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 \(x\):
- \(2x - 1 = 0 \Rightarrow x = \frac{1}{2}\)
- \(x + 1 = 0 \Rightarrow x = -1\)
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:
- \(\left(-1, 4\right)\)
- \(\left(\frac{1}{2}, -\frac{1}{2}\right)\)
Checking the provided options, the correct answers are:
- (-1, 4)
- (-0.5, 0.5) and (1, -2)
- (0.5, -0.5) and (-1, 4)
- (0.5, -0.5)
The correct response is thus:
3. (0.5, -0.5) and (-1, 4)
1 answer