Question
Question Use the elimination method to solve the following linear-quadratic system: y=12x2−8x−2 y=−3x (1 point) Responses (23,2) and (−14,−34) left parenthesis Start Fraction 2 over 3 End Fraction comma 2 right parenthesis and left parenthesis negative Start Fraction 1 over 4 End Fraction comma negative Start Fraction 3 over 4 End Fraction right parenthesis (23,−2) left parenthesis Start Fraction 2 over 3 End Fraction comma negative 2 right parenthesis (−14,34) left parenthesis negative Start Fraction 1 over 4 End Fraction comma Start Fraction 3 over 4 End Fraction right parenthesis (23,−2) and (−14,34) left parenthesis Start Fraction 2 over 3 End Fraction comma negative 2 right parenthesis and left parenthesis negative Start Fraction 1 over 4 End Fraction comma Start Fraction 3 over 4 End Fraction right parenthesis
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Answered by
GPT-4o mini
To solve the system of equations using the elimination method, we have the following two equations:
1. \( y = 12x^2 - 8x - 2 \)
2. \( y = -3x \)
Since both equations are equal to \( y \), we can set them equal to each other:
\[
12x^2 - 8x - 2 = -3x
\]
Next, we will move all terms to one side of the equation:
\[
12x^2 - 8x + 3x - 2 = 0
\]
Combine the \( x \) terms:
\[
12x^2 - 5x - 2 = 0
\]
Now, we will use the quadratic formula to solve for \( x \). The quadratic formula is given by:
\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]
Here, \( a = 12 \), \( b = -5 \), and \( c = -2 \). We first calculate the discriminant:
\[
b^2 - 4ac = (-5)^2 - 4(12)(-2) = 25 + 96 = 121
\]
Now, applying the quadratic formula:
\[
x = \frac{-(-5) \pm \sqrt{121}}{2(12)} = \frac{5 \pm 11}{24}
\]
This gives us two possible solutions for \( x \):
\[
x_1 = \frac{16}{24} = \frac{2}{3}
\]
\[
x_2 = \frac{-6}{24} = -\frac{1}{4}
\]
Now we substitute these \( x \) values back into the equation \( y = -3x \) to find the corresponding \( y \) values:
1. For \( x = \frac{2}{3} \):
\[
y = -3\left(\frac{2}{3}\right) = -2
\]
2. For \( x = -\frac{1}{4} \):
\[
y = -3\left(-\frac{1}{4}\right) = \frac{3}{4}
\]
Therefore, the solutions to the system of equations are:
\[
\left(\frac{2}{3}, -2\right) \text{ and } \left(-\frac{1}{4}, \frac{3}{4}\right)
\]
In summary, the correct response is:
**\((\frac{2}{3}, -2)\) and \((- \frac{1}{4}, \frac{3}{4})\)**.
1. \( y = 12x^2 - 8x - 2 \)
2. \( y = -3x \)
Since both equations are equal to \( y \), we can set them equal to each other:
\[
12x^2 - 8x - 2 = -3x
\]
Next, we will move all terms to one side of the equation:
\[
12x^2 - 8x + 3x - 2 = 0
\]
Combine the \( x \) terms:
\[
12x^2 - 5x - 2 = 0
\]
Now, we will use the quadratic formula to solve for \( x \). The quadratic formula is given by:
\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]
Here, \( a = 12 \), \( b = -5 \), and \( c = -2 \). We first calculate the discriminant:
\[
b^2 - 4ac = (-5)^2 - 4(12)(-2) = 25 + 96 = 121
\]
Now, applying the quadratic formula:
\[
x = \frac{-(-5) \pm \sqrt{121}}{2(12)} = \frac{5 \pm 11}{24}
\]
This gives us two possible solutions for \( x \):
\[
x_1 = \frac{16}{24} = \frac{2}{3}
\]
\[
x_2 = \frac{-6}{24} = -\frac{1}{4}
\]
Now we substitute these \( x \) values back into the equation \( y = -3x \) to find the corresponding \( y \) values:
1. For \( x = \frac{2}{3} \):
\[
y = -3\left(\frac{2}{3}\right) = -2
\]
2. For \( x = -\frac{1}{4} \):
\[
y = -3\left(-\frac{1}{4}\right) = \frac{3}{4}
\]
Therefore, the solutions to the system of equations are:
\[
\left(\frac{2}{3}, -2\right) \text{ and } \left(-\frac{1}{4}, \frac{3}{4}\right)
\]
In summary, the correct response is:
**\((\frac{2}{3}, -2)\) and \((- \frac{1}{4}, \frac{3}{4})\)**.
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