Solve the system by the method of elimination. $$\{\begin{array}{l}{x}^2+y^2=5\\ x^2-y^2=3\end{array}$$

A) $(4,1),(-4,1)$
B) $(2,1),(-2,1)$
C) $(4,-1),(4,1),(-4,-1),(-4,1)$
D) $(2,-1),(2,1),(-2,-1),(-2,1)$
E) no solution exists

We can use the method of elimination by adding the two equations together. This will eliminate the $y^2$ term:

$$\begin{array}{rl}(x^2+y^2)+(x^2-y^2)& =5+3\\ 2x^2& =8\\ x^2& =4\\ x& =\pm 2\end{array}$$

Now we can substitute this value of $x$ into either of the original equations to find the corresponding value of $y$. Substituting into the first equation gives:

$$\begin{array}{rl}{x}^2+y^2& =5\\ (2{)}^2+y^2& =5\\ y^2& =1\\ y& =\pm 1\end{array}$$

Therefore, the solution is $(2,1)$, $(-2,1)$, $(2,-1)$, and $(-2,-1)$, which correspond to choice D.