How do you solve these 2 equations?

Question

$$xy = 1/6$$ $$y+x = 5xy$$ I tried solving them using all methods - substitution, elimination and graphing - but can't get the solutions

Answer 1

You can solve this equation by using substitution of variables.

Using the first equation $xy=1/6$, we can rewrite it as $y=\frac1{6x}$.

Plugging this in to the second equation, we get $$x+\frac1{6x}=5\frac{x}{6x}$$

Simplify the right-hand side to $\frac56$ and multiply both sides by $x$ to obtain

$$x^2+\frac16=\frac56x \ \ \ \ \to \ \ \ \ \ x^2-\frac56x+\frac16=0$$

Using the quadratic formula, we now have $$x=\frac{\frac56 \pm \sqrt{\frac{25}{36}-\frac46}}{2}=\frac5{12} \pm \frac12\sqrt{\frac1{36}}=\frac5{12} \pm \frac1{12}$$

Our solutions for $x$ are $\frac13$ and $\frac12$. Using $y=\frac1{6x}$, we get the coordinate pairs to be $(\frac13,\frac12)$ and $(\frac12,\frac13)$.

Answer 2

\begin{array}{c} xy = \frac 16 \\ x + y = 5xy = \frac 56 \end{array}

Let $x = \frac{5}{12} + t$ and $y = \frac{5}{12}-t$. Then $x+y=\frac 56$.

\begin{align} xy &= \frac 16 \\ \bigg(\frac{5}{12} + t \bigg) \bigg(\frac{5}{12} - t\bigg) &= \frac 16 \\ \frac{25}{144} - t^2 &= \frac{24}{144} \\ t^2 &= \frac{1}{144} \\ t &= \pm \frac{1}{12} \\ \end{align}

$\frac{5}{12} + \frac{1}{12} = \frac 12$ and $\frac{5}{12} - \frac{1}{12} = \frac 13$

So $(x,y) = \left\{\big(\frac 12,\frac 13 \big),\big(\frac 13,\frac 12 \big)\right\}$