Welcome to our community

Be a part of something great, join today!

David's question at Yahoo! Answers regarding a non-linear system of equations in two variables

  • Thread starter
  • Admin
  • #1

MarkFL

Pessimist Singularitarian
Staff member
Feb 24, 2012
13,684
St. Augustine, FL.
Here is the question:

David said:
Very hard math problem.?

Please I need help. Could someone calculate this, please.

x² + y² + x + y = 530
xy + x + y = 230

I know that x= 10, y=20, but I need how to calculate it step by step. please please please :)
I have posted a link there so the OP can view my work.
 
  • Thread starter
  • Admin
  • #2

MarkFL

Pessimist Singularitarian
Staff member
Feb 24, 2012
13,684
St. Augustine, FL.
Hello David,

We are given the two equations:

\(\displaystyle x^2+y^2+x+y=530\)

\(\displaystyle xy+x+y=230\)

If we solve the second equation for $x$, we obtain:

\(\displaystyle x=\frac{230-y}{y+1}\)

And then substituting for $x$ into the first equation, we get:

\(\displaystyle \left(\frac{230-y}{y+1}\right)^2+y^2+\frac{230-y}{y+1}+y=530\)

Multiplying through by $(y+1)^2$, and factoring a little and rearranging, there results:

\(\displaystyle (230-y)^2+y(y+1)^3+(230-y)(y+1)-530(y+1)^2=0\)

Distributing and collecting like terms, we obtain:

\(\displaystyle y^4+3y^3-527y^2-1290y+52600=0\)

Utilizing the rational roots theorem, we find that $y=10$ and $y=20$ are roots, and performing the division, we find the equation may be factored as:

\(\displaystyle (y-20)(y-10)\left(y^2+33y+263\right)=0\)

Using the quadratic formula on the quadratic factor, we find the remaining two roots:

\(\displaystyle y=\frac{-33\pm\sqrt{37}}{2}\)

And thus, using the formula for $x$ as a function of $y$ we found earlier, we find the four solutions:

\(\displaystyle \bbox[10px,border:2px solid #207498]{(x,y)=(10,20),\,(20,10),\,\left(-\frac{33+\sqrt{37}}{2},\frac{-33+\sqrt{37}}{2}\right),\,\left(\frac{-33+\sqrt{37}}{2},-\frac{33+\sqrt{37}}{2}\right)}\)