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# This is a repost of 3 hours ago

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Let $(x_1,y_1),$ $(x_2,y_2),$ $\dots,$ $(x_n,y_n)$ be the real solutions to the system \begin{align*} x + 8y &= 7, \\ x^3 + 8y^3 &= 7. \end{align*}Enter $x_1 + y_1 + x_2 + y_2 + \dots + x_n + y_n.$

Dec 5, 2020

#1
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Set the 2 equations equal to each other. You will get

x³+8y³=x+8y.

Try to solve this now.

Dec 5, 2020
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That's what I did and that's where I am confused

Dec 5, 2020
#3
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x + 8y =7
x = 7 - 8y   sub this into the 2nd equation
(7 - 8y)^3 + 8y^3 = 7, solve for y
Solve for y:
(7 - 8 y)^3 + 8 y^3 = 7

Expand out terms of the left hand side:
-504 y^3 + 1344 y^2 - 1176 y + 343 = 7

Subtract 7 from both sides:
-504 y^3 + 1344 y^2 - 1176 y + 336 = 0

The left hand side factors into a product with three terms:
-168 (y - 1)^2 (3 y - 2) = 0

Divide both sides by -168:
(y - 1)^2 (3 y - 2) = 0

Split into two equations:
(y - 1)^2 = 0 or 3 y - 2 = 0

Take the square root of both sides:
y - 1 = 0 or 3 y - 2 = 0

y = 1 or 3 y - 2 = 0

y = 1 or 3 y = 2

Divide both sides by 3:

y = 1 or y = 2/3  sub these two values into the 1st equation to get:

x = - 1  or  x =5/3

Dec 5, 2020
#4
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Wow I din't think of this that way

Dec 5, 2020