Question

What are the solutions to the system of equations?
y=−x2−5x

−6x+y=−3

Answer 1

Answer:
`\bold{\bigg((-11+√(133))/(2),\ (-11 +12√(133))/(4)\bigg)\ \text{and}\ \bigg((-11-√(133))/(2),\ (-11 +32√(133))/(4)\bigg)}`

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:
`x=(-b\pm √(b^2-4ac))/(2a)`

a = -1, b = -11, c = 3

`x=(-(-11)\pm √((-11)^2-4(-1)(3)))/(2(-1))`

`(11\pm √(121+12))/(-2)`

`(11\pm √(133))/(-2)`

`(-11\pm √(133))/(2)`

`\x=dfrac{-11+√(133)}{2}` and
`x=(-11-√(133))/(2)`

Next, solve for "y" by plugging the x-values above into the equation: y = -x² - 5x

`y=-\bigg((-11 +√(133))/(2)\bigg)^2-5\bigg((-11 +√(133))/(2)\bigg)`

`=-\bigg((121 -22√(133))/(4)\bigg)+\bigg((55 -5√(133))/(2)\bigg)`

`=(-121 +22√(133))/(4)+(110 -10√(133))/(4)`

`=(-11 +12√(133))/(4)`

and

`y=-\bigg((-11 -√(133))/(2)\bigg)^2-5\bigg((-11 -√(133))/(2)\bigg)`

`=-\bigg((121 -22√(133))/(4)\bigg)+\bigg((55 +5√(133))/(2)\bigg)`

`=(-121 +22√(133))/(4)+(110 +10√(133))/(4)`

`=(-11 +32√(133))/(4)`