Question

Use the elimination method to solve the following linear-quadratic system: y = 12x^2 - 8x - 2 & y = -3x

1. (2/3, -2)
2. (2/3, 2) and (-1/4, -3/4)
3. (2/3, -2) and (-1/4, 3/4)
4. (-1/4, 3/4)

Answer 1

To solve the system of equations using the elimination method, we have:

1. \( y = 12x^2 - 8x - 2 \) (equation 1)
2. \( y = -3x \) (equation 2)

We can set the two expressions for \( y \) equal to each other in order to eliminate \( y \):

\[
12x^2 - 8x - 2 = -3x
\]

Now, let's move all terms to one side of the equation:

\[
12x^2 - 8x + 3x - 2 = 0
\]

Combining like terms gives:

\[
12x^2 - 5x - 2 = 0
\]

Next, we can factor this quadratic equation, or we can use the quadratic formula:

The quadratic formula is given by:

\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]

Here, \( a = 12 \), \( b = -5 \), and \( c = -2 \).

First, we calculate the discriminant \( b^2 - 4ac \):

\[
b^2 - 4ac = (-5)^2 - 4(12)(-2) = 25 + 96 = 121
\]

Now, we can apply the quadratic formula:

\[
x = \frac{-(-5) \pm \sqrt{121}}{2(12)}
\]

This simplifies to:

\[
x = \frac{5 \pm 11}{24}
\]

Calculating the two possible values for \( x \):

1. \( x = \frac{5 + 11}{24} = \frac{16}{24} = \frac{2}{3} \)
2. \( x = \frac{5 - 11}{24} = \frac{-6}{24} = -\frac{1}{4} \)

Now we will find the corresponding \( y \) values by substituting these \( x \) values back into equation 2 \( y = -3x \):

1. For \( x = \frac{2}{3} \):
\[
y = -3 \left( \frac{2}{3} \right) = -2
\]
Thus, one solution is \( \left(\frac{2}{3}, -2\right) \).

2. For \( x = -\frac{1}{4} \):
\[
y = -3 \left( -\frac{1}{4} \right) = \frac{3}{4}
\]
Thus, the other solution is \( \left(-\frac{1}{4}, \frac{3}{4}\right) \).

So the solutions to the system are:

1. \( \left( \frac{2}{3}, -2 \right) \)
2. \( \left( -\frac{1}{4}, \frac{3}{4} \right) \)

Therefore, the correct answer is:

**3.** \( \left( \frac{2}{3}, -2 \right) \) and \( \left( -\frac{1}{4}, \frac{3}{4} \right) \)