y=e^x
y=4-x^2
can we solve these equations manually? how can we find the values for x?
Thank you
No, it can't be solved algebraically because we have an x as the exponent in the first equation and as a base in the second
$\displaystyle e^x = 4-x^2 \implies e^x + x^2 - 4 = 0$. Let $\displaystyle f(x) = e^x+x^2-4$ and the graphs will be equal where they intersect, thus $\displaystyle f(x)=0$
Since $\displaystyle f(1.5) > 0 \text{ and } f(1) < 0$ the root is between $\displaystyle 1 < x < 1.5$ (if you were to do this on pen and paper then you'd use an iterative solution converging on x)
Out of personal interest how would you know this particular system has two real solutions?