﻿ Solve the system x^2+xy+y^2=19 \\ x^2+y^2+x+y=18

#### Question

Solve the system

x^2+xy+y^2=19 \\ x^2+y^2+x+y=18

Collected in the board: System of equations

Steven Zheng posted 1 hour ago

#### Answer

Let a = xy, b=x+y, then the system of equations is given as

Since x^2+xy+y^2=(x+y)^2-xy=19

and

x^2+y^2+x+y = (x+y)^2-2xy+x+y=18

b^2-a=19
(1)
b^2-2a+b=18
(2)

Subtract (1) from (2)

a-b=1

Substitute a=b+1 to the first equation

b^2-b-20=0
(3)

Then, b=5 or b=-4

Correspondingly, a = 6, or a = -3

Case 1, if a=6, b=5

xy=6\\x+y=5

According to Vieta's formula

x,y is the roots of the quadratic equation z^2-5z+6=0

Solving the equation gives

\begin{cases} x=2 \\ y=3 \end{cases} \quad \text{or}\begin{cases} x=3 \\ y=2 \end{cases}

Case 2, if a=-3, b=-4

xy=-3\\x+y=-4

According to Vieta's formula

x,y is the roots of the quadratic equation z^2+4z-3=0

Solving the equation gives

\begin{cases} x=-2+\sqrt{7} \\ y=-2-\sqrt{7} \end{cases} \quad \text{or}\begin{cases} x=-2-\sqrt{7} \\ y=-2+\sqrt{7} \end{cases}

Steven Zheng posted 1 hour ago

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