[tex]\left \{ {{\frac{2x-1}{2} -\frac{y+1}{3} =-2~~\mid \cdot 6} \atop {(x+1)^{2} -2\cdot (2y-1)=x\cdot (x+1)}} \right. \\\\\\\left \{ {{3\cdot (2x-1)-2\cdot (y+1)=-12} \atop {x^{2} +2x+1-4y+2=x^{2} +x}} \right. \\\\\left \{ {{6x-3-2y-2=-12} \atop {2x-4y+3=x}} \right. \\\\\\\left \{ {{6x-2y=-12+5} \atop {x-4y=-3}} \right. \\\\\\\left \{ {{6x-2y=-7~~\mid \cdot (-2)} \atop {x-4y=-3}} \right. \\\\\\\left \{ {{-12x+4y=14} \atop {x-4y=-3}} \right. ~~\mid~~dodaje~~stronami\\\\[/tex]
[tex]-12x+4y+x-4y=14-3\\\\-11x=11~~\mid \div (-11)\\\\\\x=-1\\\\x-4y=-3~~~\land ~~x=-1~~\Rightarrow~~-1-4y=-3\\\\-1-4y=-3\\\\-4y=-3+1\\\\-4y=-2~~\mid \div (-4)\\\\y=\dfrac{1}{2} \\\\[/tex]
Odp: Rozwiązaniem układu równań jest para liczb [tex]\left \{ {{x=-1} \atop {y=\frac{1}{2} }} \right.[/tex].
Sprawdzam:
[tex]\dfrac{2x-1}{2} -\dfrac{y+1}{3} =-2 ~~\land~~x=-1~~\land~~y=\frac{1}{2}\\\\\dfrac{2\cdot (-1)-1}{2} -\dfrac{\frac{1}{2} +1}{3} =-2\\\\-\dfrac{3}{2} -\dfrac{3}{2} \cdot \dfrac{1}{3} =-2\\\\-1\dfrac{1}{2} -\dfrac{1}{2} =-2\\\\-2=-2\\\\L=P[/tex]
[tex](x+1)^{2} -2\cdot (2y-1)=x\cdot (x+1)~~\land~~x=-1~~\land~~y=\dfrac{1}{2} \\\\(-1+1)^{2} -2\cdot (2\cdot \dfrac{1}{2} -1)=(-1)\cdot (-1+1)\\\\0^{2} -2\cdot 0=(-1)\cdot 0\\\\0-0=0\\\\0=0\\\\L=P[/tex]
