As many users have not been exposed to complex numbers, I will post the solutions to the problems assuming that all variables used are real. However, please feel free to solve these problems over the complex numbers and post your solutions!

**Problem 1.** Solve

\[\left\{\begin{align}y^2&=x^3-3x^2+2x&\quad(1)\\x^2&=y^3-3y^2+2y&\quad(2)\end{align}\right.\]

**Solution.** \((1)-(2)\):

\[-(x^2-y^2)=(x^3-y^3)-(3x^2-3y^2)+(2x-2y)\]

\[-(x+y)(x-y)=(x-y)(x^2+xy+y^2)-(3x+3y)(x-y)+2(x-y)\]

\[(x-y)(x^2+xy+y^2-3x-3y+2+x+y)=0\]

\[(x-y)[x^2+(y-2)x+(y^2-2y+2)]=0\]

**Case 1.** \(x-y=0\implies x=y\). Substituting \(x=y\) into \((2)\), we have

\[y^2=y^3-3y^2+2y\]

\[y(y^2-4y+2)=0\]

**Case 1a.** \(y=0\). Since \(x=y\), \(\boxed{x=y=0}\).

**Case 1b.** \(y^2-4y+2=0\).

\[y = \frac{-(-4)\pm\sqrt{(-4)^2-4(1)(2)}}{2(1)}\]

\[=2\pm\sqrt{2}\]

Since \(x=y\), \(\boxed{x=y=2\pm\sqrt{2}}\).

**Case 2.** \(x^2+(y-2)x+(y^2-2y+2)]=0\).

\[\text{Discriminant}\,= (y-2)^2-4(1)(y^2-2y+2)\]

\[=y^2-4y+4-4y^2+8y-8\]

\[=-3y^2+4y-4\]

\[=-3\left(y^2-\frac{4}{3}y\right)-4\]

\[=-3\left[y^2-\frac{4}{3}y+\left(\frac{2}{3}\right)^2-\left(\frac{2}{3}\right)^2\right]-4\]

\[=-3\left[\left(y-\frac{2}{3}\right)^2-\frac{4}{9}\right]-4\]

\[=-3\left(y-\frac{2}{3}\right)^2+\frac{4}{3}-4\]

\[=-3\left(y-\frac{2}{3}\right)^2-\frac{8}{3}<0\]

Since we are solving this problem over the reals, Case 2 is rejected.

Hence, \(\boxed{x=y=0}\) and \(\boxed{x=y=2\pm\sqrt{2}}\).

**Problem 2.** Solve

\[\left\{\begin{align}5x^2+4y^2-10x+5y&=0&\quad(3)\\4x^2+5y^2+5x-10y&=0&\quad(4)\end{align}\right.\]

**Problem 3.** Solve

\[\left\{\begin{align}x+y-z&=4&\quad(5)\\x^2+y^2-z^2&=12&\quad(6)\\x^3+y^3-z^3&=34&\quad(7)\end{align}\right.\]

**Problem 4.** Solve

\[\left\{\begin{align}xy+yz+zx&=1&\quad(8)\\yz+zt+ty&=1&\quad(9)\\zt+tx+xz&=1&\quad(10)\\tx+xy+yt&=1&\quad(11)\end{align}\right.\]

**Problem 5.** Solve

\[\left\{\begin{align}(x+y)(x+z)&=x&\quad(12)\\(y+z)(y+x)&=2y&\quad(13)\\(z+x)(z+y)&=3z&\quad(14)\end{align}\right.\]

## Comments

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TopNewestProblem 1.Solve\[\left\{\begin{align}y^2&=x^3-3x^2+2x&\quad(1)\\x^2&=y^3-3y^2+2y&\quad(2)\end{align}\right.\]

Solution.\((1)-(2)\):\[-(x^2-y^2)=(x^3-y^3)-(3x^2-3y^2)+(2x-2y)\]

\[-(x+y)(x-y)=(x-y)(x^2+xy+y^2)-(3x+3y)(x-y)+2(x-y)\]

\[(x-y)(x^2+xy+y^2-3x-3y+2+x+y)=0\]

\[(x-y)[x^2+(y-2)x+(y^2-2y+2)]=0\]

Case 1.\(x-y=0\implies x=y\). Substituting \(x=y\) into \((2)\), we have\[y^2=y^3-3y^2+2y\]

\[y(y^2-4y+2)=0\]

Case 1a.\(y=0\). Since \(x=y\), \(\boxed{x=y=0}\).Case 1b.\(y^2-4y+2=0\).\[y = \frac{-(-4)\pm\sqrt{(-4)^2-4(1)(2)}}{2(1)}\]

\[=2\pm\sqrt{2}\]

Since \(x=y\), \(\boxed{x=y=2\pm\sqrt{2}}\).

