Chapter 7.1, Problem 38E

Solution

To calculate: The number of solutions for the system.

  $\begin{array}{c}y=x^2-3x-5\\ 1=2x-y\end{array}$

The solutions of the given system are $\left(5.7,10.4\right),\left(-0.7,-2.4\right)$ .

Given Information:

The system of equations.

  $\begin{array}{l}y=x^2-3x-5\\ 1=2x-y\end{array}$

Calculation:

Consider the given system of the equations,

Draw the graph of both the equation in the same Cartesian plane by using the graphing calculator. The number of intersection of both the equation is equal to the number of solutions.

  

From the graph, It is observed that the intersecting points of two graphs are $\left(5.7,10.4\right),\left(-0.7,-2.4\right)$ . Verify it by substituting on the given system of equation.

Substitute $\left(5.7,10.4\right)$ in $y=x^2-3x-5$

  $\begin{array}{l}10.4={5.7}^2-3\left(5.7\right)-5\\ 10.4\approx 10.39\end{array}$

Substitute $\left(5.7,10.4\right)$ in $1=2x-y$

  $\begin{array}{l}1=2\left(5.7\right)-\left(10.4\right)\\ 1=1\end{array}$

Thus, the point $\left(5.7,10.4\right)$ is the solution of the given equation.

Substitute $\left(-0.7,-2.4\right)$ in $y=x^2-3x-5$ .

  $\begin{array}{l}-2.4=-{0.7}^2-3\left(0.7\right)-5\\ -2.4\approx -2.39\end{array}$

Substitute $\left(-0.7,-2.4\right)$ in $1=2x-y$

  $\begin{array}{l}1=2\left(-0.7\right)-\left(-2.4\right)\\ 1=1\end{array}$

Thus, the point $\left(-0.7,-2.4\right)$ is the solution of the given equation.

Therefore, the obtained solution are $\left(5.7,10.4\right),\left(-0.7,-2.4\right)$ .