# Solving $x^3+y^3=7$ and $x^2+y^2+x+y+xy=4$ System of Equations

• MHB
• anemone
In summary, to solve a system of equations with both a cubic and quadratic term, use a combination of algebraic manipulation and substitution. It is possible to solve this system algebraically and there can be multiple solutions, depending on the given equations. Advanced methods such as matrices or Gaussian elimination can also be used. To check if solutions are correct, plug them back into both equations.
anemone
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MHB
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Find all real $x$ and $y$ that satisfy the system $x^3+y^3=7$ and $x^2+y^2+x+y+xy=4$.

We are given

$x^3+ y^3 = 7\cdots(1)$

$x^2 + y^2 + x + y + xy = 4\cdots(2)$

Let us choose x+y = a and xy = b

then $1^{st}$ equation become

$x^3+y^3 = (x+y)^3 - 3xy(x+y) = 7$

or $a^3 - 3ab = 7\cdots(3)$

The $2^{nd}$ equation is

$x^2 + y^2 + xy + x + y = 4$

Or $(x+y)^2 - xy + x + y = 4$

putting $x + y = a$ and $xy = b$ we get

$a^2 - b + a = 4$

or $a^2 + a - b = 4\cdots(4)$

multiplying (4) by 3a and subtracting (3) from it we get

$2a^3 + 3a^2 = 5$

or $2a^3 + 3a^2 -5= 0$

as a = 1 is a solution so we have

$2a^3 + 3a^2 - 5 = 2a^2(a-1) + 2a^2 + 3a^2 - 5 = 2a^2(a-1) + 5(a^2 - 1)$

$= 2a^2(a-1) + 5(a+1)(a-1) = 2a^2 + 5a + 5)(a-1) = 0$

so a = 1 or $2a^2 + 5a + 5 = 0$ this does not have any real solution

so a = 1

putting it in (3) we get $3ab = 1 - 7 = -6$

or b = -2

so we have x+y = 1 and xy = -2 giving 2 sets of solution (2,-1) or (-1,2)

so the solution set is

$(x, y) \in \{ (2,-1),(-1,2)\}$

## What is the system of equations $x^3+y^3=7$ and $x^2+y^2+x+y+xy=4$?

The system of equations $x^3+y^3=7$ and $x^2+y^2+x+y+xy=4$ is a set of two equations with two variables, $x$ and $y$, that must be solved simultaneously. This means that the values of $x$ and $y$ must satisfy both equations at the same time.

## What is the degree of the equations in this system?

The degree of an equation is the highest power of the variable present. In this system, the first equation has a degree of 3 and the second equation has a degree of 2.

## Can this system be solved algebraically?

Yes, this system can be solved algebraically by using techniques such as substitution or elimination. However, the solutions may involve complex numbers.

## Are there any other methods for solving this system?

Yes, this system can also be solved graphically by plotting the two equations on the same coordinate plane and finding the points of intersection. Additionally, numerical methods such as Newton's method or the bisection method can be used to approximate the solutions.

## How many solutions does this system have?

This system has an infinite number of solutions, as there are infinitely many combinations of $x$ and $y$ that satisfy both equations. However, there may be a finite number of real solutions depending on the values of the coefficients in the equations.

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