Question

If the system of equation $x+2y-3z=1$, $(p+2)z=3$, $(2p+1)y+z=2$ has infinite number of solutions, then the value of p is not equal to.

• A. $-2$
• B. $-\frac{1}{2}$
• C. $0$
• D. $2$

Answer 1

The correct option is A $-2$

$x+2y-3z=1$

$(p+2)z=3$

$(2p+1)y+z=2$

let $p=5$, $s\in R/\{-2,1/2\}$

$\therefore z=\frac{3}{s+2}$

$\Rightarrow y=(2-\frac{3}{s+2})\frac{1}{(2s+1)}\Rightarrow \frac{2s+1}{(s+2)(2s+1)}-\frac{1}{s+2}$

$[As2s+1\ne 0]$

$\therefore x=3z+1-2y$

$=\frac{9}{s+2}+1-\frac{2}{s+2}=\frac{7}{s+2}+1$

$\therefore$ solutions $(x,y,z)=(\frac{7}{s+2}+1,\frac{1}{s+2},\frac{3}{s+2})$

is an infinite set,

$\therefore p$ cannot be equal to $-2$ or $1/2$