$a\text{)}-\{\begin{array}{l}{x}^2+y^2-5\left(x+y\right)=8\\ x^2+y^2-3\left(x+y\right)=28\end{array}$  

$\underline{}$  

$-5\left(x+y\right)-\left(-3\right)\left(x+y\right)=8-28$  

$-2\left(x+y\right)=-20$  

$x+y=10$  

$x=10-y$  

Podstawiamy  $x=10-y$  do pierwszego równania:

${\left(10-y\right)}^2+y^2-5\left(10-y+y\right)=8$  

$100-20y+y^2+y^2-50=8$  

$2y^2-20y+42=0$  

$\Delta ={\left(-20\right)}^2-4\cdot 2\cdot 42=400-336=64$  

$\sqrt{\Delta }=\sqrt{64}=8$  

$y_1=\frac{20-8}{4}=\frac{12}{4}=3$  

$y_2=\frac{20+8}{4}=\frac{28}{4}=7$  

zatem:

$x_1=10-3=7$  

$x_2=10-7=3$  

więc:

$\{\begin{array}{l}x=7\\ y=3\end{array}$  

lub

$\{\begin{array}{l}x=3\\ y=7\end{array}$

---

$b\text{)}\{\begin{array}{l}{x}^2+y^2=169\\ xy+60=0\mid \cdot 2\end{array}$  

$-\{\begin{array}{l}{x}^2+y^2=169\\ 2xy+120=0\end{array}$   

$\underline{}$  

$x^2-2xy+y^2-120=169$  

${\left(x-y\right)}^2=289$  

${\left(x-y\right)}^2=\sqrt{289}$  

$x-y=17\text{lub}x-y=-17$  

$x=17+y\text{lub}x=-17+y$  

 

Podstawiamy  $x=17+y$  do pierwszego równania:

${\left(17+y\right)}^2+y^2=169$  

$289+34y+y^2+y^2=169$  

$2y^2+34y+120=0$  

$\Delta ={34}^2-4\cdot 2\cdot 120=1156-960=196$  

$\sqrt{\Delta }=\sqrt{196}=14$  

$y_1=\frac{-34-14}{4}=\frac{-48}{4}=-12$  

$y_2=\frac{-34+14}{4}=\frac{-20}{4}=-5$  

zatem:

$x_1=17-12=5$  

$x_2=17-5=12$  

więc:

$\{\begin{array}{l}x=5\\ y=-12\end{array}$  

lub

$\{\begin{array}{l}x=12\\ y=-5\end{array}$

 

Podstawiamy  $x=-17+y$  do pierwszego równania:

${\left(-17+y\right)}^2+y^2=169$  

$289-34y+y^2+y^2=169$  

$2y^2-34y+120=0$  

$\Delta ={\left(-34\right)}^2-4\cdot 2\cdot 120=1156-960=196$  

$\sqrt{\Delta }=\sqrt{196}=14$  

$y_1=\frac{34-14}{4}=\frac{20}{4}=5$  

$y_2=\frac{34+14}{4}=\frac{48}{4}=12$  

zatem:

$x_1=-17+5=-12$  

$x_2=-17+12=-5$  

więc:

$\{\begin{array}{l}x=-12\\ y=5\end{array}$  

lub

$\{\begin{array}{l}x=-5\\ y=12\end{array}$

---

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

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

zatem:

$x+y=2\text{lub}x+y=-2$  

$x=2-y\text{lub}x=-2-y$  

Podstawiamy  $x=2-y$  do drugiego równania:

$2-y+y+\left(2-y\right)y+3,25=0$  

$2+2y-y^2+3,25=0$  

$-y^2+2y+5,25=0$  

$\Delta =2^2-4\cdot \left(-1\right)\cdot 5,25=4+21=25$  

$\sqrt{\Delta }=5$  

$y_1=\frac{-2-5}{-2}=\frac{7}{2}$  

$y_2=\frac{-2+5}{-2}=-\frac{3}{2}$  

więc:

$x_1=2-\frac{7}{2}=-\frac{3}{2}$  

$x_2=2-\left(-\frac{3}{2}\right)=\frac{7}{2}$  

więc:

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

lub

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

Podstawiamy  $x=-2-y$  do drugiego równania:

$-2-y+y+\left(-2-y\right)y+3,25=0$  

$-2-2y-y^2+3,25=0$  

$-y^2-2y+1,25=0$  

$\Delta =2^2-4\cdot \left(-1\right)\cdot 1,25=4+5=9$  

$\sqrt{\Delta }=3$  

$y_1=\frac{2-3}{-2}=\frac{1}{2}$  

$y_2=\frac{2+3}{-2}=-\frac{5}{2}$  

więc:

$x_1=-2-\frac{1}{2}=-\frac{5}{2}$  

$x_2=-2-\left(-\frac{5}{2}\right)=\frac{1}{2}$  

więc:

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

lub

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