Chapter 7.1, Problem 27E

Solution

To calculate: The solution of the system of the equation.

  $\begin{array}{l}y=1+2x-x^2\\ y=1-x\end{array}$

The solution is $\left(0,1\right),\left(3,-2\right)$ .

Given Information:

The given system of equation is defined as,

  $\begin{array}{l}y=1+2x-x^2\\ y=1-x\end{array}$

And the graph is,

  

Calculation:

Consider the given information,

By using the graphing calculator make the graph of the equations.

  

From the graph, it is observed that the intersecting points of two graphs are $\left(0,1\right),\left(3,-2\right)$ . Verify it by substituting on the given system of equation.

Substitute 0 for x and 1 for y in the equation $y=1+2x-x^2$

  $\begin{array}{c}1=1+2\left(0\right)-0^2\\ =1+0-0\\ =1\end{array}$

Substitute 0 for x and 1 for y in the equation $y=1-x$ .

  $\begin{array}{c}1=1-0\\ 1=1\end{array}$

Thus, the point $\left(0,1\right)$ is the solution of the given equation.

Substitute 3 for x and $-2$ for y in the equation $y=1+2x-x^2$ .

  $\begin{array}{c}-2=1+2\left(3\right)-3^2\\ =1+6-9\\ =7-9\\ =-2\end{array}$

Substitute 3 for x and $-2$ for y in the equation $y=1-x$ .

  $\begin{array}{c}-2=1-3\\ =-2\end{array}$

Thus, the point $\left(3,-2\right)$ is the solution of the given equation.

Therefore, the obtained solution is $\left(0,1\right),\left(3,-2\right)$ .