To find the solution for the two equations \(y = \frac{1}{2}x + 2\) and \(y = -\frac{1}{4}x + 8\), we can follow these steps:
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Graph the equations:
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For the first equation, \(y = \frac{1}{2}x + 2\):
- The y-intercept is 2 (which is the point (0, 2)).
- The slope is \(\frac{1}{2}\), which means for every 2 units you move to the right (increase x by 2), you move up 1 unit (increase y by 1).
- Another point can be found at \(x = 2\): \[y = \frac{1}{2}(2) + 2 = 1 + 2 = 3\] So another point is (2, 3).
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For the second equation, \(y = -\frac{1}{4}x + 8\):
- The y-intercept is 8 (which is the point (0, 8)).
- The slope is \(-\frac{1}{4}\), meaning for every 4 units you move to the right (increase x by 4), you move down 1 unit (decrease y by 1).
- Another point can be found at \(x = 4\): \[y = -\frac{1}{4}(4) + 8 = -1 + 8 = 7\] So another point is (4, 7).
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Plot the points and draw the lines:
- Plot the points (0, 2) and (2, 3) for the first line and draw a line through them.
- Plot the points (0, 8) and (4, 7) for the second line and draw a line through them.
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Find the intersection point:
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To find the solution algebrically, set the two equations equal to each other: \[ \frac{1}{2}x + 2 = -\frac{1}{4}x + 8 \]
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Multiply every term by 4 to eliminate the fractions: \[ 4 \left(\frac{1}{2}x\right) + 4(2) = 4\left(-\frac{1}{4}x\right) + 4(8) \] \[ 2x + 8 = -x + 32 \]
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Add \(x\) to both sides: \[ 2x + x + 8 = 32 \] \[ 3x + 8 = 32 \]
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Subtract 8 from both sides: \[ 3x = 24 \]
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Divide by 3: \[ x = 8 \]
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Substitute \(x\) back into one of the original equations to find \(y\):
- Using the first equation: \[ y = \frac{1}{2}(8) + 2 = 4 + 2 = 6 \]
So, the solution to the system of equations is \((8, 6)\).
Summary:
- The intersection point is \((8, 6)\).
- The lines representing the two equations intersect at this point.