Solve the system of equations $x(x+y) - 3 = y$ and $x^2 + y^2 = 5$

The second equation is straightforward: circle, but how do we link this equation with the first? It is a system from the entrance examination in university. The list of topics for this exam is limited to school knowledge and essential calculus: skill to solve basic derivatives and integrals.

I think I solved this system with graphics. The second equation is a circle, and I try to understand the graphic in the first equation and find an intersection after that. I tried this in Desmos, which has four solutions, but how can I find specific values?

$$\begin{equation*} \begin{cases} x(x+y)-3=y\\ x^2+y^2=5 \end{cases} \end{equation*}$$

For example, I get $$y = (x^2 - 3)/ (1-x)$$ in first equation and I try put this in second equations and only one thing I get is hard equation with 4th degree of $$x$$ which I dont know how to solve that. The situation is similar with $$x$$ instead $$y$$.

New contributor
Firefly is a new contributor to this site. Take care in asking for clarification, commenting, and answering. Check out our Code of Conduct.
• Welcome to MSE. Your question is phrased as an isolated problem, without any further information or context. This does not match many users' quality standards, so it may attract downvotes, or be closed. To prevent that, please edit the question. This will help you recognize and resolve the issues. Concretely: please provide context, and include your work and thoughts on the problem. These changes can help in formulating more appropriate answers. yesterday
• Hi, may you eliminate $x^2$ by substitution ?
– EDX
yesterday
• for example, I get y = (x^2 - 3)/ (1-x) in first equation and I try put this in second equations and only one thing I get is hard equation with 4th degree of x which I dont know how to solve that. The situation is similar with x instead y yesterday

$$x(x+y)-3=y \iff y=\frac{-x^2+3}{x-1} , x\neq1$$

For $$x^2+y^2=5$$ substitute $$y$$ with $$\frac{-x^2+3}{x-1}$$

$$x^2+\left(\frac{-x^2+3}{x-1}\right)^2=5 \implies 2x^4-2x^3-5x^2+9=5(x-1)^2\\\implies (x-2)(x^3+x^2-3x-1)=0,$$

so $$x_1=2$$ or $$x^3+x^2-3x-1=0$$.

$$x^3+x^2-3x-1=0\implies -1-3(t-\frac{1}{3})+(t-\frac{1}{3})^2+(t-\frac{1}{3})^3 ,\text{where t=x+\frac{1}{3} }\\t^3-\frac{10}{3}t+\frac{2}{27}=0$$ Change coordinates by substituting $$t=u+\frac{λ} {u}$$ where $$λ$$ is a constanct value that will be determined later: $$\frac{2}{27}-\frac{10}{3}(u+\frac{λ}{y})+(y+\frac{λ}{u})\implies u^6+u^4(3λ-\frac{10}{3}+\frac{2u^3}{27})+u^2(3λ^2-\ \frac{10λ}{3})+λ^3=0$$

Now substituting $$λ=\frac{10} {9}$$ and then $$z=u^3$$ , yielding a quadratic equation in the variable $$z$$ that can be solved using the quadratic formula.

$$x_2=\frac{(3i\sqrt{111} -1)^{2/3}+10}{3\sqrt[3]{2i\sqrt{111}-1}}-\frac{1}{3}\\ x_3=\frac{10(-1)^{2/3}-\sqrt[3]{-1}(3i\sqrt{111}-1)^{2/3}}{3\sqrt[3]{2i\sqrt{111}-1}}\\x_4=\frac{1}{3}\sqrt[3]{\frac{i}{i+3\sqrt{111}}}(\sqrt[3]{-1}(3i\sqrt{111}-1)^{2/3}-10)-\frac{1}{3}$$

Now substitute each one of those $$4$$ values of $$x$$ to $$x(x+y)-3=y$$ and you get four pairs $$(x,y)$$ that are too long to right the closed forms of.

$$(x,y)=(2,-1)\lor \dots$$

Symbolab can solve this system and the approximate solutions are shown here .

• Powerful solution... In other words, do you think it can be a misprint in the condition of this task? I`m not sure that it can be on exam if compare this with other tasks. Other task more simple and has simple solution without such big decimal numbers if you take a look on other tasks yesterday
• Since $y$ is already used as part of the problem formulation, it probably shouldn’t be used again in the transformation applied to the cubic in $t$. yesterday
• You are correct I'll edit it now , It was not meant to be the $y$ we were solving for from the beginning.
– ράτ
yesterday
• @FireflyMath Yes, It is 99% a misprint , it is pretty unlikely that this would be in any level exam . Maybe there is a simple solution that I could not find but I think that is unlikely too.
– ράτ
yesterday