Question

Question 6
(Multiple Choice Worth 2 Points)
(Systems of Linear Equations MC)

Which point is a solution to the system of linear equations?

y = −x + 1
3x − y = 3
three halves comma negative one half
(1, 0)
negative one half comma three halves
(0, 1)

Answer 1

To determine which point is a solution to the system of equations, we will substitute each given point into both equations and see if they satisfy both.

The equations are:
1. \( y = -x + 1 \)
2. \( 3x - y = 3 \)

Let's evaluate each point:

1. **Point (3/2, -1/2)**:
- For the first equation:
\[
y = -x + 1 \implies -\frac{1}{2} = -\frac{3}{2} + 1 \implies -\frac{1}{2} = -\frac{3}{2} + \frac{2}{2} \implies -\frac{1}{2} = -\frac{1}{2} \quad \text{(True)}
\]
- For the second equation:
\[
3x - y = 3 \implies 3\left(\frac{3}{2}\right) - \left(-\frac{1}{2}\right) = 3 \implies \frac{9}{2} + \frac{1}{2} = 3 \implies \frac{10}{2} = 3 \quad \text{(False)}
\]

2. **Point (1, 0)**:
- For the first equation:
\[
y = -x + 1 \implies 0 = -1 + 1 \implies 0 = 0 \quad \text{(True)}
\]
- For the second equation:
\[
3x - y = 3 \implies 3(1) - 0 = 3 \implies 3 = 3 \quad \text{(True)}
\]

3. **Point (-1/2, 3/2)**:
- For the first equation:
\[
y = -x + 1 \implies \frac{3}{2} = -\left(-\frac{1}{2}\right) + 1 \implies \frac{3}{2} = \frac{1}{2} + 1 \quad \text{(False)}
\]

4. **Point (0, 1)**:
- For the first equation:
\[
y = -x + 1 \implies 1 = -0 + 1 \implies 1 = 1 \quad \text{(True)}
\]
- For the second equation:
\[
3x - y = 3 \implies 3(0) - 1 = 3 \implies -1 = 3 \quad \text{(False)}
\]

After checking all points, the only point that satisfies both equations is:

**(1, 0)**.