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

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The discussion focuses on solving a non-linear system of equations involving two variables, x and y. The equations provided are x² + y² + x + y = 530 and xy + x + y = 230. A step-by-step solution is outlined, starting with isolating x in the second equation and substituting it into the first. This leads to a fourth-degree polynomial in y, which is factored to find roots, revealing y = 10 and y = 20 as solutions. The final results yield four pairs of solutions for (x, y), including (10, 20) and (20, 10).
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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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