System of equation doesn't want to get solved

Question

I've been trying to solve the following system of equations for hours, to no avail:

$$(1)\;\frac{x}{y}+\frac{y}{x}=\frac{10}{3}\\\\$$

$$(2)\;x^2+y^2=8$$

Can anyone give me a hint?

Answer 1

Hint:

The first equation is equivalent to $x^2+y^2=\dfrac{10}3xy$.

Given the second equation, we have $\dfrac{10}3xy=8, $ so $\color{blue}{xy=\dfrac{12}5}$.

Also, $(x+y)^2=x^2+y^2+2xy=8+\dfrac{24}5=\dfrac{64}5, $ so $\color{blue}{x+y=\pm\dfrac8{\sqrt5}}$.

Can you take it from here?

Answer 2

As Peter Foreman already mentioned, given this equation system

$$\begin{equation*}\ x/y+y/x=10/3 \\ x^2+y^2=8\end{equation*}$$

we can multiply the first equation by $xy$ and get the following equivalent equation system

$$\begin{equation*}\ x^2+y^2=(10/3)xy \\ x^2+y^2=8\end{equation*}$$

From this we see that $$x^2+y^2=(10/3)xy=8$$ and $$(10/3)xy=8$$ or $$\color{red}{y = (12/5)/x}$$

Pluging $y = (12/5)/x$ into $(x/y)+(y/x)=(10/3)$ we get $$x/((12/5)/x)+((12/5)/x)/x=10/3$$ or this: $$(5/12)x^2+(12/5)/x^2=10/3$$

Multiplying by $x^2$, we get: $$(5/12)x^4+(12/5)=(10/3)x^2$$ or this: $$(5/12)x^4-(10/3)x^2+(12/5)=0$$

Multiplying by $(12/5)$, we get: $$x^4-8x^2+(12/5)^2=0$$ Substituting $u = x^2$, we get: $$u^2-8u+(12/5)^2=0$$

Completing the square, we get: $$u^2-8u+(12/5)^2=(u+(-4))^2-4^2+(12/5)^2=0$$

This leads to $$x_{1,2,3,4}=\pm\sqrt{u_{1,2}} = \pm\sqrt{4\pm\sqrt{4^2-(12/5)^2}}=\begin{cases}6/\sqrt{5} &\rightarrow \color{red}{y_1 = (12/5)/(6/\sqrt{5})} = 2/\sqrt{5} \\ -6/\sqrt{5} &\rightarrow \color{red}{y_2 = (12/5)/(-6/\sqrt{5})} = -2/\sqrt{5}\\2/\sqrt{5} &\rightarrow \color{red}{y_3 = (12/5)/(2/\sqrt{5})}=6/\sqrt{5}\\ -2/\sqrt{5}&\rightarrow \color{red}{y_4 = (12/5)/(-2/\sqrt{5})}=-6/\sqrt{5}\end{cases}$$

Another way was suggested by rsadhvika can be done this way:

Start with the first equation of the original equation system $$x/y+y/x=10/3$$ Set $y = \alpha x$, then we get $$x/(\alpha x)+(\alpha x)/x=10/3$$ or $$1/\alpha + \alpha =10/3$$ Multiply by $\alpha$ and it follows $$1+ \alpha^2 =(10/3) \alpha$$ or $$ \alpha^2-(10/3) \alpha+1 =0$$ Completing the square: $$ \alpha^2-(10/3) \alpha+1 =(\alpha+(-5/3))^2-(5/3)^2+1=0$$ This leads to $$ \alpha_{1,2} = 5/3\pm\sqrt{(5/3)^2-1} = \begin{cases}\alpha_1 = 3 \\ \alpha_2 = 1/3 \end{cases}$$

The equations $\color{blue}{y = \alpha_1 x= 3x}$ and $\color{blue}{y= \alpha_2 x= x/3}$ can be substituted seperately for $y$ in the original equation system and they will lead to the same results!

A third method was suggested by J. W. Tanner

Because we have $xy=12/5$ and $x^2+y^2=8$, we use these to evaluate $$(x+y)^2=x^2+2 \cdot xy+y^2=x^2+y^2+2\cdot xy=8+2\cdot (12/5)=8+24/5=40/5+24/5=64/5$$

This means that $(x+y)^2=64/5$ and it follows that $x+y=\pm 8/\sqrt{5}$. From that we get $\color{orange}{y= 8/\sqrt{5}-x} $ or $\color{orange}{y= -8/\sqrt{5} -x}$ that we can again use to eliminate $y$ in the original equation system and to solve for $x$. With these equations for $y$ we will also get the same results at the end!

In all methods we try to express $y$ in terms of $x$. We try to find an equation $y = f(x)$ with $y$ on the left and $x$ on the right. Then we go with our minds back to the original equation system and plug the $f(x)$ expression for the $y$ each time we encounter an $y$ in the original equation system. This way of doing things gives us equations that only contain $x$. We gain equations with one unknown variable $x$ instead of equations with two variables, instead of equations with $x$ and $y$. Equations with just one variable are usually easier to solve.