Blog: `x^2 + y = 4, e^x - y = 0` Solve the system graphically or algebraically. Explain your choice of method.

Wednesday, July 1, 2015

$\displaystyle{{x}}^{{2}}+{y}={4},{{e}}^{{x}}-{y}={0}$ Solve the system graphically or algebraically. Explain your choice of method.

You need to solve the system, hence, selecting to solve it algebraically, yields:

$\displaystyle{{x}}^{{2}}+{y}={4}$

$\displaystyle{{e}}^{{x}}-{y}={0}$

Replacing $\displaystyle{{e}}^{{x}}$ for $\displaystyle{y}$ in the first equation yields:

$\displaystyle{{x}}^{{2}}+{{e}}^{{x}}={4}$

$\displaystyle{{e}}^{{x}}={4}-{{x}}^{{2}}$

Now, you need to solve for x the equation, using graphical method, hence, you need to trace the curves for the equations $\displaystyle{y}={{e}}^{{x}}$ and $\displaystyle{y}={4}-{{x}}^{{2}},$ and then you need to find out where these curves intercept each other. Notice that the red curve representing $\displaystyle{y}={{e}}^{{x}}$ and the black curve representing $\displaystyle{y}={4}-{{x}}^{{2}}$ , intercept each other at a point located between [1,2], much closer to the value x = 1.