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- Feb 14, 2012

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Find all real $x$ and $y$ that satisfy the system $x^3+y^3=7$ and $x^2+y^2+x+y+xy=4$.

- Thread starter anemone
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- Thread starter
- Admin
- #1

- Feb 14, 2012

- 3,802

Find all real $x$ and $y$ that satisfy the system $x^3+y^3=7$ and $x^2+y^2+x+y+xy=4$.

- Mar 31, 2013

- 1,331

$x^3+ y^3 = 7\cdots(1)$

$x^2 + y^2 + x + y + xy = 4\cdots(2)$

Let us choose x+y = a and xy = b

then $1^{st}$ equation become

$x^3+y^3 = (x+y)^3 - 3xy(x+y) = 7$

or $a^3 - 3ab = 7\cdots(3)$

The $2^{nd}$ equation is

$x^2 + y^2 + xy + x + y = 4$

Or $(x+y)^2 - xy + x + y = 4$

putting $ x + y = a$ and $xy = b$ we get

$a^2 - b + a = 4$

or $a^2 + a - b = 4\cdots(4)$

multiplying (4) by 3a and subtracting (3) from it we get

$2a^3 + 3a^2 = 5$

or $2a^3 + 3a^2 -5= 0$

as a = 1 is a solution so we have

$2a^3 + 3a^2 - 5 = 2a^2(a-1) + 2a^2 + 3a^2 - 5 = 2a^2(a-1) + 5(a^2 - 1)$

$= 2a^2(a-1) + 5(a+1)(a-1) = 2a^2 + 5a + 5)(a-1) = 0$

so a = 1 or $2a^2 + 5a + 5 = 0$ this does not have any real solution

so a = 1

putting it in (3) we get $3ab = 1 - 7 = -6$

or b = -2

so we have x+y = 1 and xy = -2 giving 2 sets of solution (2,-1) or (-1,2)

so the solution set is

$(x, y) \in \{ (2,-1),(-1,2)\}$