To solve the system of equations using the elimination method, we have:
- \( y = 12x^2 - 8x - 2 \) (equation 1)
- \( y = -3x \) (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 \( x \):
- \( x = \frac{5 + 11}{24} = \frac{16}{24} = \frac{2}{3} \)
- \( x = \frac{5 - 11}{24} = \frac{-6}{24} = -\frac{1}{4} \)
Now we will find the corresponding \( y \) values by substituting these \( x \) values back into equation 2 \( y = -3x \):
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For \( x = \frac{2}{3} \): \[ y = -3 \left( \frac{2}{3} \right) = -2 \] Thus, one solution is \( \left(\frac{2}{3}, -2\right) \).
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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:
- \( \left( \frac{2}{3}, -2 \right) \)
- \( \left( -\frac{1}{4}, \frac{3}{4} \right) \)
Therefore, the correct answer is:
3. \( \left( \frac{2}{3}, -2 \right) \) and \( \left( -\frac{1}{4}, \frac{3}{4} \right) \)
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