system of equations

**jmedsy** · January 25th 2010, 06:05 PM

solve the system of equations where x,y are variables and a,b are constants:

x + y = a
(x^2)*y + x*(y^2) = b

**Sudharaka** · January 25th 2010, 06:21 PM

Dear jmedsy,

$x+y=a$ -------A

$(x+y)xy = axy$

$axy=b$ --------B

By solving A and B you could easily get the answer.

**Soroban** · January 25th 2010, 06:57 PM

Hello, jmedsy!

Exactly where is your difficulty?
Straight substitution will solve it.

Solve the system of equations where $x,y$ are variables and $a,b$ are constants:

. . . $\begin{array}{cccc}x + y &= &a & [1] \\<br /> x^2y + xy^2 &=& b & [2] \end{array}$

From [1], we have: . $y \:=\:a-x\;\;[3]$

Substitute into [2]: . $x^2(a-x) + x(a-x)^2 \:=\:b$

Expand: . $ax^2 - x^3 + a^2x - 2ax^2 + x^3 \:=\:b \quad\Rightarrow\quad ax^2 - a^2x + b \:=\:0$

Quadratic Formula: . $\boxed{x \;=\;\frac{a^2 \pm\sqrt{a^4-4ab}}{2a}}$

Substitute into [3]: . $y \;=\;a - \frac{a^2\pm\sqrt{a^4-4ab}}{2a} \quad\Rightarrow\quad\boxed{ y \;=\;\frac{a^2 \mp\sqrt{a^4-4ab}}{2a}}$

**jmedsy** · January 25th 2010, 07:04 PM

I get this answer. When I test the solutions in maple, they work in the first equation but not for the second.

nevermind i'm doing some really bizarre algebra. everything is ok now, thanks for the help

**HallsofIvy** · January 26th 2010, 04:26 AM

Mistaken post

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