Here's the question you clicked on:
urlilmami
method of substitution 3a=5-b and 3a=2-b
i think that's just a wrong question, dude
use the method of substitution to solve the system of liner equations. 3a=5-b and 3a=2-b
\[\large{3a=5-b\space ; 3a=2-b}\] \[\large{\color{red}{5-b=2-b}}\] \[\large{5-2=-b+b}\] \[\large{\color{blue}{3\ne 0}}\] hence no solutions
I thought you had to find the solution for one equation first
what you say ok \[\large{\color{blue}{b=5-3a}}\] \[\large{\color{red}{3a=2-b}}\] \[\large{\color{green}{3a=2-5+3a}}\] \[\large{\color{orange}{3a=-3+3a}}\] \[\large{\color{brown}{0\ne -3}}\] no soln in this also
lol @mathslover your posts really trigger epileptic shocks =))))