Krdytkyu

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

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 $$x+frac1{6x}=5frac{x}{6x}$$

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 $$x=frac{frac56 pm sqrt{frac{25}{36}-frac46}}{2}=frac5{12} pm frac12sqrt{frac1{36}}=frac5{12} pm frac1{12}$$

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

1) Substitute xy=$frac{1}{6}$ into the 2nd equation to get x+y=$frac{5}{6}$

2) Solving xy=$frac{1}{6}$ and x+y=$frac{5}{6}$ is equivalent to finding the zeros of the function f(z)=$z^2$+$frac{5}{6}$z+$frac{1}{6}$=0, so using the quadratic formula, x=$-frac{1}{3}$ & y=$-frac{1}{2}$ or y=$-frac{1}{3}$ & x=$-frac{1}{2}$.