$\text{a)}\{\begin{array}{l}\left(y-x\right)\left(y-3x\right)=0\\ x^2+y^2=50\end{array}$  

$\{\begin{array}{l}y-x=0\text{lub}y-3x=0\\ x^2+y^2=50\end{array}$  

$\{\begin{array}{l}y-x=0\\ x^2+y^2=50\end{array}\text{lub}\{\begin{array}{l}y-3x=0\\ x^2+y^2=50\end{array}$  

$\{\begin{array}{l}y=x\\ x^2+y^2=50\end{array}\text{lub}\{\begin{array}{l}y=3x\\ x^2+y^2=50\end{array}$  

Podstawiamy wyznaczone wartości y do drugiego równania w układzie.

$\{\begin{array}{l}y=x\\ x^2+x^2=50\end{array}\text{lub}\{\begin{array}{l}y=3x\\ x^2+{\left(3x\right)}^2=50\end{array}$  

$\{\begin{array}{l}y=x\\ 2x^2=50\mid :2\end{array}\text{lub}\{\begin{array}{l}y=3x\\ x^2+9x^2=50\end{array}$  

$\{\begin{array}{l}y=x\\ x^2=25\end{array}\text{lub}\{\begin{array}{l}y=3x\\ 10x^2=50\mid :10\end{array}$  

$\{\begin{array}{l}y=x\\ x^2=25\end{array}\text{lub}\{\begin{array}{l}y=3x\\ x^2=5\end{array}$  

$\{\begin{array}{l}y=x\\ x=-5\text{lub}x=5\end{array}\text{lub}\{\begin{array}{l}y=3x\\ x=-\sqrt{5}\text{lub}x=\sqrt{5}\end{array}$  

$\{\begin{array}{l}x=-5\\ y=x\end{array}\text{lub}\{\begin{array}{l}x=5\\ y=x\end{array}\text{lub}\{\begin{array}{l}x=-\sqrt{5}\\ y=3x\end{array}\text{lub}\{\begin{array}{l}x=\sqrt{5}\\ y=3x\end{array}$  

Podstawiamy wyznaczone wartości x do drugiego równania w układzie.

$\{\begin{array}{l}x=-5\\ y=-5\end{array}\text{lub}\{\begin{array}{l}x=5\\ y=5\end{array}\text{lub}\{\begin{array}{l}x=-\sqrt{5}\\ y=-3\sqrt{5}\end{array}\text{lub}\{\begin{array}{l}x=\sqrt{5}\\ y=3\sqrt{5}\end{array}$  

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$\text{b)}\{\begin{array}{l}{x}^2+2xy+y^2=4\\ x+y+xy+3,25=0\end{array}$  

$\{\begin{array}{l}{\left(x+y\right)}^2=4\\ x+y+xy+3,25=0\end{array}$  

$\{\begin{array}{l}x+y=-2\text{lub}x+y=2\\ x+y+xy+3,25=0\end{array}$  

$\{\begin{array}{l}x+y=-2\\ x+y+xy+3,25=0\end{array}\text{lub}\{\begin{array}{l}x+y=2\\ x+y+xy+3,25=0\end{array}$  

$\{\begin{array}{l}x=-y-2\\ x+y+xy+3,25=0\end{array}\text{lub}\{\begin{array}{l}x=-y+2\\ x+y+xy+3,25=0\end{array}$  

Podstawiamy wyznaczone wartości x do drugiego równania w układzie.

$\{\begin{array}{l}x=-y-2\\ -y-2+y+\left(-y-2\right)y+3,25=0\end{array}\text{lub}\{\begin{array}{l}x=-y+2\\ -y+2+y+\left(-y+2\right)y+3,25=0\end{array}$  

$\{\begin{array}{l}x=-y-2\\ \underset{\Delta =4+5=9,\sqrt{\Delta }=3}{-y^2-2y+1,25=0}\end{array}\text{lub}\{\begin{array}{l}x=-y+2\\ \underset{4+21=25,\sqrt{\Delta }=5}{-y^2+2y+5,25=0}\end{array}$  

$\{\begin{array}{l}x=-y-2\\ y=\frac{2-3}{-2}=\frac{-1}{-2}=\frac{1}{2}\text{lub}y=\frac{2+3}{-2}=\frac{5}{-2}=-\frac{5}{2}\end{array}\text{lub}\{\begin{array}{l}x=-y+2\\ y=\frac{-2-5}{-2}=\frac{-7}{-2}=\frac{7}{2}\text{lub}y=\frac{-2+5}{-2}=\frac{3}{-2}=-\frac{3}{2}\end{array}$  

$\{\begin{array}{l}x=-y-2\\ y=\frac{1}{2}\end{array}\text{lub}\{\begin{array}{l}x=-y-2\\ y=-\frac{5}{2}\end{array}\text{lub}\{\begin{array}{l}x=-y+2\\ y=\frac{7}{2}\end{array}\text{lub}\{\begin{array}{l}x=-y+2\\ y=-\frac{3}{2}\end{array}$  

Podstawiamy wyznaczone wartości y do pierwszego równania w układzie.

