#1**+1 **

Find real numbers x and y such that x + y = 6 and x^3 + y^3 = 60.

**Hello Guest!**

\(x+y=6\\ y=6-x\\ x^3+y^3=60\\ x^3+(6-x)^3=60\\ x^3+(36-12x+x^2)(6-x)=60\)

\(x^3+216-36x-72x+12x^2+6x^2-x^3=60\\ 18x^2-108x+216=60\)

\(18x^2-108x+156=0\)

\(x = {-108 \pm \sqrt{11664-4\cdot 18\cdot 156} \over 2\cdot 18}\\ x=\frac{-108\pm 20.785}{36}\)

\(x_1=-2.423\\ x_2=-3.577\) \(y_1=8.423\\ y_2=9.577\)

!

asinus Sep 26, 2020

#2**0 **

x + y = 6.......................(1)

x^3 + y^3 =60.............(2)

y =6 - x

Solve for x:

(6 - x)^3 + x^3 = 60

Expand out terms of the left hand side:

18 x^2 - 108 x + 216 = 60

Divide both sides by 18:

x^2 - 6 x + 12 = 10/3

Subtract 12 from both sides:

x^2 - 6 x = -26/3

Add 9 to both sides:

x^2 - 6 x + 9 = 1/3

Write the left hand side as a square:

(x - 3)^2 = 1/3

Take the square root of both sides:

x - 3 = 1/sqrt(3) or x - 3 = -1/sqrt(3)

Add 3 to both sides:

x = 3 + 1/sqrt(3) or x - 3 = -1/sqrt(3)

Add 3 to both sides

**x = 3 + 1/sqrt(3) and y = 3 - 1/sqrt(3) OR y = 3 + 1/sqrt(3) and x = 3 - 1/sqrt(3) **

Guest Sep 26, 2020