Sistemi frazionari

{6·x·y = 1

{(3·x + 1)/(2 - 2·y) = 2

Per sostituzione:

x = 1/(6·y)

(3·(1/(6·y)) + 1)/(2 - 2·y) = 2

(2·y + 1)/(4·y - 4·y^2) = 2

(2·y + 1)/(4·y - 4·y^2) - 2 = 0

(8·y^2 - 6·y + 1)/(4·y·(1 - y)) = 0

posto: 

4·y·(1 - y) ≠ 0----> y ≠ 1 ∧ y ≠ 0

8·y^2 - 6·y + 1 = 0

risolvo ed ottengo:

y = 1/4 ∨ y = 1/2

per 

y = 1/4 : x = 1/(6·(1/4))---> x = 2/3

y=1/2 : x = 1/(6·(1/2))---> x = 1/3

Soluzione sistema: [x = 1/3 ∧ y = 1/2, x = 2/3 ∧ y = 1/4]

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$\small \begin{Bmatrix}{6xy} & {=} &{1}\\ {\dfrac{3x+1}{2-2y}} & {=} & {2}\end{Bmatrix}$

lavoriamo sulla 2° equazione:

$\small \begin{Bmatrix}{6xy} & {=} &{1}\\ {\dfrac{3x+1}{2(1-y)}} & {=} & {2}\end{Bmatrix}$

$\small \begin{Bmatrix}{6xy} & {=} &{1}\\ {3x+1} & {=} & {2·2(1-y)}\end{Bmatrix}$

$\small \begin{Bmatrix}{6xy} & {=} &{1}\\ {3x+1} & {=} & {4(1-y)}\end{Bmatrix}$

$\small \begin{Bmatrix}{6xy} & {=} &{1}\\ {3x+1} & {=} & {4-4y}\end{Bmatrix}$

$\small \begin{Bmatrix}{6xy} & {=} &{1}\\ {3x} & {=} & {4-4y-1}\end{Bmatrix}$

$\small \begin{Bmatrix}{6xy} & {=} &{1}\\ {3x} & {=} & {3-4y}\end{Bmatrix}$

$\small \begin{Bmatrix}{6xy} & {=} &{1}\\ {\dfrac{\cancel3x}{\cancel3}} & {=} & {\dfrac{3-4y}{3}}\end{Bmatrix}$

$\small \begin{Bmatrix}{6xy} & {=} &{1}\\ {x} & {=} & {\dfrac{3}{3}-\dfrac{4y}{3}}\end{Bmatrix}$

$\small \begin{Bmatrix}{6xy} & {=} &{1}\\ {x} & {=} & {1-\dfrac{4y}{3}}\end{Bmatrix}$

ora lavoriamo sulla 1° equazione sostituendo la x come segue:

$\small \begin{Bmatrix}{6(1-\dfrac{4}{3}y)·y} & {=} &{1}\\ {x} & {=} & {1-\dfrac{4y}{3}}\end{Bmatrix}$

$\small \begin{Bmatrix}{(6-\dfrac{\cancel{24}^8}{\cancel3_1}y)·y} & {=} &{1}\\ {x} & {=} & {1-\dfrac{4y}{3}}\end{Bmatrix}$

$\small \begin{Bmatrix}{(6-8y)·y} & {=} &{1}\\ {x} & {=} & {1-\dfrac{4y}{3}}\end{Bmatrix}$

$\small \begin{Bmatrix}{6y-8y^2} & {=} &{1}\\ {x} & {=} & {1-\dfrac{4y}{3}}\end{Bmatrix}$

$\small \begin{Bmatrix}{-8y^2+6y-1} & {=} &{0}\\ {x} & {=} & {1-\dfrac{4y}{3}}\end{Bmatrix}$

$\small \begin{Bmatrix}{8y^2-6y+1} & {=} &{0}\\ {x} & {=} & {1-\dfrac{4y}{3}}\end{Bmatrix}$

equazione di 2° grado completa, risolviamo con i seguenti dati:

$\small a= 8; b= -6; c= 1;$

$\small \Delta= b^2-4ac = (-6)^2-4·8·1 = 36-32 = 4;$

applica la formula risolutiva:

$\small y_{1,2}= \dfrac{-b\pm\sqrt{\Delta}}{2a} = \dfrac{-(-6)\pm\sqrt4}{2·8} = \dfrac{6\pm2}{16}$

$\small y_1= \dfrac{6-2}{16} = \dfrac{4}{16} = \dfrac{1}{4};$

$\small y_2= \dfrac{6+2}{16} = \dfrac{8}{16} = \dfrac{1}{2};$

ora sostituisci la y trovata nella 2° equazione:

$\small x_1= 1-\dfrac{4}{3}y;$

$\small x_1 = 1-\dfrac{\cancel4^1}{3}·\dfrac{1}{\cancel4_1}$

$\small x_1= 1-\dfrac{1}{3}$

$\small x_1= \dfrac{3-1}{3} = \dfrac{2}{3};$

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$\small x_2= 1-\dfrac{4}{3}y;$

$\small x_2 = 1-\dfrac{\cancel4^2}{3}·\dfrac{1}{\cancel2_1}$

$\small x_2= 1-\dfrac{2}{3}$

$\small x_2= \dfrac{3-2}{3} = \dfrac{1}{3};$

risultati:

$\small x= \dfrac{2}{3} ∧ \dfrac{1}{3};$

$\small y= \dfrac{1}{4} ∧ \dfrac{1}{2};$