Chapter 11.6, Problem 16AYU

In Problems 5-24, graph each equation of the system. Then solve the system to find the points of intersection.

$\{\begin{array}{l}{x}^2\text{ + }{y}^2\text{ = }10\\ y\text{ = }x\text{ + 2}\end{array}$

To determine

To graph: The equation in the system and solve to find the points of intersection.

Answer to Problem 16AYU

Explanation of Solution

Given:

$x^2 + y^2 = 10$    -----Eq 1

$y = x + 2$    -----Eq 2

Graph:

Interpretation:

Solving for $x$ and $y$ from Eq 1 and Eq2:

Substituting Eq 2 in Eq 1,

$x^2 + {\left(x + 2\right)}^2 = 10$

$= > 2x^2 + 4x - 6 = 0$

$= > x^2 + 2x – 3 = 0$

Solving the above quadratic equation, $x = 1$ and $x = \text{-}3$.

For $x = 1$ we get $y = 3$ and

For $x = \text{-}3$ we get $y = \text{-}1$.

The solution to the set of equation is $(1, 3)$ and $\left(- 3, - 1\right)$.

The same can be seen from the graph where the curves for the equations $x^2 + y^2 = 10$ and $y = x + 2$ intersect at $(1, 3)$ and $\left(- 3, - 1\right)$.