Answer: [tex]\bold{\bigg(\dfrac{-11+\sqrt{133}}{2},\ \dfrac{-11 +12\sqrt{133}}{4}\bigg)\ \text{and}\ \bigg(\dfrac{-11-\sqrt{133}}{2},\ \dfrac{-11 +32\sqrt{133}}{4}\bigg)}[/tex]
Step-by-step explanation:
y = -x² - 5x
-6x + y = -3
Use substitution method:
-6x + (-x² - 5x) = -3 substituted "y" with "-x² - 5x"
-x² -11x = -3 simplified left side (added like terms)
-x² - 11x + 3 = 0 added 3 to both sides
Solve using quadratic formula: [tex]x=\dfrac{-b\pm \sqrt{b^2-4ac}}{2a}[/tex]
a = -1, b = -11, c = 3
[tex]x=\dfrac{-(-11)\pm \sqrt{(-11)^2-4(-1)(3)}}{2(-1)}[/tex]
[tex]\dfrac{11\pm \sqrt{121+12}}{-2}[/tex]
[tex]\dfrac{11\pm \sqrt{133}}{-2}[/tex]
[tex]\dfrac{-11\pm \sqrt{133}}{2}[/tex]
[tex]\x=dfrac{-11+\sqrt{133}}{2}[/tex] and [tex]x=\dfrac{-11-\sqrt{133}}{2}[/tex]
Next, solve for "y" by plugging the x-values above into the equation: y = -x² - 5x
[tex]y=-\bigg(\dfrac{-11 +\sqrt{133}}{2}\bigg)^2-5\bigg(\dfrac{-11 +\sqrt{133}}{2}\bigg)[/tex]
[tex]=-\bigg(\dfrac{121 -22\sqrt{133}}{4}\bigg)+\bigg(\dfrac{55 -5\sqrt{133}}{2}\bigg)[/tex]
[tex]=\dfrac{-121 +22\sqrt{133}}{4}+\dfrac{110 -10\sqrt{133}}{4}[/tex]
[tex]=\dfrac{-11 +12\sqrt{133}}{4}[/tex]
and
[tex]y=-\bigg(\dfrac{-11 -\sqrt{133}}{2}\bigg)^2-5\bigg(\dfrac{-11 -\sqrt{133}}{2}\bigg)[/tex]
[tex]=-\bigg(\dfrac{121 -22\sqrt{133}}{4}\bigg)+\bigg(\dfrac{55 +5\sqrt{133}}{2}\bigg)[/tex]
[tex]=\dfrac{-121 +22\sqrt{133}}{4}+\dfrac{110 +10\sqrt{133}}{4}[/tex]
[tex]=\dfrac{-11 +32\sqrt{133}}{4}[/tex]
