Question

a)  $\left \{ {{2x-y²=1} \atop {3x+y=4}} \right.$
b)  

Answer 1

Olá,

isole y na equação ii e substitua-o na equação i:

$\begin{cases}2x-y^2=1~~(i)\\ 3x+y=4~~(ii)\end{cases}\\\\ y=4-3x\\\\ 2x-(4-3x)^2=1\\ 2x-(16-24x+9x^2)=1\\ 2x-16+24x-9x^2=1\\ 9x^2-26x+17=0\\\\ \Delta=b^2-4ac\\ \Delta=(-26)^2-4\cdot9\cdot17\\ \Delta=676-612\\ \Delta=64\\\\ x= \dfrac{-b\pm \sqrt{\Delta} }{2a}= \dfrac{-(-26)\pm \sqrt{64} }{2\cdot9}= \dfrac{26\pm8}{18}$

$\begin{cases}x'= \dfrac{26-8}{18}= \dfrac{18}{18}=1\\\\ x''= \dfrac{26+8}{18}= \dfrac{34}{18}= \dfrac{17}{9} \end{cases}$

Para x=1, y valerá:

$3x+y=4\\ 3\cdot1+y=4\\ 3+y=4\\ y=4-3\\ y=1$

Para x=17/9, y valerá:

$3x+y=4\\\\ 3\cdot \dfrac{17}{9}+y=4\\\\ \dfrac{51}{9}+y=4\\\\ y=4- \dfrac{51}{9}\\\\ y=- \dfrac{5}{3}$

Portanto a solução do sistema do 2º grau acima é:

$\large\boxed{\boxed{\boxed{S_{x,y}=\left\{\left(1,1\right);\left( \dfrac{17}{9},- \dfrac{5}{3}\right)\right\}}}}.\\.$

Tenha ótimos estudos ;D