Question

Решите систему уравнений. Очень надо. Решить без придуриваний, иначе забаню. {(2х+у-1)(2х-у+1)=0
{3у-4х=2

Answer 1

$\left \{ {{(2x+y-1)(2x-y+1)=0} \atop {3y-4x=2}} \right.\\\\\left \{ {{\left[\begin{array}{ccc}2x+y-1=0\\2x-y+1=0\end{array}\right } \atop {3y-4x=2}} \right.\\\\\\\left \{ {{\left[\begin{array}{ccc}y=1-2x\\y=2x+1\end{array}\right } \atop {3y-4x=2}} \right.\\\\\\1)\left \{ {{y=1-2x} \atop {3(1-2x)-4x=2}} \right.\\\\\left \{ {{y=1-2x} \atop {3-6x-4x=2}} \right. \\\\\left \{ {{y=1-2x} \atop {-10x=-1}} \right.\\\\\left \{ {{x=0,1} \atop {y=1-2*0,1}} \right.$

$\left \{ {{x_{1}=0,1 } \atop {y_{1}=0,8 }} \right.\\\\\\2)\left \{ {{y=2x+1} \atop {3(2x+1)-4x=2}} \right.\\\\\left \{ {{y=2x+1} \atop {6x+3-4x=2}} \right.\\\\\left \{ {{y=2x+1} \atop {2x=-1}} \right. \\\\\left \{ {{x=-0,5} \atop {y=2*(-0,5)+1}} \right. \\\\\left \{ {{x_{2} =-0,5} \atop {y_{2}=0 }} \right. \\\\Otvet:\boxed{(0,1 \ ; \ 0,8) \cup \ (-0,5 \ ; \ 0)}$

Answer 2

$\displaystyle \left \{ {{(2x+y-1)(2x-y+1)=0} \atop {3y-4x=2}} \right.$

$\displaystyle \left \{ {{(2*\frac{3y-2}{4}+y-1)(2*\frac{3y-2}{4}-y+1)=0} \atop {x=\frac{3y-2}{4} }} \right.$

$\displaystyle \left \{ {{(\frac{3y-2}{2}+y-1)(\frac{3y-2}{2}-y+1)=0} \atop {x=\frac{3y-2}{4} }} \right.$

$\displaystyle \left \{ {{(3y-2+2y-2)(3y-2-2y+2)=0} \atop {x=\frac{3y-2}{4} }} \right.$

$\displaystyle \left \{ {{(5y-4)*y=0} \atop {x=\frac{3y-2}{4} }} \right.$

y₁= 0                              y₂= 4/5

x₁ = (3*0-2)/4 = -0,5      x₂= (3*0,8-2)/4 = 0,1

ответ: (-0,5; 0),(0,1 ; 0,8)