$a\text{)}\{\begin{array}{l}x-y=1\\ y=x^2-6x+9\end{array}$  

$\{\begin{array}{l}x-{\left(x-3\right)}^2=1\\ y={\left(x-3\right)}^2\end{array}$  

$\{\begin{array}{l}x-\left(x^2-6x+9\right)=1\\ y={\left(x-3\right)}^2\end{array}$  

$\{\begin{array}{l}-x^2+7x-9=1\\ y={\left(x-3\right)}^2\end{array}$

$\{\begin{array}{l}-x^2+7x-10=0\\ y={\left(x-3\right)}^2\end{array}$

Rozwiążmy równanie: 

$-x^2+7x-10=0$  

$\Delta =7^2-4\cdot \left(-1\right)\cdot \left(-10\right)=49-40=9,\sqrt{\Delta }=3$  

$x_1=\frac{-7-3}{-2}=\frac{-10}{-2}=5$  

$x_2=\frac{-7+3}{-2}=\frac{-4}{-2}=2$  

zatem:

$y_1={\left(5-3\right)}^2=2^2=4$  

$y_2={\left(2-3\right)}^2={\left(-1\right)}^2=1$  

więc:

$\{\begin{array}{l}{x}_1=5\\ y_1=4\end{array}$  

lub

$\{\begin{array}{l}{x}_2=2\\ y_2=1\end{array}$  

Interpretacja geometryczna:

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$b\text{)}\{\begin{array}{l}2x-y=0\\ y=-2{\left(x-1\right)}^2+6\end{array}$  

$\{\begin{array}{l}y=2x\\ y=-2{\left(x-1\right)}^2+6\end{array}$  

$\{\begin{array}{l}y=2x\\ 2x=-2{\left(x-1\right)}^2+6\end{array}$  

$\{\begin{array}{l}y=2x\\ 2x=-2\left(x^2-2x+1\right)+6\end{array}$

$\{\begin{array}{l}y=2x\\ 2x=-2x^2+4x-2+6\end{array}$

$\{\begin{array}{l}y=2x\\ 0=-2x^2+2x+4\mid :2\end{array}$

$\{\begin{array}{l}y=2x\\ 0=-x^2+x+2\end{array}$

Rozwiążmy równanie:

$-x^2+x+2=0$  

$\Delta =1^2-4\cdot \left(-1\right)\cdot 2=1+8=9,\sqrt{\Delta }=3$  

$x_1=\frac{-1-3}{-2}=\frac{-4}{-2}=2$  

$x_2=\frac{-1+3}{-2}=\frac{2}{-2}=-1$  

zatem:

$y_1=2\cdot 2=4$  

$y_2=2\cdot \left(-1\right)=-2$  

więc:

$\{\begin{array}{l}{x}_1=2\\ y_1=4\end{array}$  

lub

$\{\begin{array}{l}{x}_2=-1\\ y_2=-2\end{array}$  

Interpretacja geometryczna:

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$c\text{)}\{\begin{array}{l}y=-0,5x^2-2x+3\\ -2x+y=11\end{array}$  

$\{\begin{array}{l}y=-0,5x^2-2x+3\\ y=11+2x\end{array}$  

$\{\begin{array}{l}y=-0,5x^2-2x+3\\ -0,5x^2-2x+3=11+2x\end{array}$  

$\{\begin{array}{l}y=-0,5x^2-2x+3\\ -0,5x^2-4x-8=0\mid \cdot \left(-2\right)\end{array}$

$\{\begin{array}{l}y=-0,5x^2-2x+3\\ x^2+8x+16=0\end{array}$

Rozwiążmy równanie:

$x^2+8x+16=0$  

$\Delta =8^2-4\cdot 1\cdot 16=64-64=0$  

$x=\frac{-8}{2}=-4$  

zatem:

$y=-0,5\cdot 4^2-2\cdot \left(-4\right)+3=-0,5\cdot 16+8+3=-8+8+3=3$  

więc:

$\{\begin{array}{l}x=-4\\ y=3\end{array}$  

Interpretacja geometryczna:

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$d\text{)}\{\begin{array}{l}y=\frac{1}{4}{x}^2+x\mid \cdot 4\\ -5x+2y=-4\mid \cdot 2\end{array}$  

$\{\begin{array}{l}4y=x^2+4x\\ -10x+4y=-8\end{array}$  

$\{\begin{array}{l}4y=x^2+4x\\ 4y=-8+10x\end{array}$  

$\{\begin{array}{l}4y=x^2+4x\\ x^2+4x=-8+10x\end{array}$  

$\{\begin{array}{l}4y=x^2+4x\\ x^2-6x+8=0\end{array}$  

Rozwiążmy równanie:

$x^2-6x+8=0$  

$\Delta ={\left(-6\right)}^2-4\cdot 1\cdot 8=36-32=4,\sqrt{\Delta }=2$  

$x_1=\frac{6-2}{2}=\frac{4}{2}=2$  

$x_2=\frac{6+2}{2}=\frac{8}{2}=4$  

zatem:

$y_1=\frac{1}{4}\cdot 2^2+2=\frac{1}{4}\cdot 4+2=1+2=3$  

$y_2=\frac{1}{4}\cdot 4^2+4=\frac{1}{4}\cdot 16+4=4+4=8$  

więc:

$\{\begin{array}{l}{x}_1=2\\ y_1=3\end{array}$  

lub

$\{\begin{array}{l}{x}_2=4\\ y_2=8\end{array}$  

Interpretacja geometryczna: