Certainly! To solve the system of equations, we need to find the values of [tex]\( x \)[/tex] and [tex]\( y \)[/tex] that satisfy both equations simultaneously. The given system of equations is:
[tex]\[
\begin{cases}
y = 2x^2 \\
y = -3x - 1
\end{cases}
\][/tex]
### Step-by-Step Solution:
1. Set the equations equal to each other: Since both expressions equal [tex]\( y \)[/tex], we can set them equal to one another:
[tex]\[
2x^2 = -3x - 1
\][/tex]
2. Move all terms to one side to set up a quadratic equation: We will bring all terms to one side of the equation to set the equation to 0:
[tex]\[
2x^2 + 3x + 1 = 0
\][/tex]
3. Factor the quadratic equation: To solve for [tex]\( x \)[/tex], we need to factor the quadratic expression [tex]\( 2x^2 + 3x + 1 \)[/tex]. The factors will be in the form [tex]\((ax + b)(cx + d) = 0\)[/tex]. Let's find [tex]\( a, b, c, \)[/tex] and [tex]\( d \)[/tex] so that:
[tex]\[
2x^2 + 3x + 1 = (2x + 1)(x + 1)
\][/tex]
To verify this factored form, we can use the distributive property:
[tex]\[
(2x + 1)(x + 1) = 2x^2 + 2x + x + 1 = 2x^2 + 3x + 1
\][/tex]
4. Set each factor to 0 and solve for [tex]\( x \)[/tex]:
[tex]\[
2x + 1 = 0 \quad \text{or} \quad x + 1 = 0
\][/tex]
Solving these equations:
[tex]\[
2x + 1 = 0 \quad \Rightarrow \quad x = -\frac{1}{2}
\][/tex]
[tex]\[
x + 1 = 0 \quad \Rightarrow \quad x = -1
\][/tex]
5. Substitute back to find [tex]\( y \)[/tex]: Using the original equations, substitute [tex]\( x \)[/tex] back into either equation to find the corresponding [tex]\( y \)[/tex]-values. It's convenient to use [tex]\( y = 2x^2 \)[/tex] for substitution.
For [tex]\( x = -1 \)[/tex]:
[tex]\[
y = 2(-1)^2 = 2 \cdot 1 = 2
\][/tex]
For [tex]\( x = -\frac{1}{2} \)[/tex]:
[tex]\[
y = 2\left(-\frac{1}{2}\right)^2 = 2 \cdot \frac{1}{4} = \frac{1}{2}
\][/tex]
### Solutions:
Thus, the solutions to the system of equations are:
[tex]\[
(x, y) = (-1, 2) \quad \text{and} \quad \left(-\frac{1}{2}, \frac{1}{2}\right)
\][/tex]
So, the points where the equations intersect and both [tex]\( x \)[/tex] and [tex]\( y \)[/tex] satisfy the system are:
[tex]\[
(-1, 2) \quad \text{and} \quad \left(-\frac{1}{2}, \frac{1}{2}\right)
\][/tex]
