Question

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.

Answer 1

`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=(-b \pm √(b^2-4ac))/(2a)=(-31 \pm √(31^2-4 \cdot 6 \cdot (-30)))/(2 \cdot 6)=(-31 \pm √(1681))/(12)=(-31 \pm 41)/(12) \\ x=(-31 -41)/(12) \ \hbox{or} \ x=(-31+41)/(12) \\ x=-6 \ \hbox{or} \ x=(5)/(6)`


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

Answer 2

`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=(5)/(6) \vee x=-6\\\\ y=-12\cdot(5)/(6) +15\vee y=-12\cdot(-6)+15\\ y=-10 +15\vee y=72+15\\ y=5 \vee y=87\\\\ x=(5)/(6) \wedge y=5\\ x=-6 \wedge y=87`