To solve the system of equations by graphing, we'll first rewrite both equations:
- \( y = -x + 1 \)
- \( y = -3x + 7 \)
Next, we can find points to graph each line.
Graphing the first equation: \( y = -x + 1 \)
To find points, we can substitute values for \( x \):
- If \( x = 0 \): \[ y = -0 + 1 = 1 \quad \Rightarrow \quad (0, 1) \]
- If \( x = 1 \): \[ y = -1 + 1 = 0 \quad \Rightarrow \quad (1, 0) \]
- If \( x = -1 \): \[ y = -(-1) + 1 = 2 \quad \Rightarrow \quad (-1, 2) \]
Graphing the second equation: \( y = -3x + 7 \)
Similarly, substitute values for \( x \):
- If \( x = 0 \): \[ y = -3(0) + 7 = 7 \quad \Rightarrow \quad (0, 7) \]
- If \( x = 1 \): \[ y = -3(1) + 7 = 4 \quad \Rightarrow \quad (1, 4) \]
- If \( x = 2 \): \[ y = -3(2) + 7 = 1 \quad \Rightarrow \quad (2, 1) \]
Now plot the points on a graph:
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For the first line \( y = -x + 1 \):
- Plot points \( (0, 1) \), \( (1, 0) \), and \( (-1, 2) \).
- Connect these points to draw the line.
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For the second line \( y = -3x + 7 \):
- Plot points \( (0, 7) \), \( (1, 4) \), and \( (2, 1) \).
- Connect these points to draw the line.
Finding the intersection point:
After graphing both lines, look for the point where they intersect.
From the calculations:
- The first line \( y = -x + 1 \) has a y-intercept of 1 and a slope of -1.
- The second line \( y = -3x + 7 \) has a y-intercept of 7 and a steeper slope of -3.
The point of intersection occurs where both equations are satisfied.
Solution
To find it algebraically: Set the two equations equal to each other: \[ -x + 1 = -3x + 7 \] Solving for \( x \): \[ 3x - x = 7 - 1 \ 2x = 6 \ x = 3 \] Now plug \( x \) back into one of the equations to find \( y \): \[ y = -3 + 1 = -2 \]
Thus, the solution to the system of equations is: \[ \boxed{(3, -2)} \]
You can verify that both equations yield the same \( y \) value when \( x = 3 \).