Question

Solve the following system of equations and show all work.

y = 2x2
y = –3x −1

Answer 1

For this case we have a system of two equations with two unknowns:

Matching we have:

$2x ^ 2 = -3x-1\\2x ^ 2 + 3x + 1 = 0$

We have a quadratic equation of the form $ax ^ 2 + bx + c = 0$

Where:

$a = 2\\b = 3\\c = 1$

Applying the quadratic formula we have:

$x = \frac {-b \pm \sqrt {b ^ 2-4 (a) (c)}} {2a}$

Substituting:

$x = \frac {-3 \pm \sqrt {3 ^ 2-4 (2) (1)}} {2 * 2}\\x = \frac {-3 \pm \sqrt {9-8}} {4}\\x = \frac {-3 \pm \sqrt {1}} {4}\\x = \frac {-3 \pm1} {4}$

We have two roots:

$x_ {1} = \frac {-3 + 1} {4} = - \frac {2} {4} = - \frac {1} {2}\\x_ {2} = \frac {-3-1} {4} = - \frac {4} {4} = - 1$

Now, we look for the values of y, substituting in any of the equations:

$For\ x_ {1} = - \frac {1} {2}\then\ y_ {1} = - 3 (- \frac {1} {2}) - 1 = \frac {3} {2} -1 = \frac {1} {2}\\For\ x_ {2} = - 1\ then\ y_ {2} = - 3 (-1) -1 = 3-1 = 2$

ANswer:

$For\ x_ {1} = - \frac {1} {2}\then\ y_ {1} = - 3 (- \frac {1} {2}) - 1 = \frac {3} {2} -1 = \frac {1} {2}\\For\ x_ {2} = - 1\ then\ y_ {2} = - 3 (-1) -1 = 3-1 = 2$

Answer 2

Answer:

Two solutions:

1) x=-1 and y=2: Point=(-1,2)

2) x=-1/2 and y=1/2: Point=(-1/2,1/2)=(-0.5,0.5)

Step-by-step explanation:

y=2x^2

y=-3x-1

Solution:

y=y

Equaling the equations:

$2x^{2}=-3x-1$

This is a quadratic equation. Equaling to zero: Adding 3x and 1 both sides of the equation:

$2x^{2}+3x+1=-3x-1+3x+1\\ 2x^{2}+3x+1=0$

Using the quadratic formula:

$ax^{2}+bx+c=0; a=2, b=3, c=1$

$x=\frac{-b+-\sqrt{b^2-4ac}}{2a}\\x=\frac{-3+-\sqrt{3^2-4(2)(1)}}{2(2)}\\ x=\frac{-3+-\sqrt{9-8}}{4}\\ x=\frac{-3+-\sqrt{1}}{4}\\ x=\frac{-3+-1}{4}$

Two solutions:

$\left \{ {{x_{1} =\frac{-3-1}{4}} \atop {x_{2} =\frac{-3+1}{4}}} \right.$

$\left \{ {{x_{1} =\frac{-4}{4}} \atop {x_{2} =\frac{-2}{4}}} \right.$

$\left \{ {{x_{1} =-1} \atop {x_{2} =-\frac{1}{2}}} \right.$

For x=-1:

$y=2(-1)^{2}\\ y=2(1)\\ y=2$

First solution x=-1 and y=2: Point=(x,y)→Point=(-1,2)

For x=-(1/2):

$y=2(-\frac{1}{2})^{2}\\ y=2(\frac{1^{2}}{2^{2}})\\ y=2(\frac{1}{4})\\ y=\frac{2*1}{4}\\ y=\frac{2}{4}\\ y=\frac{\frac{2}{2}}{\frac{4}{2}}\\ y=\frac{1}{2}$

Second solution x=-1/2 and y=1/2: Point=(x,y)→Point=(-1/2,1/2)