Alternative methods for solving a system of one linear one non linear simultaneous equations Take the equations $$x+y=5$$ $$x^2 + y^2 =13$$
The most basic method to solve this system is to first express the linear equation in terms of one of the variables and then sub that into the non-linear equation.
But I am curious if there are other methods to solve such a system ?
 A: We have $$(x+y)^2=13+2xy,$$ which gives
$$xy=6$$ and by the Viete's theorem $x$ and $y$ are roots of the equation:
$$t^2-5t+6=0$$ or
$$(t-2)(t-3)=0,$$ which gives the answer:
$$\{(2,3),(3,2)\}$$
A: You can use some symmetries (but I'm not sure if that makes any difference)
$$
2 x y =
(x + y) ^2 - (x^2 + y^2) =
25 - 13 = 12,
$$
express the difference
$$
(x - y)^2 =
(x^2 + y^2) - 2 x y =
1,
$$
and get a system of linear equations
$$
\begin{aligned}
x + y &= 5,\\
x - y &= \pm 1,
\end{aligned}
$$
that yields $x = 3$ and $y = 2$ or $x=2$ and $y=3$
A: The quadratic equation can be used.
Given:
x + y = 5, then y = 5 -4

Given
x^2 + y^2 = 13
then x^2 + (4-x)^2 = 13
and x^2 + x^2 - 10x + 25 -13 = 0
2x^2 + (-10x) + 12 = 0

Then the co-factors are a = 2, b = -10, c = 12
y = [-b (+-) sqrt(b^2 - 4ac)]/[2a]  <-- Quadratic Formula
y = [-(-10) (+-) sqrt((-10)^2 - 4(2*12))]/(2*2)
y = [10 (+-) sqrt(100-96)]/4
y = [10 + 2]/4  and  y = [10-2]/4
y = 12/4  and  y = 8/4
y = 3  and  y = 2
given x + y = 5
When y = 3, x + 3 = 5, x = 5-3, x = 2
when y = 2, x+2 = 5, x = 5-2, x = 3
Answers: x = 3, y = 2 and x = 2, y = 3
Try your answers in all of the original equations and against any given or implied restrictions to make sure they work. They do! (Always check for 'extraneous' answers.)
A: $\DeclareMathOperator{\lcm}{lcm}$
Compute the Gröbner basis of your system.  Let us start by writing this with zeroes on the right of the equals signs.  \begin{align*}
0 &= x+y-5  \\
0 &= x^2 + y^2 - 13  \text{.}  
\end{align*}
We pick a variable ordering.  Let us choose $x < y$.  (The given system is unchanged by the exchange of the variables $x$ and $y$, so we get the same computation, but with the variables swapped, if we choose the other ordering.)  We compute the first $s$-polynomial.  We need the LCM of the leading terms
$$  \lcm(x, x^2) = x^2  $$
and using this we get
\begin{align*}
0 &= \frac{x^2}{x}(x+y-5) - \frac{x^2}{x^2}(x^2 + y^2 - 13)  \\
  &= x^2 + xy - 5x -(x^2 + y^2 - 13)  \\
  &= xy - y^2 -5x + 13  \text{.}
\end{align*}
Now $\lcm(xy, x) = xy$ and
\begin{align*}
0 &= \frac{xy}{x}(x+y-5) - \frac{xy}{xy}(xy - y^2 -5x + 13)  \\
  &= xy + y^2 - 5y -(xy - y^2 -5x + 13)  \\
  &= 2y^2 +5x -5y -13
\end{align*}
and since we have a relation for $x$ and $y$ both of degree $1$, \begin{align*}
0 &= 2y^2 +5x - 5y - 13 -5(x+y-5)  \\
  &= 2y^2 +5x - 5y - 13 -5x -5y + 25  \\
  &= 2y^2 -10y + 12   \\
  &= 2(y^2 - 5y + 6)  \text{,}
\end{align*}
and since twice a thing is zero means the thing is zero, we have
$$  y^2 - 5y + 6 = 0  \text{.}  $$
Our collection of expressions which evaluate to zero is then (sorting by decreasing total degree, then according to the order we picked for the variables)
\begin{align*}
x^2 + y^2 - 13 &= 0  \\
xy - y^2 -5x + 13 &= 0  \\
y^2 - 5y + 6 &= 0  \\
x+y-5 &= 0  \text{.}
\end{align*}
Notice that in degree $2$ we slowly decreased the degree of the dependence on $x$ until we were left with a polynomial in $y$ alone.  Solving that polynomial, $y = 2$ or $y = 3$.  Then the collection becomes (by specializing the value of $y$ and appending a final equation for that value of $y$) either
\begin{align*}
x^2 - 9 &= 0  \\
-3x + 9 &= 0  \\
0 &= 0  \\
x-3 &= 0  \\
y -2 &= 0  \text{,}
\end{align*}
giving the solution $(x,y) = (3,2)$, or
\begin{align*}
x^2 - 4 &= 0  \\
-2x + 4 &= 0  \\
0 &= 0  \\
x-2 &= 0  \\
y - 3 &= 0  \text{,}
\end{align*}
giving the solution $(x,y) = (2,3)$.
A: In general, the set of equations:
$$\sum_{k=1}^{N}x_k^p = S_p$$
for $1\leq p\leq N$, can be solved by considering the function:
$$f(x) = -\sum_{p=1}^N\log\left(1-\frac{x_p}{x}\right) \tag{1}$$
The expansion of $f(x)$ around infinity is given by:
$$f(x) = \sum_{r=1}^{\infty}\frac{S_r}{r x^r}$$
We can thus write down $f(x)$ to order $x^{-2}$ as:
$$f(x) = \frac{5}{x} + \frac{13}{2 x^2} + \mathcal{O}\left(x^{-3}\right)\tag{2}$$
From (1) it follows that  $x^2 \exp\left[-f(x)\right]$ is a second degree polynomial that has the solutions as its roots. Using (2) it follows that:
$$\exp\left[-f(x)\right] = 1 - \frac{5}{x}  +  \frac{6}{x^2} +  \mathcal{O}\left(x^{-3}\right)$$
It thus follows that:
$$(x-x_1)(x-x_2) = x^2 - 5 x + 6$$
So, the solutions are $x_1=2$ and $x_2 = 3$ and vice versa.
A: Let $u^2-5u+c$ be the polynomial whose roots are $x$ and $y$, i.e. 
$$u^2-5u+c=(u-x)(u-y)=u^2-u(x+y)+xy.$$
Then
\begin{align*}
x^2-5x+c&=0\\
y^2-5y+c&=0.
\end{align*}
Adding the two equations and using the facts given we get 
$$13-25+2c=0 \implies c=6.$$
Thus we have $u^2-5u+6$ as our polynomial, so $x=2,y=3$ or vice versa.
A: Let's use some geometry. 
I tried as simple approach as I was able to muster
We can do it because it's easy to see that the only possible solutions will always contain x > 0 and y > 0: if it's not so then at least one of them will be greater than 5 as follows from the first equation and then it's square is greater than 25 which contradicts with the second equation. Let it be x <= y for simplicity.
Your equations tell this picture:

