{ x + y = 3 3 x + 1 + 3 y = 18 {\displaystyle {\begin{cases}x+y=3\\3^{x+1}+3^{y}=18\end{cases}}}
נציב:
3 x + 1 + 3 3 − x = 18 3 x + 1 + 3 3 − x = 3 2 ∗ 2 / : 3 3 x + 3 2 − x = 3 ∗ 2 / : 3 3 x + 3 2 3 x = ∗ 2 3 2 x + 3 2 = 3 ∗ 2 ∗ 3 x 3 x = t t 2 − 6 t + 9 = 0 ( t − 3 ) 2 = 0 t = 3 3 x = 3 x = 1 y ( 1 ) = 3 − 1 = 2 ( 1 , 2 ) {\displaystyle {\begin{aligned}3^{x+1}+3^{3-x}=18\\3^{x+1}+3^{3-x}=3^{2}*2/:3\\3^{x}+3^{2-x}=3*2/:3\\3^{x}+{\frac {3^{2}}{3^{x}}}=*2\\3^{2x}+3^{2}=3*2*3^{x}\\3^{x}=t\\t^{2}-6t+9=0\\(t-3)^{2}=0\\t=3\\3^{x}=3\\x=1\\y(1)=3-1=2\\(1,2)\end{aligned}}}
