{k}+1}{9}=\frac{5}{8\mathrm{k}-1} \\ {\left(\mathrm{k}+1\right)}^2=9\mathrm{and}\left(\mathrm{k}+1\right)\left(8\mathrm{k}-1\right)=45 \\ {\mathrm{k}}^2+2\mathrm{k}+1-9=0\mathrm{and}8{\mathrm{k}}^2+8\mathrm{k}-\mathrm{k}-1-45=0 \\ {\mathrm{k}}^2+2\mathrm{k}-8=0\mathrm{and}8{\mathrm{k}}^2+7\mathrm{k}-46=0 \\ {\mathrm{k}}^2+4\mathrm{k}-2\mathrm{k}-8=0\mathrm{and}8{\mathrm{k}}^2-16\mathrm{k}+23\mathrm{k}-46=0 \\ \mathrm{k}\left(\mathrm{k}+4\right)-2\left(\mathrm{k}+4\right)=0\mathrm{and}8\mathrm{k}\left(\mathrm{k}-2\right)+23\left(\mathrm{k}-2\right)=0 \\ \left(\mathrm{k}-2\right)\left(\mathrm{k}+4\right)=0\mathrm{and}\left(8\mathrm{k}+23\right)\left(\mathrm{k}-2\right)=0 \\ \mathrm{Common}\mathrm{value}\mathrm{is}\mathrm{k}-2=0 \\ \mathrm{So},\mathrm{k}=2\mathrm{ANS}... \\ \left(2\right)2\mathrm{x}+3\mathrm{y}=7 \\ \left(\mathrm{k}-1\right)\mathrm{x}+\left(\mathrm{k}+2\right)\mathrm{y}=3\mathrm{k} \\ \mathrm{For}\mathrm{infinite}\mathrm{solutions}, \\ \frac{2}{\mathrm{k}-1}=\frac{3}{\mathrm{k}+2}=\frac{7}{3\mathrm{k}} \\ 2\mathrm{k}+4=3\mathrm{k}-3\mathrm{and}9\mathrm{k}=7\mathrm{k}+14 \\ \mathrm{k}=7\mathrm{and}\mathrm{k}=7 \\ \mathrm{So},\mathrm{k}=7\mathrm{ANS}... \end{gathered}$$