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I was talking to a co-worker this morning and he stated that he was able to make both of these statements true using only $1, 3, 5,$ or $ 7$. He then proceeded to tell me:

  • $1 = 5$
  • $2 = 4 + 7$
  • $3 = 1$
  • $4 = 5$
  • $5 = 6 - 3$
  • $6 = 4$
  • $7 = 5 + 1$
  • $8 = 2 - 3 + 5$

While solving for $x$ and $y$ make both statements true.

Solve for $x$. $$x = \frac{1 \cdot 3}{7 + 3^7} + 2y^2$$ Solve for $y$. $$y = 4x^3 - 5x^2 + 6x^1 - 8$$

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  • $\begingroup$ Hey! Congrats on 3,000! :D keep up the good puzzles. $\endgroup$ – QuantumTwinkie Oct 1 '18 at 20:23
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Not 100% sure this is how this is supposed to be solved but this is what I did.
For x:

$x = \frac{1\cdot3}{7+3^7} +2y^2$
$x = \frac{5\cdot1}{(5+1)+1^7} +(4+7)y^2$
$x = \frac{5}{((6-3)+1)+1} +(5+5+1)y^2$
$x = \frac{5}{5} +((6-3)+(6-3)+1)y^2$
$x = 1 +((4-1)+(4-1)+1)y^2$
$x = 1 +((5-5)+(5-5)+1)y^2$
$x = 1 +1y^2$
Finally turn the $y^2$ to a y by changing the 2 to 1 by the same process that the coefficient of y was turned from a 2 to a 1 giving you : $x = 1+y$

For y:

$y = 4x^3-5x^2+6x^1-8$
$y = 4x^1 +6x^1-(6-3)x^2 -(2-3+5)$
$y = 4x^1 +4x^1-(4-3)x^2 -(2-3+(6-3))$
$y = 8x-(5-3)x^2 -(8-6))$
$y = 2x-3x+5x-((6-3)-3)x^2 -(2-3+5-6))$
$y = 2x-x+6x-3x-0x^2 -(4-6))$
$y = 4x-(5-6))$
$y = 5x-(6-3-6))$
$y = 6x-3x-(6-3-4))$
$y =3x-(6-1-4))$
$y =1x-(1))$
which has equality with $x = y +1$

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