Question

(50 PONTOS)

$\left \{ \begin{array} {l} 2x+y=5\\ x+y^2=10 \end{array} \right.$

Answer 1

$\large\begin{array}{l} \textsf{Resolver o sistema:}\\\\ \left\{\! \begin{array}{rrrrrc} \mathsf{2x}&\!\!+\!\!&\mathsf{y}&\!\!=\!\!&\mathsf{5}&\quad\mathsf{(i)}\\ \mathsf{x}&\!\!+\!\!&\mathsf{y^2}&\!\!=\!\!&\mathsf{10}&\quad\mathsf{(ii)} \end{array} \right. \end{array}$


$\large\begin{array}{l} \textsf{Multiplicando a equa\c{c}\~ao (ii) por }\mathsf{(-2):}\\\\ \left\{\! \begin{array}{rrrrrc} \mathsf{2x}&\!\!+\!\!&\mathsf{y}&\!\!=\!\!&\mathsf{5}&\quad\mathsf{(i)}\\ \mathsf{-2x}&\!\!-\!\!&\mathsf{2y^2}&\!\!=\!\!&\mathsf{-20}&\quad\mathsf{(iii)} \end{array} \right.\\\\\ \textsf{sendo }\mathsf{(iii)=(-2)\cdot (ii).}\end{array}$


$\large\begin{array}{l} \textsf{Agora, some as equa\c{c}\~oes (i) e (iii) membro a membro:}\\\\ \mathsf{\diagup\!\!\!\!\! 2x+y-\diagup\!\!\!\!\! 2x-2y^2=5-20}\\\\ \mathsf{y-2y^2=-15}\\\\ \mathsf{0=-15-y+2y^2}\\\\ \mathsf{2y^2-y-15=0} \end{array}$


$\large\begin{array}{l} \textsf{Vou resolver a equa\c{c}\~ao acima via fatora\c{c}\~ao por agrupamento.}\\\\ \textsf{Reescreva convenientemente }\mathsf{-y}\textsf{ como }\mathsf{-6y+5y},\textsf{ e fatore:}\\\\ \mathsf{2y^2-6y+5y-15=0}\\\\ \mathsf{2y(y-3)+5(y-3)=0}\\\\ \mathsf{(y-3)(2y+5)=0} \end{array}$

$\large\begin{array}{l} \begin{array}{rcl} \mathsf{y-3=0}&~\textsf{ ou }~&\mathsf{2y+5=0}\\\\ \mathsf{y=3}&~\textsf{ ou }~&\mathsf{2y=-5}\\\\ \mathsf{y=3}&~\textsf{ ou }~&\mathsf{y=-\,\frac{5}{2}} \end{array} \end{array}$


$\large\begin{array}{l} \bullet~~\textsf{Para }\mathsf{y=3,}\textsf{ obtemos}\\\\ \mathsf{2x+3=5}\qquad\textsf{(pela equa\c{c}\~ao (i))}\\\\ \mathsf{2x=5-3}\\\\ \mathsf{2x=2}\\\\ \mathsf{x=\frac{2}{2}}\\\\ \mathsf{x=1}\\\\\\ \textsf{Uma solu\c{c}\~ao para o sistema \'e}\\\\ \mathsf{(x,\,y)=(1,\,3).} \end{array}$


$\large\begin{array}{l} \bullet~~\textsf{Para }\mathsf{y=-\,\frac{5}{2},}\textsf{ obtemos}\\\\ \mathsf{2x-\dfrac{5}{2}=5}\qquad\textsf{(pela equa\c{c}\~ao (i))}\\\\ \mathsf{2x=5+\dfrac{5}{2}}\\\\ \mathsf{2x=\dfrac{10}{2}+\dfrac{5}{2}}\\\\ \mathsf{2x=\dfrac{15}{2}}\\\\ \mathsf{x=\dfrac{15}{2}\cdot \dfrac{1}{2}}\\\\ \mathsf{x=\dfrac{15}{4}}\\\\\\ \textsf{Outra solu\c{c}\~ao para o sistema \'e}\\\\ \mathsf{(x,\,y)=\left(\frac{15}{4},\,-\,\frac{5}{2}\right).} \end{array}$


$\large\begin{array}{l} \textsf{Conjunto solu\c{c}\~ao: }\mathsf{S=\left\{(1,\,3),\,\left(\frac{15}{4},\,-\,\frac{5}{2}\right) \right \}} \end{array}$


Caso tenha problemas para visualizar a resposta, experimente abrir pelo navegador: http://brainly.com.br/tarefa/7471875


$\large\begin{array}{l} \textsf{D\'uvidas? Comente.}\\\\\\ \textsf{Bons estudos! :-)} \end{array}$


Tags: sistema equação segundo grau quadrática adição substituição fatoração por agrupamento

Answer 2

Explicação passo-a-passo:

$2x + y = 5 \\x + {y}^{2} = 10 \\ \\ y = 5 - 2x \\ x + {y}^{2} = 10 \\ \\ x + (5 - 2x {)}^{2} = 10 \\ \\ x = \frac{15}{4} \\ x = 1 \\ \\ y = 5 - 2 \times \frac{15}{4} \\ y = 5 - 2 \times 1 \\ \\ y = - \frac{5}{2} \\ y = 3 \\ \\ ( x_{1} ,y_{1}) = ( \frac{15}{4} , - \frac{5}{2} ) \\ ( x_{2} ,y_{2}) = (1,3)$

Verificação:

$2 \times \frac{15}{4} - \frac{5}{2} = 5 \\ \frac{15}{4} + ( - \frac{5}{2} {)}^{2} = 10 \\ \\ 2 \times 1 + 3 = 5 \\ 1 + {3}^{2} = 10 \\ \\ 5 = 5 \\ 10 = 10$

• Espero ter ajudado.