To find the solution to the system of linear equations represented by the lines \( f(x) \) and \( g(x) \), we need to first derive the equations of each line based on the points provided.
For \( f(x) \):
The points on the line \( f(x) \) are (-3, 3), (0, 2), and (3, 1). We first determine the slope (m) using two points:
Using (-3, 3) and (0, 2): \[ m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{2 - 3}{0 - (-3)} = \frac{-1}{3} = -\frac{1}{3} \]
To write the equation of the line in point-slope form: \[ y - y_1 = m(x - x_1) \] Using the point (0, 2): \[ y - 2 = -\frac{1}{3}(x - 0) \implies y = -\frac{1}{3}x + 2 \]
For \( g(x) \):
The second line \( g(x) \) passes through the points (-3, 0) and (0, 2). We also calculate the slope: \[ m = \frac{2 - 0}{0 - (-3)} = \frac{2}{3} \]
Using point-slope form for the line \( g(x) \): Using point (0, 2): \[ y - 2 = \frac{2}{3}(x - 0) \implies y = \frac{2}{3}x + 2 \]
Finding the intersection:
We now have the equations:
- \( f(x): y = -\frac{1}{3}x + 2 \)
- \( g(x): y = \frac{2}{3}x + 2 \)
Setting them equal to find the intersection: \[ -\frac{1}{3}x + 2 = \frac{2}{3}x + 2 \]
Subtracting 2 from both sides: \[ -\frac{1}{3}x = \frac{2}{3}x \]
Adding \(\frac{1}{3}x\) to both sides: \[ 0 = \frac{2}{3}x + \frac{1}{3}x = \frac{3}{3}x \] \[ 0 = x \]
Substituting \( x = 0 \) into either equation to find \( y \): From \( f(0) \): \[ y = -\frac{1}{3}(0) + 2 = 2 \]
Thus the intersection point (the solution to the system of equations) is: (0, 2)
Conclusion:
The solution to the system of linear equations is \( (0, 2) \).
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