## Algebra 2 (1st Edition)

Adding twice the first and negative one times the second equation we get: $2x^2-20-x-8=0\\2x^2-x-28=0\\(2x+7)(x-4)=0$ Thus $x=-3.5$ or $x=4$ Plugging this into the first equation we get: if $x=-3.5$: $12.25+2y^2-10=0$ but this has no solution. if $x=4$: $16+2y^2-10=0$, but this has no solution Thus there are no solutions.