# Question #1a1c4

May 4, 2017

${x}^{2} + {y}^{2} = 26$
$x y = 5$

#### Explanation:

$x + y = 6$ called @
$x - y = 4$ called @@
then @-@@
get$2 y = 2$ so $y = 1$
then $x = 5$
then ${x}^{2} + {y}^{2} = 26$
$x y = 5$

May 4, 2017

${x}^{2} + {y}^{2} = 26$
$x y = 5$

#### Explanation:

Method 1
Since $x + y = 6$, ${\left(x + y\right)}^{2} = 36$. Expand the left-hand side to get ${x}^{2} + 2 x y + {y}^{2} = 36$.

Since $x - y = 4$, ${\left(x - y\right)}^{2} = 16$. Expand the left-hand side to get ${x}^{2} - 2 x y + {y}^{2} = 16$.

Now, add the first equation to the second equation: $\left({x}^{2} + 2 x y + {y}^{2}\right) + \left({x}^{2} - 2 x y + {y}^{2}\right) = 36 + 16$. This gives $2 {x}^{2} + 2 {y}^{2} = 52$, or ${x}^{2} + {y}^{2} = 26$.

Now, subtract the second equation from the first equation: $\left({x}^{2} + 2 x y + {y}^{2}\right) - \left({x}^{2} - 2 x y + {y}^{2}\right) = 36 - 16$. This gives $4 x y = 20$, or $x y = 5$.

Method 2
Add $x + y = 6$ to $x - y = 4$ to get $\left(x + y\right) + \left(x - y\right) = 6 + 4$, or $x = 5$. Subtract $x - y = 4$ from $x + y = 6$ to get $\left(x + y\right) - \left(x - y\right) = 6 - 4$, or $y = 1$.

Thus, ${x}^{2} + {y}^{2} = {5}^{2} + {1}^{2} = 26$ and $x y = 5 \cdot 1 = 5$.