Question

The number of solutions of the equation
$3x+3y-z=5,x+y+z=3,2x+2y-z=3$

• A. infinite
• B. $1$
• C. $0$
• D. two

Answer 1

Answer: (A)

infinite

$3x+3y-z=5$

$x+y+z=3$

$2x+2y-z=3$

Let $x+y=a$

The given system becomes

$3a-z=5$ …….$\left(1\right)$

$a+z=3$ ……….$\left(2\right)$

$2a-z=3$ ……..$\left(3\right)$

Adding $\left(1\right)$ and $\left(2\right)$ gives $4a=8$ $\Rightarrow a=2$

From $\left(2\right)$, we have $2+z=3$ $\Rightarrow z=1$

From $\left(3\right)$, we have $2a=3+z=3+1=4$

$\Rightarrow a=2$

So, $a=2$ and $z=1$

i.e., $z+y=2$ and $z=1$

If $x=k$, they $y=2-k$ and $z=1$

i.e, thus are infinite solutions possible for the given system of equations.