Matematyka
Rozwiązanie zadanego układu równań metodą podstawiania:
$\begin{cases}2x+1=3y-3
\\ 4x-2y+3=0
\end{cases}$
$-3y+2x=-4-2y+4x=-3
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i: Ostateczne rozwiązanie to:
4x = 2y - 3
x = y/2 - 3/4
2•(y /2-3/4) - 3y = -4
- 2y = -5/2
- 4y = -5
4y = 5
y = 5/4
x = y/2-3/4
y = 5/4 x = (/2)(5/4)-3/4 = -1/8
{x,y} = {-1/8,5/4}