91Ó°ÊÓ

We are given a system of equations,

$\left\{\begin{array}{l}3x+y=0\\ x+2y=-5\end{array}\right.$

We need to determine whether the point $(1,-3)$is a solution to the system or not.

Substitute $1$for $x$and $-3$for $y$in the equation $3x+y=0$we get,

$\begin{array}{rcl}3×1+(-3)& =& 0\\ 3-3& =& 0\\ 0& =& 0\end{array}$

This is a true statement. So, the point$(1,-3)$satisfies the first equation.

Substitute $1$for $x$and $-3$for $y$in the second equation.

$x+2y=-5$

$\begin{array}{rcl}1+2×(-3)& =& -5\\ 1-6& =& -5\\ -5& =& -5\end{array}$

This is a true statement, so $(1,-3)$satisfies the second equation.

As $(1,-3)$satisfies both the equation so it is a solution of the system$\left\{\begin{array}{l}3x+y=0\\ x+2y=-5\end{array}\right.$

For the given system$\left\{\begin{array}{l}3x+y=0\\ x+2y=-5\end{array}\right.$we need to determine whether the point$(0,0)$is a solution of the system or not.

Substitute $0$for $x$and $0$for $y$in the equation $3x+y=0$we get,

localid="1644572937857" $\begin{array}{rcl}3×0+0& =& 0\\ 0+0& =& 0\\ 0& =& 0\end{array}$

As the statement is true so the point$(0,0)$satisfies the first equation.

Substitute $0$for $x$and $0$for $y$in the equation $x+2y=-5$we get,

localid="1644572948618" $\begin{array}{rcl}0+2×0& =& -5\\ 0+0& =& -5\\ 0& =& -5\end{array}$

As the statement is false, so $(0,0)$does not satisfy the second equation.

Thus, the point$\left(0,0\right)$is not the solution of the system.