Given the question:
After a bunch of manipulation of the above equations we find the solution to be 19/4. Yet could the above be solved by simultaneous equations?
From $1)$ we can conclude that $x = 1 - y$.
Substituting this into $2)$ we get a quadratic $2y^2 - 2y - 1 = 0$.
However, when we solve for $y$ we get two possible solutions. Hence how do we proceed from here? Why doesn't simultaneous equations work to solve the above problem?
On
First note that we can get $$2xy = (x+y)^2 - x^2-y^2 = 1 - 2 = -1$$
We can then apply binomial theorem again to obtain $x^3+y^3$ and $x^5+y^5$ $$ \begin{align*} x^3 + y^3 &= (x+y)^3 - 3x^2y - 3xy^2 \\ &= 1 + \frac{3}{2}(x+y) = \frac{5}{2} \\ x^5 + y^5 &= (x+y)^5 - 5x^4y - 10x^3y^2 - 10x^2y^3 - 5xy^4 \\ &= 1 + \frac{5}{2}(x^3+y^3) - \frac{5}{2}(x+y) \\ &= 1 + \frac{25}{4} - \frac{5}{2} = \frac{19}{4} \end{align*} $$
On
$$1=(x+y)^2= x^2+2xy+y^2 = 2+2xy$$$$ \implies xy = -1/2$$
$$4=(x^2+y^2)^2 =x^4+2x^2y^2 +y^4$$$$\implies x^4+y^4= 7/2$$
and $$2=(x+y)(x^2+y^2) = x^3+xy(x+y)+y^3$$ $$\implies x^3+y^3 = 5/2 $$ So
$$7/2=(x+y)(x^4+y^4) = x^5+y^5 + xy(x^3+y^3)$$ $$\implies x^5+y^5 =19/4$$
On
We might also think of the first two equations as describing the intersections of a line with intercepts $ \ (0 \ , \ 1) \ \ $ and $ \ (1 \ , \ 0 ) \ $ and the circle of radius $ \ \sqrt2 \ $ centered on the origin. These intersections are symmetric about the line $ \ y \ = \ x \ \ . $
Since the axis intercepts are within the circle, the intersection points occur in the second and fourth quadrants, so their coordinates are $ \ (-X \ , \ Y) \ \ $ and $ \ (Y \ , \ -X) \ \ . \ $ The numerical result for $ \ x^5 + y^5 \ \ $ will thus be the same for both points.
We can solve the two equations simultaneously then as $$ Y \ \ = \ \ 1 \ - \ X \ \ \Rightarrow \ \ Y^2 \ \ = \ \ (1 - X)^2 \ \ = \ \ 2 \ - \ X^2 \ \ \Rightarrow \ \ 2X^2 - 2X - 1 \ \ = \ \ 0 \ \ , $$ for which the solutions are $ \ X \ = \ \frac12 \pm \frac{\sqrt3}{2} \ \ , \ $ corresponding to the fourth- and second-quadrant solutions.
From this, we obtain $$ X^2 \ \ = \ \ X \ + \ \frac12 \ \ = \ \ 1 \pm \frac{\sqrt3}{2} \ \ , \ \ Y^2 \ \ = \ \ 2 \ - \ X^2 \ \ = \ \ 1 \ \mp \frac{\sqrt3}{2} $$ $ \Rightarrow \ \ \mathbf{X^5 \ + \ Y^5} \ \ = \ \ X·(X^2)^2 \ + \ Y·(Y^2)^2 $ $$ = \ \ \left( \frac12 \pm \frac{\sqrt3}{2} \right)·\left( 1 \pm \frac{\sqrt3}{2} \right)^2 \ + \ \left( \frac12 \mp \frac{\sqrt3}{2} \right)·\left( 1 \mp \frac{\sqrt3}{2} \right)^2 $$ $$ = \ \ \left( \frac12 \pm \frac{\sqrt3}{2} \right)·\left( \frac74 \pm \sqrt3 \right) \ + \ \left( \frac12 \mp \frac{\sqrt3}{2} \right)·\left( \frac74 \mp \sqrt3 \right) $$ [all of the "middle terms" involving $ \ \sqrt3 \ $ cancel!] $$ = \ \ \frac12·\frac74 \ + \ \frac32 \ + \ \frac12·\frac74 \ + \ \frac32 \ \mathbf{= \ \frac{19}{4} } \ \ . $$
The green curve in the graph represents this last equation.
On
Why doesn't simultaneous equations work to solve the above problem?
It does work. You just need to carry the solution far enough.
To start, note that $x$ and $y$ are symmetric. In other words, you can replace $x \rightarrow y$ and $y \rightarrow x$ and you'd still be solving the same problem. Whatever value $x, y$ have, the same values for $y, x$ will also solve the problem. Hence you should expect there to be two possible solutions for both $x$ and $y$.
If you solve the quadratic, you find:
$y = 1/2(1+\sqrt{3})$ or $y = 1/2(1-\sqrt{3})$
Putting these solutions into $x + y = 1$, you get
if $y = 1/2(1+\sqrt{3})$ then $x = 1/2(1-\sqrt{3})$, and if $y = 1/2(1-\sqrt{3})$ then $x = 1/2(1+\sqrt{3})$.
The solutions are interchangeable, as expected from the discussion above about $x$ and $y$ being symmetric.
From here it's just a matter of putting these expressions into $x^5 + y^5$. You only need to do this calculation once, because $x$ and $y$ are symmetric. Taking the fifth power is tedious but straightforward with pen and paper, and trivial if you're allowed a calculator. The result is the answer.
(The other answers give more elegant ways to solve the problem.)
HINT:
$$(x^2+y^2)\left((x+y)^3-3xy(x+y)\right)=\color{red}{(x^5+y^5)}+(xy)^2(x+y)$$ where, $xy=\frac{(x+y)^2-(x^2+y^2)}{2}$.