Algebraically solve the system of equations shown below. Note that you can use either factoring or the quadratic formula to find the X – coordinates, but the quadratic formula is probably easier.

[tex]6x^2+19x-15=-12x+15\\ 6x^2+31x-30=0\\ 6x^2+36x-5x-30=0\\ 6x(x+6)-5(x+6)=0\\ (6x-5)(x+6)=0\\ x=\frac{5}{6} \vee x=-6\\\\ y=-12\cdot\frac{5}{6} +15\vee y=-12\cdot(-6)+15\\ y=-10 +15\vee y=72+15\\ y=5 \vee y=87\\\\ x=\frac{5}{6} \wedge y=5\\ x=-6 \wedge y=87[/tex]

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naǫ naǫ · Answer 2 · 2014-11-15T12:43:30-05:00

[tex]y=6x^2+19x-15 \\ y=-12x+15 \\ \\ 6x^2+19x-15=-12x+15 \\ 6x^2+19x+12x-15-15=0 \\ 6x^2+31x-30=0 \\ \\ a=6 \\ b=31 \\ c=-30 \\ \\ x=\frac{-b \pm \sqrt{b^2-4ac}}{2a}=\frac{-31 \pm \sqrt{31^2-4 \cdot 6 \cdot (-30)}}{2 \cdot 6}=\frac{-31 \pm \sqrt{1681}}{12}=\frac{-31 \pm 41}{12} \\ x=\frac{-31 -41}{12} \ \hbox{or} \ x=\frac{-31+41}{12} \\ x=-6 \ \hbox{or} \ x=\frac{5}{6}[/tex]

[tex]y=-12 \cdot (-6)+15 \ \hbox{or} \ y=-12 \cdot \frac{5}{6}+15 \\y=87 \ \hbox{or} \ y=5 \\ \\ \hbox{the answer:} \\ \boxed{x=-6, \ y=87} \ \hbox{or} \ \boxed{x=\frac{5}{6}, \ y=5}[/tex]