David's Math: Solving a Non-Linear System in 2 Variables

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SUMMARY

The discussion focuses on solving a non-linear system of equations involving two variables, specifically the equations x² + y² + x + y = 530 and xy + x + y = 230. The solution process involves substituting one equation into the other, resulting in a fourth-degree polynomial. The rational roots theorem identifies y = 10 and y = 20 as roots, leading to the factorization of the polynomial and the application of the quadratic formula to find additional solutions. The final solutions for the pairs (x, y) are (10, 20), (20, 10), and two additional complex solutions derived from the quadratic factor.

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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.
 
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Hello David,

We are given the two equations:

$$x^2+y^2+x+y=530$$

$$xy+x+y=230$$

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

$$x=\frac{230-y}{y+1}$$

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

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

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

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

Distributing and collecting like terms, we obtain:

$$y^4+3y^3-527y^2-1290y+52600=0$$

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:

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

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

$$y=\frac{-33\pm\sqrt{37}}{2}$$

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

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

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