# 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.

• Just to make sure, is that exactly the question? Are you sure you haven't made a typo?
– Nick
Nov 12, 2014 at 12:54
• "A simpler method": Does it get simpler than just substituting and solving a cubic equation? You're confusing "simple" and "not tedious". It's like solving explicitly a $18 \times 18$ linear equation; it's simple, but annoying to do. Nov 12, 2014 at 14:54
• Is it possible the problem assumes knowledge of the general formula of solutions of cubic equations? I heard rumor students used to memorize this one. Nov 12, 2014 at 18:03
• What was the section of the textbook covering? It could be a typo of a problem with a simple solution that the section covered. Or it could have been a section on Newton's method for all we know. Nov 13, 2014 at 3:46
• @DanielV it was a typo - see below for the correct question. Nov 13, 2014 at 6:51

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.

• OK, thanks, I tried that and this gives rise to three even more involved real roots bit.ly/1xxb5ts. Nov 12, 2014 at 13:10
• Graphing the functions shows two roots. Nov 12, 2014 at 13:15
• @dustin I think people are up-voting the final sentence - that there is no "easy" solution, but if one has the patience then it is doable by hand. Nov 12, 2014 at 13:19
• @user1729 - no not two roots, you got your equation mistyped there when it shows $\frac{3x}{4}y$. Nov 12, 2014 at 13:21
• @Nicky Yeah, I just spotted that! However, a new link still shows two real roots. Or have I made another mistake? (Certainly, the second equation looks like a hyperbola, but I forget the exact form...but if it is, should be able to sketch and see that there are two solutions.) Nov 12, 2014 at 13:25

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).

• Perhaps you should mention this in the original question (i.e., edit your question) Nov 13, 2014 at 10:18
• -1: This is an answer to a different question. It sounds like you have no problem with this one, so there's no point asking and answering it here. People liked your original question, so let's just go with that one! Nov 13, 2014 at 17:32
• @Garrett I definitely agree the original is more interesting, I just wanted to include it if anyone was interested. Nov 14, 2014 at 2:43

$$2x + 3y = 6 + 5x$$$$x^2 - 2y^2 - \frac{3x}{4y} + 6xy = 60$$

Let $\chi=17x+5$ and $\gamma=17y-29$
Then $x=\frac1{17}\!\left(\chi-5\right)$ and $y=\frac1{17}\!\left(\gamma+29\right)$

$$\frac2{17}\!\left(\chi-5\right) + \frac3{17}\!\left(\gamma+29\right) = 6 + \frac5{17}\!\left(\chi-5\right)$$ $$\left(\frac1{17}\!\left(\chi-5\right)\right)^2 - 2\cdot\left(\frac1{17}\!\left(\gamma+29\right)\right)^2 - \frac{\frac3{17}\!\left(\chi-5\right)}{\frac4{17}\!\left(\gamma+29\right)} + 6\left(\gamma+29\right)\!\!\left(\chi-5\right) = 60$$

By expanding, collecting terms and multiplying with constants these can be quite easily be changed into

$$\gamma=\chi$$ $$\frac{\chi^2}{289} - \frac{2\gamma^2}{289} +\frac{50276\chi}{289} - \frac{8554\gamma}{289} - \frac{3\chi-15}{4\gamma+116} + 6\chi\gamma-\frac{253087}{289} = 60$$

Since $\gamma=\chi$ we can subtitute one for another, to avoid confusion I will now use $\lambda=\gamma=\chi$. This also reduces the equation to $\lambda=\lambda$ which reduces this to single equality.

$$\frac{1733\lambda^2}{289} + \frac{41722\lambda}{289}-\frac{3\lambda-15}{4\lambda+116} = \frac{270427}{289}$$

I'm tired of 289

$$1 733\lambda^2 + 41 722\lambda-\frac{867\lambda-4 335}{4\lambda+116} = 270 427$$

Multiply with $4\lambda+116$

$$6 932\lambda^3+367 916\lambda^2+3 757 177\lambda-31 373 867=0$$

Divide with $6 932$

$$\lambda^3+\frac{91 979\lambda^2}{1 733}+\frac{3 757 177\lambda}{6 932}-\frac{31 373 867}{6 932} = 0$$

Subtitute $\lambda=v-\frac{91 979}{5 199}$

$$\left(v-\frac{91 979}{5 199}\right)^3+\frac{91 979\left(v-\frac{91 979}{5 199}\right)^2}{1 733}+\frac{3 757 177 \left(v-\frac{91 979}{5 199}\right)}{6 932}-\frac{31 373 867}{6 932} = 0$$

Expand this and divide with $36 039 468$

$$v^3 - \frac{14 306 982 541}{36 039 468}v - \frac{427 215 759 480 560}{140 526 895 599} = 0$$

A wild large numbers appear.
Darksonn used variables, it's super effective.

$$p=-\frac{14 306 982 541}{36 039 468}$$ $$q=-\frac{427 215 759 480 560}{140 526 895 599}$$

$$v^3 + pv + q=0$$

Perform the substitution $v=w-\frac p{3w}$

$$\left(w-\frac p{3w}\right)^3+pw-\frac{p^2}{3w}+q=0$$

Expand the equation

$$w^3 - \frac{p^3}{27w^3} + q = 0$$

Let $u=w^3$ and multiply by $u$

$$u^2 + qu - \frac{p^3}{27} = 0$$

We pick one of the roots, in the end it dosen't matter which one. If you don't believe me, try yourself.

