Question

Find the value of Alpha so that the following linear equation have no solution. (3 alpha +1)x+3y-2=0 and alpha +1 X +Alpha - 2y-5

Answer 1

Answer:

The given system of linear equation is:

1.  (3 A +1)x +3 y -2=0

2. (A+1)x+A-2 y-5=0

(A+1)x-2 y+A-5=0, Here  A= alpha

The two system of linear equation

1. a x + b y=c

2. p x + q y =r

have no solution,  

$\frac{a}{p}=\frac{b}{q}\neq\frac{c}{r}$

$\frac{(3 A+1)}{A+1}=\frac{3}{-2}=\frac{-2}{A-5}\\\\ -6A-2=3A+3 \\\\ 9A=-5\\\\A=\frac{-5}{9}$

or

4≠3 A -15

3A≠4+15

3A≠19

$A\neq\frac{19}{3}$

$A=\frac{-5}{9}$

Answer 2

a=qb=rc

\begin{gathered}\frac{(3 A+1)}{A+1}=\frac{3}{-2}=\frac{-2}{A-5}\\\\ -6A-2=3A+3 \\\\ 9A=-5\\\\A=\frac{-5}{9}\end{gathered}A+1(3A+1)=−23=A−5−2−6A−2=3A+39A=−5A=9−5

or

4≠3 A -15

3A≠4+15

3A≠19

A\neq\frac{19}{3}A=319

A=\frac{-5}{9}A=9−5