Question

Identify a system of linear equations with one sollution by graphing or writing the equations in slope-intercept form

Answer 1

To identify a system of linear equations with one solution, we need two equations that intersect at exactly one point.

Let's consider the following two linear equations:

1. \( y = 2x + 1 \)
2. \( y = -\frac{1}{2}x + 3 \)

Next, let's analyze these equations for their slopes and y-intercepts:

1. The first equation, \( y = 2x + 1 \), has a slope (m) of 2 and a y-intercept (b) of 1. This means the line rises steeply.

2. The second equation, \( y = -\frac{1}{2}x + 3 \), has a slope of -1/2 and a y-intercept of 3. This means the line falls gently.

### Graphing the equations:
- For \( y = 2x + 1 \):
- When \( x = 0 \), \( y = 1 \) (point: (0, 1))
- When \( x = 1 \), \( y = 3 \) (point: (1, 3))

- For \( y = -\frac{1}{2}x + 3 \):
- When \( x = 0 \), \( y = 3 \) (point: (0, 3))
- When \( x = 6 \), \( y = 0 \) (point: (6, 0))

### Finding the intersection:
To determine the point where these two lines intersect, we can set the equations equal to each other:

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

Now, let's solve for \( x \):

1. Add \( \frac{1}{2}x \) to both sides:
\[
2x + \frac{1}{2}x + 1 = 3
\]

This simplifies to:
\[
\frac{5}{2}x + 1 = 3
\]

2. Subtract 1 from both sides:
\[
\frac{5}{2}x = 2
\]

3. Multiply both sides by \( \frac{2}{5} \):
\[
x = \frac{4}{5}
\]

Now substitute \( x \) back to find \( y \):

For \( y = 2x + 1 \):
\[
y = 2 \left( \frac{4}{5} \right) + 1 = \frac{8}{5} + 1 = \frac{13}{5}
\]

So, the point of intersection is \( \left( \frac{4}{5}, \frac{13}{5} \right) \).

### Conclusion:
The system of linear equations given by \( y = 2x + 1 \) and \( y = -\frac{1}{2}x + 3 \) intersects at one point, thus representing a system of equations with one unique solution: \( \left( \frac{4}{5}, \frac{13}{5} \right) \).

You can graph these two equations to visually confirm they intersect at exactly this point.