UKŁADY RÓWNAŃ: METODA Podstawiania 5 (x+y)+5=2 (2x+y) 2 (x-1)=y+1 Proszę o szybkie wykonanie daje naj!!!

{5 (x+y)+5=2 (2x+y) {2 (x-1)=y+1 {5x+5y+5=4x+2y {2x-2=y+1 {5x-4x=2y-5y-5 {2x-2=y+1 {x=-3y-5 {2 (-3y-5)-2=y+1 {x=-3y-5 {-6y-10-2=y+1 {x=-3y-5 {-6y-y=1+12 {x=-3y-5 {-7y=13 | :(-7) {x = -3y-5 {y = -13/7 {x = -3×(-13/7) -5 {y = - 1 6/7 {x = 39/7 - 5 {y = - 1 6/7 {x = 5 4/7 - 5 {y = -1 6/7 {x = 4/7 {y = -1 6/7

[latex] left { {{5(x+y)+5=2(2x+y)} atop {2(x-1)=y+1}} ight. \ \ left { {{5x+5y+5=4x+2y} atop {2x-2=y+1}} ight. \ \ left { {{5x+5y-4x-2y=-5} atop {2x-y=1+2}} ight. \ \ left { {{x+3y=-5} atop {2x-y=3}} ight. \ x=-5-3y \ 2(-5-3y)-y=3 \ -10-6y-y=3 \ -10-7y=3 \ -7y=3+10 \ -7y=13/:(-7) \ y=- frac{13}{7} \ \ x=-5-3y \ x=-5-3*(- frac{13}{7}) \ x=-5+ frac{39}{7} \ x=-5+5 frac{4}{7} \ x= frac{4}{7} \ \ [/latex] [latex] left { {{x= frac{4}{7} } atop {y=- frac{13}{7} }} ight. [/latex]