[Risolto] Sistemi lineari

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$\small \begin{Bmatrix}
\dfrac{2x-3}{5}+\dfrac{x+1}{2} & = & \dfrac{-3x+4y}{10} \\
\dfrac{3x-2}{3}-x & = & 1+2y
\end{Bmatrix}$

$\small \begin{Bmatrix}
2(2x-3)+5(x+1) & = & -3x+4y \\
\cancel{3x}-2\cancel{-3x} & = & 3+6y
\end{Bmatrix}$

$\small \begin{Bmatrix}
4x-6+5x+5 & = & -3x+4y \\
-6y &= & 3+2
\end{Bmatrix}$

$\small \begin{Bmatrix}
9x-1 & = & -3x+4y \\
-6y &= & 5
\end{Bmatrix}$

$\small \begin{Bmatrix}
9x+3x-4y & = & 1 \\
6y &= & -5
\end{Bmatrix}$

$\small \begin{Bmatrix}
12x-4y & = & 1 \\
\dfrac{\cancel6y}{\cancel6} &= & \dfrac{-5}{6}
\end{Bmatrix}$

$\small \begin{Bmatrix}
12x-4y & = & 1 \\
y &= & -\dfrac{5}{6}
\end{Bmatrix}$

sostituisci la $\small y$ trovata nella 1° equazione:

$\small \begin{Bmatrix}
12x-4·\left(-\dfrac{5}{6}\right) & = & 1 \\
y &= & -\dfrac{5}{6}
\end{Bmatrix}$

$\small \begin{Bmatrix}
12x+\dfrac{\cancel{20}^{10}}{\cancel6_3} & = & 1 \\
y &= & -\dfrac{5}{6}
\end{Bmatrix}$

$\small \begin{Bmatrix}
12x+\dfrac{10}{3} & = & 1 \\
y &= & -\dfrac{5}{6}
\end{Bmatrix}$

$\small \begin{Bmatrix}
36x+10 & = & 3 \\
y &= & -\dfrac{5}{6}
\end{Bmatrix}$

$\small \begin{Bmatrix}
36x & = & 3-10 \\
y &= & -\dfrac{5}{6}
\end{Bmatrix}$

$\small \begin{Bmatrix}
36x & = & -7 \\
y &= & -\dfrac{5}{6}
\end{Bmatrix}$

$\small \begin{Bmatrix}
\dfrac{\cancel{36}x}{\cancel{36}} & = & \dfrac{-7}{36} \\
y &= & -\dfrac{5}{6}
\end{Bmatrix}$

$\small \begin{Bmatrix}
x & = & -\dfrac{7}{36} \\
y &= & -\dfrac{5}{6}
\end{Bmatrix}$

{(2·x - 3)/5 + (x + 1)/2 = (- 3·x + 4·y)/10

{(3·x - 2)/3 - x = 1 + 2·y

---------------------------------

{2·(2·x - 3) + 5·(x + 1) - (- 3·x + 4·y) = 0

{(3·x - 2) - 3·x - 3 - 6·y = 0

-------------------------------

{(4·x - 6) + (5·x + 5) - (- 3·x + 4·y) = 0

{- 6·y - 5 = 0

----------------------------------

{12·x - 4·y = 1

{- 6·y - 5 = 0

Dalla seconda: y = - 5/6

sostituendo nella prima:

12·x - 4·(- 5/6) = 1

12·x = 1 - 10/3

12·x = - 7/3----> x = - 7/36

Soluzione sistema determinato: [x = - 7/36 ∧ y = - 5/6]