Case 2.\(x^2+(y-2)x+(y^2-2y+2)]=0\).\[\text{Discriminant}\,= (y-2)^2-4(1)(y^2-2y+2)\]

\[=y^2-4y+4-4y^2+8y-8\]

\[=-3y^2+4y-4\]

\[=-3\left(y^2-\frac{4}{3}y\right)-4\]

\[=-3\left[y^2-\frac{4}{3}y+\left(\frac{2}{3}\right)^2-\left(\frac{2}{3}\right)^2\right]-4\]

\[=-3\left[\left(y-\frac{2}{3}\right)^2-\frac{4}{9}\right]-4\]

\[=-3\left(y-\frac{2}{3}\right)^2+\frac{4}{3}-4\]

\[=-3\left(y-\frac{2}{3}\right)^2-\frac{8}{3}<0\]

Since we are solving this problem over the reals, Case 2 is rejected.

Hence, \(\boxed{x=y=0}\) and \(\boxed{x=y=2\pm\sqrt{2}}\). – Victor Loh · 2 years, 1 month ago

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What we actually have, is \( f(f(x)) = x \), and so we are interested in the cases where \( f(x) = f^{-1} (x) \). If you can accurately plot this graph, you can then find all such solutions to this equation.

Otherwise, take the approach as Victor did, where you subtract the 2 equations and factor out \( (x - y) \). If you believe that the only solutions occur when \( x = y \), then you just need to show that the dividend has no (real) roots. – Calvin Lin Staff · 2 years, 1 month ago

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– Victor Loh · 2 years, 1 month ago

Do you think I should continue to post the solutions as comments? Depending on the number of upvotes of each solution, the solutions will not be arranged in chronological order.Log in to reply

– Victor Loh · 2 years, 1 month ago

I guess I'll post the solution of each problem below the problem itself.Log in to reply

I think that posting the solution below the problem makes it disruptive to seeing the entire list. I would prefer if you simply posted the solution as comments, or at the end of the note. – Calvin Lin Staff · 2 years, 1 month ago

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This is a nice set of problems to work on. The tricky part is justifying that one has indeed found all of the solutions. Try solving these problems over the reals and over the complex numbers, and see how your solution differs. The first problem is a great example! – Calvin Lin Staff · 2 years, 1 month ago

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– Rahul Saha · 2 years, 1 month ago

Subtracting the equations in \(1\),we get a factorization.One of the factors yields a solution,and looking at the discriminant of the other one yields no solutions(real).Log in to reply

– Victor Loh · 2 years, 1 month ago

Indeed, justifying that one has indeed found all of the solutions is the tricky part. As many other users have not been exposed to complex numbers, I will post the solutions to the problems assuming that all variables used are real. However, other users may feel free to solve these problems over the complex numbers and post their own solutions.Log in to reply

And of course, we can then ask, what are the solutions over the quarternions? Over the matrices? It never ends :) – Calvin Lin Staff · 2 years, 1 month ago

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(3).

x+y=4+z \(x^{2}+y^{2}=12+z^{2}\)

After this we can find xy, and hence \(x^{3}+y^{3}\) in terms of z. Comparing with the third equation will give a cubic equation. – Joel Tan · 2 years, 1 month ago

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Problem 1: y^2=x(x-1)(x-2) and x^2=y(y-1)(y-2). x=y=0. (if there is only one solution) Problem 2: x=y=0. (if that is the only solution) Problem 3: (x,y,z)-->(2,3,1) or (3,2,1) Problem 4: t=x=y=z=sqrt(1/3) Problem 5: x=y=z=0 I assume there is only 1 solution for all. very dangerous. Anyway, I'm actually 12, and i'm going RI too! Hope to see you soon, Victor and Srivathsan Veeramani. – Aloysius Ng · 2 years, 1 month ago

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and\( x \neq y \).With these questions, you should think beyond finding the obvious answer, and trying to ensure that you could find all possible solutions. – Calvin Lin Staff · 2 years, 1 month ago

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– Srivathsan Veeramani · 2 years, 1 month ago

Good luck aloysius to get into RI quite surprised you knowthis at a young age I only learnt this recently but I feel that these sums will come in smo round 1 which does not requre methods and sometimes it is quite obvious like qs 1 that x=y and I am too lazy to write the method.anyway you have solved it correctly and congrats.Good luck and have fun in RI.Log in to reply

FIRST ONE IS X=Y fourth one is x=y=z=t – Srivathsan Veeramani · 2 years, 1 month ago

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