# system of equations

• Jan 25th 2010, 06:05 PM
jmedsy
system of equations
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
• Jan 25th 2010, 06:21 PM
Sudharaka
Dear jmedsy,

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

$(x+y)xy = axy$

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

By solving A and B you could easily get the answer.
• Jan 25th 2010, 06:57 PM
Soroban
Hello, jmedsy!

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

Quote:

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

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

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

• Jan 25th 2010, 07:04 PM
jmedsy
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
• Jan 26th 2010, 04:26 AM
HallsofIvy
Mistaken post