Question

y=2/3 x + 1 and y=−2/3 x + 1

Answer 1

how would i graph that

Answer 2

The two equations you've provided are linear equations in slope-intercept form (y = mx + b), where \( m \) is the slope and \( b \) is the y-intercept.

1. **Equation 1**: \( y = \frac{2}{3}x + 1 \)
- Slope (\( m \)): \( \frac{2}{3} \)
- Y-intercept (\( b \)): \( 1 \)

2. **Equation 2**: \( y = -\frac{2}{3}x + 1 \)
- Slope (\( m \)): \( -\frac{2}{3} \)
- Y-intercept (\( b \)): \( 1 \)

### Graphing the Equations
- Both lines have the same y-intercept (1), which means they intersect the y-axis at the point (0, 1).
- The first line (with a positive slope) will rise to the right, while the second line (with a negative slope) will fall to the right.

### Finding the Intersection Point
To find the intersection point of the two equations, you can set the right-hand sides equal to each other:

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

Now, simplify and solve for \( x \):

1. Subtract 1 from both sides:
\[
\frac{2}{3}x = -\frac{2}{3}x
\]

2. Add \( \frac{2}{3}x \) to both sides:
\[
\frac{2}{3}x + \frac{2}{3}x = 0
\]
\[
\frac{4}{3}x = 0
\]

3. Divide by \( \frac{4}{3} \):
\[
x = 0
\]

Now substitute \( x = 0 \) back into either equation to find \( y \):
\[
y = \frac{2}{3}(0) + 1 = 1
\]

Thus, both lines intersect at the point \( (0, 1) \).

### Summary
- The equations represent two lines that intersect at the point \( (0, 1) \).
- The first line rises and the second line falls, both intersecting the y-axis at the same point.