Question

Solve the system of linear equations by graphing.

y equals negative one half times x plus 2

y equals one half times x minus 3

negative 5 comma negative one half
5 comma negative one half
negative one half comma 5

Answer 1

To solve the system of linear equations by graphing, we will first rewrite the equations in slope-intercept form (if they are not already) and then graph them.

1. The first equation is:

\[
y = -\frac{1}{2}x + 2
\]

- The slope is \(-\frac{1}{2}\) and the y-intercept is \(2\).
- To graph, start at the point \((0, 2)\) on the y-axis. From there, for every 2 units you move down (due to the negative slope), you move 1 unit to the right.

2. The second equation is:

\[
y = \frac{1}{2}x - 3
\]

- The slope is \(\frac{1}{2}\) and the y-intercept is \(-3\).
- To graph this line, start at the point \((0, -3)\) on the y-axis. From there, for every 2 units you move up, you move 1 unit to the right.

Now, let’s graph both lines:

1. For the first equation \(y = -\frac{1}{2}x + 2\):
- Point 1: (0, 2)
- Point 2: Use x = 2 to find y:
\[
y = -\frac{1}{2}(2) + 2 = -1 + 2 = 1 \implies (2, 1)
\]

2. For the second equation \(y = \frac{1}{2}x - 3\):
- Point 1: (0, -3)
- Point 2: Use x = 2 to find y:
\[
y = \frac{1}{2}(2) - 3 = 1 - 3 = -2 \implies (2, -2)
\]

After plotting these points and drawing the lines, you will look for the intersection point of the two lines.

Next, let’s analyze the given options for the potential solution point:

- \((-5, -\frac{1}{2})\)
- \( (5, -\frac{1}{2})\)
- \((- \frac{1}{2}, 5)\)

Since we cannot visually graph the equations here, we can also substitute these potential solution points back into both equations to see if they satisfy both.

1. For \( (-5, -\frac{1}{2}) \):
\[
y = -\frac{1}{2}(-5) + 2 \implies y = \frac{5}{2} + 2 = \frac{5}{2} + \frac{4}{2} = \frac{9}{2} \text{ (not equal to } -\frac{1}{2})
\]

2. For \( (5, -\frac{1}{2}) \):
\[
y = -\frac{1}{2}(5) + 2 \implies y = -\frac{5}{2} + 2 = -\frac{5}{2} + \frac{4}{2} = -\frac{1}{2} \text{ (satisfies the first equation)}
\]
\[
y = \frac{1}{2}(5) - 3 \implies y = \frac{5}{2} - 3 = \frac{5}{2} - \frac{6}{2} = -\frac{1}{2} \text{ (satisfies the second equation)}
\]

3. For \( (-\frac{1}{2}, 5) \):
\[
y = -\frac{1}{2}(-\frac{1}{2}) + 2 \implies y = \frac{1}{4} + 2 = \frac{1}{4} + \frac{8}{4} = \frac{9}{4} \text{ (not equal to } 5)
\]

The only point that satisfies both equations is:

**Solution: \((5, -\frac{1}{2})\)**.