$\{\begin{array}{l}x=-\frac{1}{2}-\frac{4}{2}\\ y=\frac{1}{2}\end{array}\text{lub}\{\begin{array}{l}x=-\left(-\frac{5}{2}\right)-\frac{4}{2}\\ y=-\frac{5}{2}\end{array}\text{lub}\{\begin{array}{l}x=-\frac{7}{2}+\frac{4}{2}\\ y=\frac{7}{2}\end{array}\text{lub}\{\begin{array}{l}x=-\left(-\frac{3}{2}\right)+\frac{4}{2}\\ y=-\frac{3}{2}\end{array}$  

$\{\begin{array}{l}x=-\frac{5}{2}\\ y=\frac{1}{2}\end{array}\text{lub}\{\begin{array}{l}x=\frac{1}{2}\\ y=-\frac{5}{2}\end{array}\text{lub}\{\begin{array}{l}x=-\frac{3}{2}\\ y=\frac{7}{2}\end{array}\text{lub}\{\begin{array}{l}x=\frac{7}{2}\\ y=-\frac{3}{2}\end{array}$  

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$\text{c)}\{\begin{array}{l}3\left(x-y\right)-\left(x^2-y^2\right)=0\\ xy=2\end{array}$  

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

$\{\begin{array}{l}\left(x-y\right)\left[3-\left(x+y\right)\right]=0\\ xy=2\end{array}$  

$\{\begin{array}{l}\left(x-y\right)\left(3-x-y\right)=0\\ xy=2\end{array}$  

$\{\begin{array}{l}x-y=0\text{lub}3-x-y=0\\ xy=2\end{array}$  

$\{\begin{array}{l}x-y=0\\ xy=2\end{array}\text{lub}\{\begin{array}{l}3-x-y=0\\ xy=2\end{array}$  

$\{\begin{array}{l}x=y\\ xy=2\end{array}\text{lub}\{\begin{array}{l}x=3-y\\ xy=2\end{array}$  

Podstawiamy wyznaczone wartości x do drugiego równania w układzie.

$\{\begin{array}{l}x=y\\ y\cdot y=2\end{array}\text{lub}\{\begin{array}{l}x=3-y\\ \left(3-y\right)\cdot y=2\end{array}$  

$\{\begin{array}{l}x=y\\ y^2=2\end{array}\text{lub}\{\begin{array}{l}x=3-y\\ 3y-y^2=2\end{array}$  

$\{\begin{array}{l}x=y\\ y^2=2\end{array}\text{lub}\{\begin{array}{l}x=3-y\\ \underset{\Delta =9-8=1,\sqrt{\Delta }=1}{y^2-3x+2=0}\end{array}$  

$\{\begin{array}{l}x=y\\ y=-\sqrt{2}\text{lub}y=\sqrt{2}\end{array}\text{lub}\{\begin{array}{l}x=3-y\\ y=\frac{3-1}{2}=\frac{2}{2}=1\text{lub}y=\frac{3+1}{2}=\frac{4}{2}=2\end{array}$  

$\{\begin{array}{l}x=y\\ y=-\sqrt{2}\end{array}\text{lub}\{\begin{array}{l}x=y\\ y=\sqrt{2}\end{array}\text{lub}\{\begin{array}{l}x=3-y\\ y=1\end{array}\text{lub}\{\begin{array}{l}x=3-y\\ y=2\end{array}$  

Podstawiamy wyznaczone wartości y do pierwszego równania w układzie.

$\{\begin{array}{l}x=-\sqrt{2}\\ y=-\sqrt{2}\end{array}\text{lub}\{\begin{array}{l}x=\sqrt{2}\\ y=\sqrt{2}\end{array}\text{lub}\{\begin{array}{l}x=3-1\\ y=1\end{array}\text{lub}\{\begin{array}{l}x=3-2\\ y=2\end{array}$  

$\{\begin{array}{l}x=-\sqrt{2}\\ y=-\sqrt{2}\end{array}\text{lub}\{\begin{array}{l}x=\sqrt{2}\\ y=\sqrt{2}\end{array}\text{lub}\{\begin{array}{l}x=2\\ y=1\end{array}\text{lub}\{\begin{array}{l}x=1\\ y=2\end{array}$  

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$\text{d)}\{\begin{array}{l}{x}^2-9y^2=0\mid \cdot \left(-1\right)\\ x^2+y^2=10\end{array}$  

$\{\begin{array}{l}-x^2+9y^2=0\\ x^2+y^2=10\end{array}$  

Dodajemy równania stronami.

