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

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  • $\begingroup$ What were "all those methods" that you tried? $\endgroup$ – Parcly Taxel Nov 18 '18 at 6:07
  • $\begingroup$ Substitution, elimination and graphing. $\endgroup$ – Jullian Santos Nov 18 '18 at 6:11
  • $\begingroup$ Take the first equation and insert it into the second $\endgroup$ – Fakemistake Nov 18 '18 at 6:11
  • $\begingroup$ Actually I can't use the graphing so what I meant was substitution and elimination. $\endgroup$ – Jullian Santos Nov 18 '18 at 6:12
  • $\begingroup$ Remember if the amount of variables and amount of equations are equal, you can always solve that system of equations. Just substitute the value of any one variable from one equation in another. $\endgroup$ – PradyumanDixit Nov 18 '18 at 6:12
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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 $(\frac12,\frac13)$.

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    $\begingroup$ In the quadratic equation, it's easier to compute if we just first multiply both sides with $6$ to get $6x^2 -5x +1=0$, and get only integers to compute with. The determinant becomes $(-5)^2 - 4\cdot 6 = 1$ so we get a nice result. $\endgroup$ – Henno Brandsma Nov 18 '18 at 6:27
  • $\begingroup$ Thanks for the answer! $\endgroup$ – Jullian Santos Nov 18 '18 at 6:27
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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, $\;x=-\frac{1}{3}\;\text{and}\; y=-\frac{1}{2}\;$ or $\;y=-\frac{1}{3}\;\text{and}\;x=-\frac{1}{2}.$

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  • $\begingroup$ Would you check your quadratic equation? The signs. $\endgroup$ – user376343 Dec 29 '18 at 10:00
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\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 $(x,y) = \left\{\big(\frac 12,\frac 13 \big),\big(\frac 13,\frac 12 \big)\right\}$

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