Answered

Westonci.ca is the premier destination for reliable answers to your questions, provided by a community of experts. Join our platform to get reliable answers to your questions from a knowledgeable community of experts. Get quick and reliable solutions to your questions from a community of experienced experts on our platform.

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.

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 Quadra class=

Sagot :

konrad509
[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]
naǫ
[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]
We appreciate your visit. Hopefully, the answers you found were beneficial. Don't hesitate to come back for more information. Thank you for visiting. Our goal is to provide the most accurate answers for all your informational needs. Come back soon. Thank you for visiting Westonci.ca, your go-to source for reliable answers. Come back soon for more expert insights.