# Some Really Confusing Math

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 $$y$$. $$y = 4x^3 - 5x^2 + 6x^1 - 8$$

• Hey! Congrats on 3,000! :D keep up the good puzzles. – QuantumTwinkie Oct 1 '18 at 20:23

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 $$x = y +1$$