problema con i sistemi lineari

{5/6·(x + y) + 3/4·(x - y) = 37

{3/7·y + x = 26

equivale a scrivere:

{(5/6·(x + y) + 3/4·(x - y) = 37)·12

{(3/7·y + x = 26)·7

Sviluppando ed ordinando si arriva alla forma normale:

{19·x + y = 444

{7·x + 3·y = 182

che risolto fornisce: [x = 23 ∧ y = 7]

248) 

Numero maggiore $= x;$

numero minore $= y;$

$\begin{Bmatrix}\dfrac{5}{6}(x+y) +\dfrac{3}{4}(x-y) &= 37 \\\dfrac{3}{7}y+x &=26\end{Bmatrix}$

$\begin{Bmatrix}10(x+y) +9(x-y) &= 444 \\3y+7x &=182\end{Bmatrix}$

$\begin{Bmatrix}10x+10y +9x-9y &= 444 \\7x+3y &=182\end{Bmatrix}$

$\begin{Bmatrix}19x+y &= 444 \\7x+3y &=182\end{Bmatrix}$

$\begin{Bmatrix} y =& 444-19x \\7x+3(444-19x) =&182\end{Bmatrix}$

$\begin{Bmatrix} y =& 444-19x \\7x+1332-57x =&182\end{Bmatrix}$

$\begin{Bmatrix} y =& 444-19x \\-50x =&182-1332\end{Bmatrix}$

$\begin{Bmatrix} y =& 444-19x \\-50x =&-1150\end{Bmatrix}$

$\begin{Bmatrix} y =& 444-19x \\\dfrac{-50x}{-50} =&\dfrac{-1150}{-50}\end{Bmatrix}$

$\begin{Bmatrix} y =& 444-19x \\\dfrac{\cancel{-50}x}{\cancel{-50}} =&\dfrac{-1150}{-50}\end{Bmatrix}$

$\begin{Bmatrix} y =& 444-19x \\x =& 23 \end{Bmatrix}$

$\begin{Bmatrix} y =& 444-19·23 \\x =& 23 \end{Bmatrix}$

$\begin{Bmatrix} y =& 444-437 \\x =& 23 \end{Bmatrix}$

$\begin{Bmatrix} y =& 7 \\x =& 23 \end{Bmatrix}$

per cui:

numero maggiore $= x = 23;$

numero minore $= y= 7.$