Rozwiąż układ równań dla 2x+1=3y-3 i 4x-2y+3=0

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}