Résoudre le système par substitution [tex] \left \{ {{2x-y=-5} \atop {8x+3y=8}} \right. [/tex]

-y = -2x-5

8x-3(-2x-5) = 8
8x-(-6x-15) = 8
8x+6x+15 = 8
14x = 8-15
14x = -7
x = -1/2

2x-y = -5
-1-y =-5
-y = -5+1
y = 4
-y = -4
y = 4

Bonjour Marinabose

[tex]\left\{\begin{matrix}2x-y=-5\\8x+3y=8\end{matrix}\right. \\\\\left\{\begin{matrix}y=2x+5\\8x+3y=8\end{matrix}\right.\\\\\left\{\begin{matrix}y=2x+5\\8x+3(2x+5)=8\end{matrix}\right. \\\\\left\{\begin{matrix}y=2x-5\\8x+6x+15=8\end{matrix}\right.\\\\\left\{\begin{matrix}y=2x-5\\14x=8-15\end{matrix}\right.[/tex]

[tex]\\\\\left\{\begin{matrix}y=2x-5\\14x=-7\end{matrix}\right.\\\\\left\{\begin{matrix}y=2x-5\\x=-\dfrac{7}{14}\end{matrix}\right.\\\\\left\{\begin{matrix}y=2x-5\\x=-\dfrac{1}{2}\end{matrix}\right.\\\\\left\{\begin{matrix}y=2\times(-\dfrac{1}{2})-5\\\\x=-\dfrac{1}{2}\end{matrix}\right.[/tex]

[tex]\\\\\left\{\begin{matrix}y=-1+5\\\\x=-\dfrac{1}{2}\end{matrix}\right.\\\\\\\left\{\begin{matrix}y=4\\\\x=-\dfrac{1}{2}\end{matrix}\right.[/tex]

Par conséquent, 

[tex]\boxed{S=\{(-\dfrac{1}{2},4)\}}[/tex]