To find the solution to the system of equations given by the red line \(y = 3 + 2\) (which simplifies to \(y = 5\)) and the blue line \(y = -x + 6\), we can determine the intersection point of the two lines on the graph.
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Identify the equations:
- The red line has a constant \(y\) value of \(5\) (horizontal line).
- The blue line has a negative slope and \(y\)-intercept of \(6\).
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Plotting the lines:
- The red line will be a horizontal line at \(y = 5\).
- The blue line will decrease as \(x\) increases, starting at \(y = 6\) when \(x = 0\) and crossing the \(y = 5\) line at some \(x\) value.
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Finding the intersection:
- Set \(y = 5\) (from the red line) into the blue line's equation: \[ 5 = -x + 6 \] Rearranging gives: \[ -x = 5 - 6 \ -x = -1 \ x = 1 \]
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Coordinates of the intersection point:
- The intersection point can then be written as the ordered pair \((1, 5)\).
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Conclusion:
- The solution to the system of equations is the point where the two lines intersect. Since the red line is constant at \(y = 5\) and the blue line is calculated to intersect at \(x = 1\), the solution to the system is: \[ \boxed{(1, 5)} \]
This is how we determine the solution: by finding where the lines intersect on the graph and confirming that the coordinates of that intersection satisfy both equations.