Answer
$(-5,25),(2,4)$
Work Step by Step
$y= x^{2}$ Equation $(1)$
$3x+y=10$ Equation $(2)$
Substitute $y=x^{2}$ in Equation $(2)$
$3x+y=10$
$3x+x^{2}=10$
$x^{2}+3x-10=0$
By factoring,
$(x+5)(x-2)=0$
$x = -5$ or $x=2$
Substitute $ x$ values in Equation $(1)$
Let $x = -5$
$y= x^{2}$
$y = (-5)^{2}$
$y = 25$
Let $x =2$
$y= x^{2}$
$y = (2)^{2}$
$y = 4$
$(-5,25),(2,4)$ satisfy both the given equations.
So, solutions are $(-5,25),(2,4)$
