Alternative methods for solving a system of one linear one non linear simultaneous equations

Solution 1:

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.

Solution 2:

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: $$\{(2,3),(3,2)\}$$

Solution 3:

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 $y=3$

Solution 4:

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.)