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Question
-x + 2y > 6
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To graph \(-x + 2y>6\), follow these steps:
Step1: Rewrite the inequality in slope - intercept form (\(y = mx + b\))
Start with \(-x + 2y>6\). Add \(x\) to both sides: \(2y>x + 6\). Then divide each term by 2: \(y>\frac{1}{2}x+3\).
Step2: Graph the boundary line
The boundary line is \(y=\frac{1}{2}x + 3\). Since the inequality is \(y>\frac{1}{2}x+3\) (not \(y\geq\frac{1}{2}x + 3\)), the boundary line should be a dashed line.
- The \(y\) - intercept (\(b\)) is 3, so plot the point \((0,3)\) on the \(y\) - axis.
- The slope (\(m\)) is \(\frac{1}{2}\), which means from the point \((0,3)\), we can rise 1 unit and run 2 units to the right to get the next point \((2,4)\), or run 2 units to the left and fall 1 unit to get the point \((- 2,2)\). Draw a dashed line through these points.
Step3: Shade the solution region
To determine which side of the line to shade, we can use a test point. A common test point is \((0,0)\) (as long as it is not on the boundary line). Substitute \(x = 0\) and \(y = 0\) into the inequality \(y>\frac{1}{2}x+3\):
\(0>\frac{1}{2}(0)+3\) simplifies to \(0 > 3\), which is false. So we shade the side of the line that does not contain the point \((0,0)\). In other words, we shade the region above the dashed line \(y=\frac{1}{2}x + 3\).