#### Answer

The solutions of the system are $(-4,-41)$ and $(\frac{1}{3},\frac{7}{3})$.

#### Work Step by Step

$-10x+y=-1 \rightarrow y=10x-1$
$y=3x^2+21x-5$
Substitute $10x-1$ for y
$10x-1=3x^2+21x-5$
$3x^2+11x-4=0$
$(3x-1)(x+4)=0$
$x+4=0$ or $3x-1=0$
$x=-4$ or $x=\frac{1}{3}$
Find corresponding y-values. Use either original equation
$y=10x-1$
$y=10(-4)-1$ or $y=10(\frac{1}{3})-1$
$y=-41$ or $y=\frac{7}{3}$
The solutions of the system are $(-4,-41)$ and $(\frac{1}{3},\frac{7}{3})$.