Chapter 6.1, Problem 52E

To determine

To calculate: The solution of system equations, $x^2+y=4$ and $e^x-y=0$ , by using graphically or algebraically.

The solution of system equations, $x^2+y=4$ and $e^x-y=0$ , are $\left(1.058,2.881\right)\text{ and }\left(-1.965,0.14\right)$ .

Calculation:

Consider, the provided equations,

$x^2+y=4$ …… (1)

And,

$e^x-y=0$ …… (2)

Now, make the graph of the provided equations by using online graphing calculator.

From above graph, it is obtained that the intersection points are,

$\left(1.058,2.881\right)$ And $\left(-1.965,0.14\right)$

Check the solution as,

Substitute $1.058$ for $x$ and $2.881$ for $y$ in above equation (1) and (2) as,

$\begin{array}{c}{\left(1.058\right)}^2+2.881=4\\ 4=4\end{array}$

And

$\begin{array}{c}{e}^{1.058}-2.881=0\\ 0=0\end{array}$

Substitute $-1.965$ for $x$ and $0.14$ for $y$ in above equation (1) and (2) as,

$\begin{array}{c}{\left(-1.965\right)}^2+0.14=4\\ 4=4\end{array}$

And

$\begin{array}{c}{e}^{\left(-1.965\right)}-0.14=0\\ 0=0\end{array}$

It’s convenient to solve the provided equation graphically because it is easy to plot logarithmic and exponential function and it is very difficult to solve these equations algebraically.

So, the solution of system equations, $x^2+y=4$ and $e^x-y=0$ , are $\left(1.058,2.881\right)\text{ and }\left(-1.965,0.14\right)$ .