Very easy! Just solve

$x+y$ = $3$ ........................[1]

$x-y$ = $1$ .........................[2]

from [1]+[2],we get, $2x$ = $4$ , or, $x$ = $2$

so, $y$ = $3-2$ = $1$

now, $xy$ = $2 \times 1$ = $2$

Let x+y=3 be equation 1.

Let x-y=1 be equation 2.

Subtract them :

x+y=3

x-y=1

2y=2

And solve:

2y=2

y=1

Now substitute y=1 into either equation 1 or 2 to solve for x.

x+y=3

x+1=3

x=3-1

x=2

Now use the values to find xy:

xy

=2(1)

=2

$x + y = 3$ ..................(1)

$x - y = 1$ ...................(2)

From equation (1)

$x + y = 3$

$⇒ x = 3 - y$ ....................(3)

Substitute $x = 3 - y$ is equation (2)

$3 - y - y = 1$

$⇒ y = 1$

Substitute the value of $y$ in equation (3)

$x = 3 - 1$

$⇒ x = 2$

Therefore $xy$ = $2$ ;[ $2 \times 1$ ]

Given $x+y=3$ and $x-y =1$

$\Rightarrow (x+y)+(x-y)=3+1=4$

$2x=4 \Rightarrow x=4 \div 2 = 2$

We know that $x=2$ . Replacing $x=2$ in:

$x+y=3 \Rightarrow y=1$ , and $x-y=1 \Rightarrow y=1$ . Therefore, $\displaystyle y=1$

So the answer of $\displaystyle xy =2\times1 =\boxed{2}$

\[ \begin { align } \begin { cases } x + y = 3 \\ x - y = 1 \end { cases } \\ 2x = 4 \\ x = 2 \\ \begin { cases } 2 + y = 3 \\ 2 - y = 1 \end { cases } \\ y = 1 \\ \therefore xy = 2 \end { align } \].

X plus y equals three. That means that x has to be 0, 1, 2, or 3. But X can't be 0 or 3 because the x minus y would not equal 1. So the answer would be 2 and 1. And 2 times 1 equals 2

Let x+y=3 be equation 1.

Let x-y=1 be equation 2.

Subtract them :

x+y=3

x-y=1

2y=2

And solve:

2y=2

y=1

Now substitute y=1 into either equation 1 or 2 to solve for x.

x+y=3

x+1=3

x=3-1

x=2

Now use the values to find xy:

xy

=2(1)

=2

Having sum and difference of two numbers you can simply obtain the two numbers this way: $(x+y)+(x-y)=2x$ $(x+y)-(x-y)=2y$

so we have:

$x = (3 + 1) / 2 = 2$ $y = (3 - 1) / 2 = 1$