To solve the given system of equations using the substitution method, we will substitute the expression for \( y \) from the second equation into the first equation.
The given equations are:
- \( y = 4x^2 - x - 1 \)
- \( y = -3x + 1 \)
Since both expressions equal \( y \), we can set them equal to each other:
\[ 4x^2 - x - 1 = -3x + 1 \]
Now, let's rearrange this equation. First, let's move all terms to one side:
\[ 4x^2 - x + 3x - 1 - 1 = 0 \]
This simplifies to:
\[ 4x^2 + 2x - 2 = 0 \]
Next, we can simplify this equation by dividing all terms by 2:
\[ 2x^2 + x - 1 = 0 \]
Now we can factor this quadratic equation. We are looking for two numbers that multiply to \( 2 \cdot -1 = -2 \) and add to \( 1 \). The numbers \( 2 \) and \( -1 \) work, so we rewrite the equation as follows:
\[ 2x^2 + 2x - x - 1 = 0 \]
Now group the terms:
\[ (2x^2 + 2x) + (-x - 1) = 0 \]
Factoring out the common terms:
\[ 2x(x + 1) - 1(x + 1) = 0 \]
Now factor by grouping:
\[ (2x - 1)(x + 1) = 0 \]
Setting each factor to zero gives:
- \( 2x - 1 = 0 \) ⟹ \( 2x = 1 \) ⟹ \( x = \frac{1}{2} \)
- \( x + 1 = 0 \) ⟹ \( x = -1 \)
Now we will find the corresponding \( y \) values for each \( x \) value using the second equation (\( y = -3x + 1 \)).
For \( x = \frac{1}{2} \):
\[ y = -3\left(\frac{1}{2}\right) + 1 = -\frac{3}{2} + 1 = -\frac{3}{2} + \frac{2}{2} = -\frac{1}{2} \]
So one solution is \( \left(\frac{1}{2}, -\frac{1}{2}\right) \).
For \( x = -1 \):
\[ y = -3(-1) + 1 = 3 + 1 = 4 \]
So another solution is \( (-1, 4) \).
Thus, the solutions to the system are:
\[ \left(\frac{1}{2}, -\frac{1}{2}\right) \quad \text{and} \quad (-1, 4) \]