Given,

\(\displaystyle{\left\lbrace\begin{array}{c} {4}{x}+{y}={3}\\{2}{x}-{3}={y}\end{array}\right.}\)

Step 2

Put x=−1 & y=7 in 4x+y=3, we get

4(−1)+7=3

\(\displaystyle\Rightarrow−{4}+{7}={3}\)

\(\displaystyle\Rightarrow{3}={3}\) which is true.

Now put x=−1 & y=7 in 2x−3=y, we get

2(−1)−3=7

\(\displaystyle\Rightarrow−{2}−{3}={7}\)

\(\displaystyle\Rightarrow−{5}={7}\) which is not true.

Hence the given ordered pair is not the solution of given system of equations.

\(\displaystyle{\left\lbrace\begin{array}{c} {4}{x}+{y}={3}\\{2}{x}-{3}={y}\end{array}\right.}\)

Step 2

Put x=−1 & y=7 in 4x+y=3, we get

4(−1)+7=3

\(\displaystyle\Rightarrow−{4}+{7}={3}\)

\(\displaystyle\Rightarrow{3}={3}\) which is true.

Now put x=−1 & y=7 in 2x−3=y, we get

2(−1)−3=7

\(\displaystyle\Rightarrow−{2}−{3}={7}\)

\(\displaystyle\Rightarrow−{5}={7}\) which is not true.

Hence the given ordered pair is not the solution of given system of equations.