Step 1

Consider a system of equation

\(\displaystyle{x}−{y}={2}\)

\(\displaystyle{x}+{y}={4}\)

Step 2

The solution of the system is

\(\displaystyle{x}={3}\)

\(\displaystyle{y}={1}\)

Step 3

We can also find the solution of system algebraically

\(\displaystyle{x}−{y}={2}\ldots\ldots.{\left({1}\right)}\)

\(\displaystyle{x}+{y}={4}\ldots\ldots.{\left({2}\right)}\)

From equation (1)

\(\displaystyle{x}−{y}={2}\)

\(\displaystyle{y}={x}−{2}\)

Put the value of y in equation (2)

\(\displaystyle{x}+{y}={4}\)

\(\displaystyle{x}+{x}−{2}={4}\)

\(\displaystyle{2}{x}−{2}={4}\)

\(\displaystyle{2}{x}={4}+{2}\)

\(\displaystyle{2}{x}={6}\)

\(\displaystyle{x}={3}\)

Put the value of x in equation (1)

\(\displaystyle{x}−{y}={2}\)

\(\displaystyle{3}−{y}={2}\)

\(\displaystyle{y}={3}−{2}\)

\(\displaystyle{y}={1}\)

Hence the solution is

\(\displaystyle{x}={3}\)

\(\displaystyle{y}={1}\)

Consider a system of equation

\(\displaystyle{x}−{y}={2}\)

\(\displaystyle{x}+{y}={4}\)

Step 2

The solution of the system is

\(\displaystyle{x}={3}\)

\(\displaystyle{y}={1}\)

Step 3

We can also find the solution of system algebraically

\(\displaystyle{x}−{y}={2}\ldots\ldots.{\left({1}\right)}\)

\(\displaystyle{x}+{y}={4}\ldots\ldots.{\left({2}\right)}\)

From equation (1)

\(\displaystyle{x}−{y}={2}\)

\(\displaystyle{y}={x}−{2}\)

Put the value of y in equation (2)

\(\displaystyle{x}+{y}={4}\)

\(\displaystyle{x}+{x}−{2}={4}\)

\(\displaystyle{2}{x}−{2}={4}\)

\(\displaystyle{2}{x}={4}+{2}\)

\(\displaystyle{2}{x}={6}\)

\(\displaystyle{x}={3}\)

Put the value of x in equation (1)

\(\displaystyle{x}−{y}={2}\)

\(\displaystyle{3}−{y}={2}\)

\(\displaystyle{y}={3}−{2}\)

\(\displaystyle{y}={1}\)

Hence the solution is

\(\displaystyle{x}={3}\)

\(\displaystyle{y}={1}\)