Odpowiedź:
metoda przeciwnych współczynników
[tex]\left \{ {{(x+1)^{2}-x^{2}=15} \atop {x-y=6}} \right.[/tex]
[tex]\left \{ {{x^{2}+2x+1-x^{2}=15}\atop {x-y=6}} \right.[/tex]
[tex]\left \{ {{2x=14} \atop {x-y=6/*(-2)}} \right.[/tex]
[tex]\left\{{{2x=14}\atop{-2x+2y=12}}\right.[/tex]
-----------------
2y=2/:2
y=1
x-1=6
x=6+1
x=7
metoda podstawiania
[tex]\left \{ {{(x+1)^{2}-x^{2} =15} \atop {x-y=6}} \right.[/tex]
[tex]\left \{{{x^{2}+2x+1-x^{2}=15}\atop{x=6+y}} \right.[/tex]
[tex]\left \{ {{2*(6+y)+1=15}\atop{x=6+y}} \right.[/tex]
[tex]\left \{{{12+2y+1=15} \atop{x=6+y}}\right.[/tex]
[tex]\left \{ {{2y=15-13}\atop{x=6+y}}\right.[/tex]
[tex]\left \{{{2y=2/:2}\atop {x=6+y}}\right.[/tex]
[tex]\left \{ {{y=1} \atop {x=6+1}} \right.[/tex]
[tex]\left \{ {{y=1} \atop {x=7}} \right.[/tex]
Szczegółowe wyjaśnienie:
