Question

RESOLVA AS EQUAÇÕES;

Answer 1

Olá Renata,

gata só vc usar o mesmo princípio dos determinantes de 2a ordem:

$\left|\begin{array}{ccc}x&x+1\\8&3\\\end{array}\right|=0\\\\\\ 3\cdot x-8(x+1)=0\\ 3x-8x-8=0\\ -5x=8\\\\ x= -\dfrac{8}{5}\\\\ \boxed{\boxed{S=\left(-\dfrac{8}{5}\right)}}$



$\left|\begin{array}{ccc}x&2\\x+2&3x+1\\\end{array}\right|=0\\\\\\ x(3x+1)-2(x+2)=0\\ 3x^2+x-2x-4=0\\ 3x^2-x-4=0\\\\ \Delta=(-1)^2-4\cdot3\cdot(-4)\\ \Delta=1+48\\ \Delta=49\\\\ x= \dfrac{-3\pm \sqrt{49} }{2\cdot3}= \dfrac{-3\pm7}{6}\begin{cases}x'= \dfrac{-3+7}{6}= \dfrac{4}{6}= \dfrac{2}{3}\\\\ x'' \dfrac{-3-7}{6}=- \dfrac{10}{6}=- \dfrac{5}{3} \end{cases}$

$\boxed{\boxed{S=\left( \dfrac{2}{3},- \dfrac{5}{3}\right) }}$


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