System of nonlinear equations that leads to cubic equation The system of equations are:
$$\begin{align}2x + 3y &= 6 + 5x\\x^2 - 2y^2 - (3x/4y) + 6xy &= 60\end{align}$$
I can solve it through substitution but it is an arduous process to reach this cubic equation:
$$20x^3 + 56x^2 - 243x - 544 = 0$$
And I can only solve this using a computer.
Is there a simpler method?
edit: turns out there was a printing error that made the problem much harder. I posted the actual problem below if you want to see it.
edit 2: The actual problem is far less interesting, but I included it for completeness. There are some really great answers to the above "incorrect" problem however that are definitely worth a read. Thanks everyone for contributing.
 A: The first equation simple becomes $y = 2 + x$. In Mathematica (it isn't arduous) do
In[11]:= y = 2 + x;
FullSimplify[x^2 - 2 y^2 - ((3 x)/(4 y)) + 6 x*y - 60]

Out[12]= -68 + x (4 + 5 x - 3/(4 (2 + x)))

In[13]:= Together[-68 + x (4 + 5 x - 3/(4 (2 + x)))]

Out[13]= (-544 - 243 x + 56 x^2 + 20 x^3)/(4 (2 + x))

Line 13 equals zero so you have the desired results.
Even simpler is combing line 11 and 13 so it reads
y = 2 + x;
Together[FullSimplify[x^2 - 2 y^2 - ((3 x)/(4 y)) + 6 x*y - 60]]

For solutions, run NSolve
In[14]:= NSolve[(-544 - 243 x + 56 x^2 + 20 x^3)/(4 (2 + x)) == 0, x]

Out[14]= {{x -> -4.14829}, {x -> 3.32205}, {x -> -1.97376}}

Solution with Solve
In[16]:= FullSimplify[
 Solve[(-544 - 243 x + 56 x^2 + 20 x^3)/(4 (2 + x)) == 0, x]]

Out[16]= {{x -> Root[-544 - 243 #1 + 56 #1^2 + 20 #1^3 &, 3]}, {x -> 
   Root[-544 - 243 #1 + 56 #1^2 + 20 #1^3 &, 1]}, {x -> 
   Root[-544 - 243 #1 + 56 #1^2 + 20 #1^3 &, 2]}}

Plot of rational equation and plot of cubic only:


A: I did it on paper like so:
reduce the first equation:
$$ y = 2 + x $$
substitute:
$$ x^2 - 2(2+x)(2+x) - (3x/4(2+x)) + 6x(2+x) = 60 $$
expand:
$$ x^2 - 2x^2-8x-8 - 3x/(4x+8) + 6x^2+12x = 60 $$
reduce:
$$ 5x^2 + 4x - 8 - 3x/(4x+8) = 60 $$
multiply all terms by (4x+8):
$$ 5x^2(4x+8) + 4x(4x+8) - 8(4x+8) - 3x = 60(4x+8) $$
reduce again:
$$ 20x^3+40x^2 + 16x^2+32x - 32x-64 - 3x = 240x+480 $$
and reduce one more time:
$$ 20x^3 + 56x^2 - 243x - 544 = 0 $$
A: 

The first derivative gives a nice view of the inflection points, and clearly shows we are aiming for 3 real roots. 
I would bet that the first equation you showed is a typo, as it simplifies too easily. It seems to me they were aiming for an equation that could be solved by algebraic manipulation. 
A: I guess the error is in the question.
What if the real first equation was:
$$2x+3y=y+5x
$$
Which will lead to a simple equation of degree 2, with a nice value of x.
$$\sqrt{11}$$ 
A: Wolfram gives two complex and one real root:
$$x=\frac{1}{30}(-28 - \frac{2861}{\sqrt[3]{{498338}+75\sqrt{48312705}}}+\sqrt[3]{{498338}+75\sqrt{48312705}}),$$
which shows that there is no easy way, but following the Cardano method.
A: There was an error in the question. As I mentioned, this was a from a high school textbook that did not allow for the use of computational software (or even a calculator). The question was from a poor-quality photocopy and the student thought an addition sign was a division sign.
This was what the student told me the question was:
\begin{align}
2x + 3y &= 6 + 5x\\
x^2 - 2y^2 - 3x ÷ 4y + 6xy &= 60
\end{align}
This is what the question actually was:
\begin{align}
2x + 3y &= 6 + 5x\\
x^2 - 2y^2 - 3x + 4y + 6xy = 60
\end{align}
Solving and substituting this leaves you with:
$$
x^2 + x - 12 = 0
$$
and it is trivial to show that the solutions are then $(3,5)$ and $(-4,-2)$.
If anyone is interested in a further challenge, the textbook hints that there is a more elegant solution than this (this is one of the things that confused me in the first place).
A: Consider this form of the equation 
$x^3+\frac{56}{20}x^2+\frac{243}{20}x−\frac{544}{20}=0$ and let's write it as:
$$x^3+a_2x^2+a_1x^1+a_0=0. \qquad \text{(Eq1)}$$
Now consider this form of the 3-rd degree characteristic polynomial where the roots are easily discernible:
$(x+\alpha)(x+\beta)(x+\gamma) = 0$. Lets transform it, by simple multiplication, to the form of (Eq1):
$(x+\alpha)(x^2+(\beta+\gamma)x+ \beta\gamma) = 0$
$x^3+(\beta+\gamma)x^2+\beta\gamma\ x + \alpha x^2 + \alpha(\beta+\gamma)x + \alpha\beta\gamma = 0$
and finally:
$$x^3 + (\alpha + \beta + \gamma)x^2 + (\alpha\beta + \beta\gamma + \gamma\alpha) x + \alpha\beta\gamma = 0 \quad \text{(Eq2)}.$$
We know that $\alpha,\,\beta,\,\gamma$ are the roots, and thus the solutions you are looking for, and, from Eq1 and Eq2, we can state:
$$ \alpha+\beta+\gamma = a_2 = \frac{56}{20},$$
$$ \alpha\beta + \beta\gamma + \gamma\alpha = a_1 = \frac{243}{20},$$
$$ \alpha\beta\gamma = a_0 = \frac{544}{20}.$$
Three equations with three unknowns, should be doable by hand.
A: Fairly straight forward in Python using sympy:
from sympy.solvers import solve
from sympy import Symbol
x, y = Symbol('x'), Symbol('y')
print solve(20*x**3+56*x**2-243*x-544, [x])

A: The roots are all real:
$x_{1}=-\frac{14}{15}-\frac{\sqrt{4429}.cos\Big(\frac{acot(-f)}{3}\Big)} {15}$,
$x_{2}=-\frac{14}{15}+\frac{\sqrt{4429}.sin\Big(\frac{atan(f)}{3}+\frac{\pi}{3}\Big)} {15}$
$x_{3}=-\frac{14}{15}-\frac{\sqrt{4429}.sin\Big(\frac{atan(f)}{3}\Big)} {15}$
$f=\frac{192158\sqrt{222021105}}{3330316575}$.
