Respuesta :
EXPLANATION
Given the system of equations:
(1) y= 2x^2-3x - 4
(2) y + x = 8
Subtract the equations:
y= 2x^2-3x - 4
-
y+x=8
y- (y+x) = 2x^2 -3x-4-8
[tex]\mathrm{Simplify}[/tex][tex]-x=2x^2-3x-12[/tex][tex]Switch\text{ sides:}[/tex][tex]2x^2-3x-12=-x[/tex][tex]\mathrm{Add\: }x\mathrm{\: to\: both\: sides}[/tex][tex]2x^2-3x-12+x=-x+x[/tex][tex]\text{Simplify}[/tex][tex]2x^2-2x-12=0[/tex]Applying the quadratic formula:
[tex]x_{1,\: 2}=\frac{-b\pm\sqrt{b^2-4ac}}{2a}[/tex][tex]\mathrm{For\: }\quad a=2,\: b=-2,\: c=-12[/tex][tex]x_{1,\: 2}=\frac{-\left(-2\right)\pm\sqrt{\left(-2\right)^2-4\cdot\:2\left(-12\right)}}{2\cdot\:2}[/tex][tex]x_{1,\: 2}=\frac{-\left(-2\right)\pm\:10}{2\cdot\:2}[/tex][tex]Separate\text{ the solutions}[/tex][tex]x_1=\frac{-\left(-2\right)+10}{2\cdot\:2},\: x_2=\frac{-\left(-2\right)-10}{2\cdot\:2}[/tex]Simplifying:
[tex]x=3,\: x=-2[/tex]Hence, the solutions are:
x= -2 and x=3
Computing the y-values:
[tex]\mathrm{For\: }y=2x^2-3x-4\mathrm{,\: subsitute\: }x\mathrm{\: with\: }3[/tex][tex]y=2\cdot\: 3^2-3\cdot\: 3-4[/tex][tex]y=5[/tex][tex]\mathrm{For\: }y=2x^2-3x-4\mathrm{,\: subsitute\: }x\mathrm{\: with\: }-2[/tex][tex]y=2\mleft(-2\mright)^2-3\mleft(-2\mright)-4[/tex][tex]y=10[/tex]Therefore, the final solutions to the system of equations are:
(x,y) = (-2,10) (x,y)=(3,5)
