Problem #38 asks us to solve the system using either graphing, substitution, or elimination. The only way that I can think of doing this is by graphing. However, is there any easy way to solve this problem by doing elimination or substitution?
Thanks!
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If you use elimination on $x^2$, it becomes $$ \begin{cases} \frac{x^2}{16} + \frac{y^2}{25} = 1 \\ \\ \frac{x^2}{16} + \frac{y^2}{4} = 1 \\ \end{cases} \implies \frac{y^2}{25} - \frac{y^2}{4} = 0 $$ from which it follows that $y^2 = 0 \implies y = 0$. Thus, evaluating each equation with $y=0$ gives the solutions $(4,0)$ and $(-4, 0)$.
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Notice that the first terms in each equation are equal. All you have to do is subtract the first equation from the second and they will cancel, leaving:
$$ \frac{y^2}{4} - \frac{y^2}{25} = 0$$
Continuing, you'll get $y^2 = 0$ and $x^2 = 16$
For the general methods, yes you can solve this using either substitution or elimination. Since $x$ and $y$ only have powers of 2, you can treat $x^2$ and $y^2$ as variables and you're back to a linear system of equations. You know how to solve that, don't you?
Hint: let $u = x^2$, and $v = y^2$. What type of system of equations results from expressing the given system in terms of $u$ and $v$ instead of $x$ and $y$? Do you know how to solve this type of system? What methods have you learned to find the solution? Once you find the solution for the auxiliary variables $u$ and $v$, how do you then express the solution set for $x$ and $y$? Be sure to check your solutions to see if they are in fact correct.