$10y^2=10\mid :10$  

$y^2=1$  

$y=-1\text{lub}y=1$  

Podstawiamy wyznaczone wartości y do dowolnego równania w układzie i wyznaczamy x.

$\{\begin{array}{l}{x}^2+y^2=10\\ y=-1\end{array}\text{lub}\{\begin{array}{l}{x}^2+y^2=10\\ y=1\end{array}$  

$\{\begin{array}{l}{x}^2+{\left(-1\right)}^2=10\\ y=-1\end{array}\text{lub}\{\begin{array}{l}{x}^2+1^2=10\\ y=1\end{array}$  

$\{\begin{array}{l}{x}^2+1=10\\ y=-1\end{array}\text{lub}\{\begin{array}{l}{x}^2+1=10\\ y=1\end{array}$  

$\{\begin{array}{l}{x}^2=9\\ y=-1\end{array}\text{lub}\{\begin{array}{l}{x}^2=9\\ y=1\end{array}$  

$\{\begin{array}{l}x=-3\text{lub}x=3\\ y=-1\end{array}\text{lub}\{\begin{array}{l}x=-3\text{lub}x=3\\ y=1\end{array}$  

$\{\begin{array}{l}x=-3\\ y=-1\end{array}\text{lub}\{\begin{array}{l}x=3\\ y=-1\end{array}\text{lub}\{\begin{array}{l}x=-3\\ y=1\end{array}\text{lub}\{\begin{array}{l}x=3\\ y=1\end{array}$  

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$\text{e)}\{\begin{array}{l}{x}^2-2xy+y^2=9\\ x^2+xy+y=9\mid \cdot \left(-1\right)\end{array}$  

$\{\begin{array}{l}{x}^2-2xy+y^2=9\\ -x^2-xy-y=-9\end{array}$  

Dodajemy równania stronami.

$-3xy+y^2-y=0$  

$y^2-3xy-y=0$  

$y\left(y-3x-1\right)=0$  

$y=0\text{lub}y-3x-1=0$  

$y=0\text{lub}y=3x+1$  

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Rozwiązujemy drugie (dowolne) równanie w układzie dla y=0.

$x^2+xy+y=9$  

$x^2+x\cdot 0+0=9$  

$x^2=9$  

$x=-3\text{lub}x=3$  

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Rozwiązujemy drugie (dowolne) równanie w układzie dla y=3x+1.

$x^2+xy+y=9$  

$x^2+x\left(3x+1\right)+3x+1=9$  

$x^2+3x^2+x+3x+1-9=0$  

$4x^2+4x-8=0\mid :4$  

$x^2+x-2=0$  

$\Delta =1+8=9,\sqrt{\Delta }=3$  

$x=\frac{-1-3}{2}=\frac{-4}{2}=-2\text{lub}x=\frac{-1+3}{2}=\frac{2}{2}=1$  

Dla wyznaczonych wartości x wyznaczamy y.

• dla x=-2 mamy:

$y=3\cdot \left(-2\right)+1=-6+1=-5$  

• dla x=1 mamy:

$y=3\cdot 1+1=3+1=4$  

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Zatem rozwiązaniami układu równań są cztery pary liczb:

$\{\begin{array}{l}x=-3\\ y=0\end{array}\text{lub}\{\begin{array}{l}x=3\\ y=0\end{array}\text{lub}\{\begin{array}{l}x=-2\\ y=-5\end{array}\text{lub}\{\begin{array}{l}x=1\\ y=4\end{array}$  

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$\text{f)}\{\begin{array}{l}{x}^2+y^2-x-2=0\mid \cdot \left(-2\right)\\ 2x^2-3x+2y^2+3=0\end{array}$  

$\{\begin{array}{l}-2x^2-2y^2+2x+4=0\\ 2x^2-3x+2y^2+3=0\end{array}$  

Dodajemy równania stronami.

$-x+7=0$  

$x=7$  

Podstawiamy x=7 do dowolnego równania w układzie i wyznaczamy y.

$\{\begin{array}{l}x=7\\ x^2+y^2-x-2=0\end{array}$  

$\{\begin{array}{l}x=7\\ 7^2+y^2-7-2=0\end{array}$  

$\{\begin{array}{l}x=7\\ 49+y^2-9=0\end{array}$  

$\{\begin{array}{l}x=7\\ 40+y^2=0\end{array}$  

$\{\begin{array}{l}x=7\\ y^2=-40\end{array}$  

Równanie y²=-40 jest sprzeczne, więc układ równań jest sprzeczny - nie ma rozwiązań.