$$u=\frac1{18}\left(\sqrt 3\cdot \sqrt{4p^3+27q^2} - 9q\right)$$

Substitute back $w^3=u$

$$w^3=\frac1{18}\left(\sqrt 3\cdot \sqrt{4p^3+27q^2} - 9q\right)$$

Take the three square roots

$$w_1=-\frac{\sqrt[3]{\sqrt{12p^3+81q^2} - 9q}}{\sqrt[3]{2} \cdot 3^{2/3}}$$ $$w_2=\frac{\sqrt[3]{\sqrt{12p^3+81q^2} - 9q}}{\sqrt[3]{2} \cdot 3^{2/3}}$$ $$w_3=\frac{(-1)^{2/3}\cdot \sqrt[3]{\sqrt{12p^3+81q^2} - 9q}}{\sqrt[3]{2} \cdot 3^{2/3}}$$

In the following, the symbol $w$ denotes any of the $3$ values above

We wan't to invert $v=w_?-\frac 9{3w_?}$, so we get

$$w=\frac12\cdot\!\left(v\pm\sqrt{v^2+12}\right)$$

This is also written as

$$v=\frac{w^2-3}w$$

We know that $v=\lambda+\frac{91979}{5199}$

$$\lambda=\frac{w^2-3}w-\frac{91979}{5199}$$

Substitute in for $w,p,q$

$$\lambda=\frac{\left(3^{\frac{1}{3}} 2^{\frac{2}{3}} {\left(9 \cdot 3^{\frac{1}{3}} 2^{\frac{2}{3}} - {\left(\frac{3}{12013156} \, \sqrt{3} \sqrt{-11345297051245155823} + \frac{427215759480560}{15614099511}\right)}^{\frac{2}{3}}\right)}\right)}{6 \, {\left(\frac{3}{12013156} \, \sqrt{3} \sqrt{-11345297051245155823} + \frac{427215759480560}{15614099511}\right)}^{\frac{1}{3}}} - \frac{91979}{5199}$$ $$\lambda=\frac{\left(3^{\frac{1}{6}} 2^{\frac{1}{3}} \\{\left(3^{\frac{2}{3}} 2^{\frac{1}{3}} {\left(15597 \, \sqrt{-34035891153735467469} + 1708863037922240\right)}^{2} - 70214429819350206466848\right)}\right)}{374738388264 \, {\left(15597 \, \sqrt{-34035891153735467469} + 1708863037922240\right)}} - \frac{91979}{5199}$$ $$\lambda=\frac{\left(3^{\frac{1}{6}} 2^{\frac{1}{3}} {\left(3^{\frac{2}{3}} 2^{\frac{1}{3}} {\left(15597 \, \sqrt{-34035891153735467469} + 1708863037922240\right)}^{2} - 70214429819350206466848\right)}\right)}{374738388264 \, {\left(15597 \, \sqrt{-34035891153735467469} + 1708863037922240\right)}} - \frac{91979}{5199}$$

At this point these values become so ugly they don't even fit on the answer area, so I'll leave doing the last simple substitution as an exercise for the reader.

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:

• I dont know why anyone is down voting. The OP ask for a simple method not analytic and stated it was solved with a computer but was arduous. This is an easy method with Mathematica which meets the specifications. Comment if you see a problem. Nov 12, 2014 at 13:00
• I didn't downvote, but I read the question as "can this be done without using a computer?" I mean, how do you know that that OP didn't do precisely what you did here? Nov 12, 2014 at 13:04
• I didn't downvote, but my interpretation of the question is to find a simple way of actually solving the system of equations, where by simple they mean without a computer and preferably at the level of pre-calculus. Nov 12, 2014 at 13:04
• I also did not down vote and I also think the OP wants to solve it without a computer. Not only that, I read your answer and I don't even see the set of solutions. What is it? Nov 12, 2014 at 13:05
• I believe you are misreading the question. They are saying that they can reach the cubic by hand (arduously), but can only then solve the cubic with a computer. The intention of the question is to find an elegant mathematical solution of the system of equations, not to find a simpler few lines of code. Nov 12, 2014 at 13:16

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: $$20x^3 + 56x^2 - 243x - 544 = 0$$

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. $$\sqrt{11}$$

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.

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.

• Why should this be doable by hand? Nov 12, 2014 at 14:25
• ... because I'm a bloody idiot m( . Nov 12, 2014 at 14:26
• The question is from a Thai textbook aimed at 14-15 year olds which is why I wasn't expecting they would need to use the cubic formula or Mathematica. On reflection, it seems I must surely have copied the question down incorrectly. I will get the question tomorrow and post it here if anyone is still interested. Thanks for the replies. Nov 12, 2014 at 14:49

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])


The roots are all real:

$$x_{1}=-\frac{14}{15}-\frac{\sqrt{4429}.cos\Big(\frac{acot(-f)}{3}\Big)} {15}$$,

$$x_{2}=-\frac{14}{15}+\frac{\sqrt{4429}.sin\Big(\frac{atan(f)}{3}+\frac{\pi}{3}\Big)} {15}$$

$$x_{3}=-\frac{14}{15}-\frac{\sqrt{4429}.sin\Big(\frac{atan(f)}{3}\Big)} {15}$$

$$f=\frac{192158\sqrt{222021105}}{3330316575}$$.