Question

Помогите пожалуйста!!!

Answer 1

$\left \{ {{y-5x=1} \atop {y^2-13x=23}} \right. \; \left \{ {{y=5x+1} \atop {(5x+1)^2-13x=23}} \right. \\\\25x^2-3x-22=0\\\\D=2209,\sqrt{D}=47\\\\x_1=\frac{3-47}{50}=-\frac{44}{50}=-\frac{22}{25}\\\\x_2=\frac{3+47}{50}=1\\\\y_1=-5\cdot \frac{22}{25}+1=\frac{-22+5}{25}=-\frac{17}{5}\\\\y_2=5\cdot 1+1=6\\\\Otvet:\; (-\frac{22}{25};-\frac{17}{5}),\; (1,6).$

Answer 2

у=1+5х
у^2-13х=23

(1+5х)^2-13х=23
1+10х+25х^2-13х-23=0
25х^2-3х-22=0
д=9-4×25×(-22)=2209
х1=(3-47)/2×25=-44/50=-22/25
х2=(3+47)/2×25=1
у1=1+5×(-22/25)=1-22/5=1-4 2/5=-3 2/5
у2=1+5×1=6
(-22/25;-3 2/5) (1;6)