Answered

Welcome to Westonci.ca, your go-to destination for finding answers to all your questions. Join our expert community today! Find reliable answers to your questions from a wide community of knowledgeable experts on our user-friendly Q&A platform. Join our platform to connect with experts ready to provide precise answers to your questions in different areas.

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]
Thank you for your visit. We're dedicated to helping you find the information you need, whenever you need it. We hope our answers were useful. Return anytime for more information and answers to any other questions you have. Discover more at Westonci.ca. Return for the latest expert answers and updates on various topics.