The areas of rectangles R are equal and also their area is equal to the area of outer square without squares X and Y all divided by 2, so R = (25 - 13)/2 = 6
Then by square symmetry we also have:

So, area of S is the area of outer square minus area R four times, thus S = 25 - 4*R = 25 - 4*6 = 1, but the side of S (which is 1 since S is a square) is also the difference between the sides of squares Y and X (which are y and x) and therefore x + 1 = y 
Remembering now our first figure and x + y = 5 we get x = 2 and y = 3.
By symmetry of course, if (x, y) is a solution, then (y, x) is too, so x = 3 and y = 2 also solves the original equations. This permutation is also easily illustrated on the figures above (as they don't change if x and y are just swapped). 
A: Another method, good for double-checking your answers on a test, is to graph the two equations on a TI-84 or similar calculator, then examine the graph to see where the lines overlap.
On the calculator, under the [y=] button, set 
Y1 = 5-x
Y2 = sqrt(13-x^2)
Y3 = -sqrt(13-x^2)
Then press [graph].
When the graph is displayed, press [2nd][trace] to get into the calculation menu.
Choose #5, intersect.
Select the lines that intersect and choose a 'guess' point that is close to the intersection. The calculator will come back with the answer, x = 3, y = 2
Do the same for the other intersection and the calculator will come back with x =2 , y = 3.
This doesn't always work where you don't have nice, text-book solutions, but when it does work, boy is it nice!
A: You can use polynomial division to eliminate a variable.
$$(x^2+y^2-13) - x(x+y-5) = y^2 -xy+5x-13$$
$$(y^2 -xy+5x-13) - (-y+5)(x+y-5) = 2y^2-10y+12$$
Solve the equation 2$y^2-10y+12=0$, then plug the values of $y$ into the linear equation.
Generalization: Given a system of polynomial equations in $2$ variables, if one of the equations has one of the variables occurring only as a linear term, then you can eliminate that variable by polynomial division to get a polynomial equation in the remaining unknown. The utility of this is somewhat suspect due to the unsolvability of many univariate polynomials.
Bigger generalizations:
Groebner bases https://en.wikipedia.org/wiki/Gr%C3%B6bner_basis
Elimination theory https://en.wikipedia.org/wiki/Elimination_theory
If you don't need exact answers, but only decimal approximations up to a specific precision, skip all this and look up Newton's method.
A: The 'ac' method can be used.
From above:
2x^2 + (-10x) + 12 = 0
a = 2, c = 12
a*c = 24
The possible factors of 24 are: (24*1),(12*2),(6*4),(3*8)
In the 'ac method' the sets of factors must multiply to 24 and sum to -10.
By examination we find that this is true of only one of the above factors, (6*4) where (-6*-4) = 24 and (-6 + -4) = -10.
2x^2 + (-10x) + 12 = 0
but, as we found, -10x = (-6x + -4x) so by substitution..
2x^2 + (-6x + -4x) + 12 = 0
Regrouping we get..
[2x^2 - 6x] + [-4x + 12] = 0
Factoring like terms out we get..
2x(x-3) + -4(x-3) = 0
Factoring (x-3) out we get..
(x-3)(2x-4) = 0
The zeros occur where x-3 = 0 and where 2x-4 = 0
x-3 = 0, x = 3 
2x - 4 = 0, 2x = 4, x = 4/2, x = 2
Given x+y = 5
when x = 3, 3+y=5, y = 5-3, y = 2
when x = 2, 2+y=5, y = 5-2, y = 3
Answers:
x = 3, y = 2 and x = 2, y = 3
