Question

Q1. Find the value of k so that the following system of equation has infinte solutions:
3x-y-5=0, 6x-2y+k=0
Q2. For which values of p, does the pair of equations given below has unique solution?
4x+py+8=0 and 2x+2y+2=0

Answer 1

Dear Student,


$$\begin{gathered} 1.3x-y-5=0and6x-2y+k=0 \\ valueofkforinfinitelymanysolution \\ \frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2} \\ \frac{3}{6}=\frac{-1}{-2}=\frac{-5}{k}=c\left(say\right) \\ c=\frac{3}{6}=\frac{1}{2} \\ \frac{-5}{k}=\frac{1}{2} \\ \Rightarrow k=-10 \\ . \\ 2.4x+py+8=0and2x+2y+2=0 \\ valueofpforuniquesolution \\ \frac{a_1}{a_2}\ne \frac{b_1}{b_2} \\ \frac{4}{2}\ne \frac{p}{2} \\ \frac{p}{2}\ne \frac{4}{2} \\ \Rightarrow \frac{p}{2}\ne 2 \\ \Rightarrow p\ne 4 \\ so,p<4\cup p>4 \end{gathered}$$
regards