To identify the number of solutions to the system of equations:
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Equation 1: \( y = \frac{1}{3}x + 2 \)
This is in slope-intercept form (y = mx + b), where the slope \( m \) is \( \frac{1}{3} \) and the y-intercept \( b \) is 2. -
Equation 2: \( -x + 3y = 6 \)
We can rewrite this in slope-intercept form by solving for \( y \):\[ -x + 3y = 6 \ 3y = x + 6 \ y = \frac{1}{3}x + 2 \]
Now, we can see that both equations are equivalent:
- From Equation 1: \( y = \frac{1}{3}x + 2 \)
- From Equation 2: \( y = \frac{1}{3}x + 2 \)
Since both equations represent the same line, there are infinitely many solutions to this system of equations, as every point on the line is a solution.
Conclusion
You can tell there are infinitely many solutions because both equations represent the same line in the coordinate plane. Therefore, they intersect at every point along that line.
Graphing
To graph the equations:
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The first equation \( y = \frac{1}{3}x + 2 \) has a y-intercept of \( 2 \) (the point \( (0, 2) \)) and a slope of \( \frac{1}{3} \) (rise over run). For every 3 units you move right (positive x-direction), you move up 1 unit (positive y-direction).
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The second equation, after re-arranging, gives you the same line. Thus, it will overlap exactly with the first line.
You can plot the following points based on the equation \( y = \frac{1}{3}x + 2 \):
- When \( x = 0 \), \( y = 2 \) (point \( (0, 2) \)).
- When \( x = 3 \), \( y = \frac{1}{3}(3) + 2 = 3 \) (point \( (3, 3) \)).
- When \( x = -3 \), \( y = \frac{1}{3}(-3) + 2 = 1 \) (point \( (-3, 1) \)).
Plotting these points along with drawing a straight line through them will represent both equations, which overlap perfectly. The graph will show a single line, confirming that there are infinitely many solutions.