Using the substitution method, solve the following linear quadratic system: y = 4x^2 - x - 1 y = -3x + 1

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)**