Chapter 7.1, Problem 53E

To determine

Tofind:The intersection point of the system of equations $\{\begin{array}{c}y=e^x\\ x-y+1=0\end{array}$ using graphical utility and verify the result by checking them in the original system.

Answer to Problem 53E

The intersection point of the system of equations $\{\begin{array}{c}y=e^x\\ x-y+1=0\end{array}$ is $\left(0,1\right)$ .

Explanation of Solution

Given information:

The given system of equations is $\{\begin{array}{c}y=e^x\\ x-y+1=0\end{array}$ .

Calculation:

Draw the graph for the equations $y=e^x$ and $x-y+1=0$ .

  

From the above graph, the intersection point is $\left(0,1\right)$ . So, the solution of the system of equations $\{\begin{array}{c}y=e^x\\ x-y+1=0\end{array}$ is $\left(0,1\right)$ .

Verify the solution by substituting the values of intersection point in the system of equations.

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

  $\begin{array}{c}y=e^x\\ 1=e^0\\ 1=1\end{array}$

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

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

From the above, it is clear that both the equations are satisfied. So, the ordered pair $\left(0,1\right)$ is a solution of the system of equations.

Therefore, the intersection point of the system of equations $\{\begin{array}{c}y=e^x\\ x-y+1=0\end{array}$ is $\left(0,1\right)$ .