Question

Determine the values of a and b so that the following system of linear equations has infinite number of
solutions;
(i). (2a-1) x +3y -5 = 0, 3x + (b-1)y -2 = 0 (ii) 2x - 3y = 7; (a + b)x ? (a + b-3)y = 4a +b

Answer 1

$$\begin{gathered} \mathrm{Dear}\mathrm{student}, \\ \left(\mathrm{i}\right) \\ (2\mathrm{a}-1)\mathrm{x}+3\mathrm{y}-5=0;3\mathrm{x}+(\mathrm{b}-1)\mathrm{y}-2=0 \\ {\mathrm{a}}_1=2\mathrm{a}-1;{\mathrm{b}}_1=3;{\mathrm{c}}_1=-5 \\ {\mathrm{a}}_2=3;{\mathrm{b}}_2=(\mathrm{b}-1);{\mathrm{c}}_2=-2 \\ \mathrm{For}\mathrm{infinite}\mathrm{number}\mathrm{of}\mathrm{solutions}: \\ \frac{{\mathrm{a}}_1}{{\mathrm{a}}_2}=\frac{{\mathrm{b}}_1}{{\mathrm{b}}_2}=\frac{{\mathrm{c}}_1}{{\mathrm{c}}_2} \\ \frac{2\mathrm{a}-1}{3}=\frac{3}{\mathrm{b}-1}=\frac{-5}{-2} \\ 1.\frac{2\mathrm{a}-1}{3}=\frac{-5}{-2}=\frac{5}{2} \\ \Rightarrow 2(2\mathrm{a}-1)=3\times 5 \\ \Rightarrow 4\mathrm{a}-2=15 \\ \Rightarrow \mathrm{a}=\frac{17}{4} \\ 2.\frac{3}{\mathrm{b}-1}=\frac{-5}{-2}=\frac{5}{2} \\ \Rightarrow 3\times 2=5(\mathrm{b}-1) \\ \Rightarrow 6=5\mathrm{b}-5 \\ \Rightarrow 11=5\mathrm{b} \\ \Rightarrow \mathrm{b}=\frac{11}{5} \\ \mathrm{For}\mathrm{remaining}\mathrm{queries}\mathrm{we}\mathrm{request}\mathrm{you}\mathrm{to}\mathrm{post}\mathrm{them}\mathrm{in}\mathrm{separate}\mathrm{threads}\mathrm{to}\mathrm{have}\mathrm{rapid}\mathrm{assistance}\mathrm{from}\mathrm{our}\mathrm{experts}. \\ \mathrm{Regards} \end{gathered}$$