To solve the system of equations
\[
y = 4x^2 - x - 1
\]
\[
y = -3x + 1
\]
we will use the substitution method. Since both equations are equal to \(y\), we can set them equal to each other:
\[
4x^2 - x - 1 = -3x + 1
\]
Next, let's move all terms to one side of the equation:
\[
4x^2 - x + 3x - 1 - 1 = 0
\]
This simplifies to:
\[
4x^2 + 2x - 2 = 0
\]
Now we can divide the entire equation by 2:
\[
2x^2 + x - 1 = 0
\]
Next, we can factor this quadratic equation. We need two numbers that multiply to \(2 \cdot -1 = -2\) and add to \(1\). The numbers \(2\) and \(-1\) will work since:
\[
2 \cdot -1 = -2 \quad \text{and} \quad 2 + (-1) = 1.
\]
Thus, we can rewrite the quadratic:
\[
2x^2 + 2x - x - 1 = 0
\]
Now, grouping the terms:
\[
2x(x + 1) - 1(x + 1) = 0
\]
Factoring by grouping gives:
\[
(2x - 1)(x + 1) = 0
\]
Setting each factor equal to zero gives us the possible solutions for \(x\):
1. \(2x - 1 = 0 \Rightarrow x = \frac{1}{2}\)
2. \(x + 1 = 0 \Rightarrow x = -1\)
Now we will find the corresponding \(y\) values for each \(x\).
**For \(x = \frac{1}{2}\):**
Substitute into \(y = -3x + 1\):
\[
y = -3\left(\frac{1}{2}\right) + 1 = -\frac{3}{2} + 1 = -\frac{3}{2} + \frac{2}{2} = -\frac{1}{2}
\]
So, one solution is \(\left(\frac{1}{2}, -\frac{1}{2}\right)\).
**For \(x = -1\):**
Substitute into \(y = -3x + 1\):
\[
y = -3(-1) + 1 = 3 + 1 = 4
\]
So, another solution is \((-1, 4)\).
Thus, the solutions to the system are:
1. \(\left(-1, 4\right)\)
2. \(\left(\frac{1}{2}, -\frac{1}{2}\right)\)
Checking the provided options, the correct answers are:
1. (-1, 4)
2. (-0.5, 0.5) and (1, -2)
3. (0.5, -0.5) and (-1, 4)
4. (0.5, -0.5)
The correct response is thus:
**3. (0.5, -0.5) and (-1, 4)**