#### Answer

$(\dfrac{1}{2},2)$, $(-1,-1)$

#### Work Step by Step

Re-arrange the given two equations and then use the substitution method to get
$x(2x+1)^2=1$
or, $2x^2+x-1=0$
This gives: $(2x-1)(x+1)=0$
when $x=\dfrac{1}{2}$ then we have $y=2(\dfrac{1}{2})+1=2$
when $x=-1$ then we have $y=2(-1)+1=-1$
Hence, our answers are: $(\dfrac{1}{2},2)$, $(-1,-1)$