QUESTION IMAGE
Question
- $y - x > 4$
$x + y > 2$
- $y < 2x + 4$
Step1: Analyze the first inequality \( y - x > 4 \)
Rewrite it in slope - intercept form (\(y=mx + b\)): \(y>x + 4\). The boundary line is \(y=x + 4\), with a slope of \(1\) and a \(y\) - intercept of \(4\). Since the inequality is \(y>x + 4\), the line should be dashed (because the inequality is strict, \(>\) not \(\geq\)) and we shade the region above the line.
Step2: Analyze the second inequality \(x + y>2\)
Rewrite it in slope - intercept form: \(y>-x + 2\). The boundary line is \(y=-x + 2\), with a slope of \(- 1\) and a \(y\) - intercept of \(2\). Since the inequality is \(y>-x + 2\), the line should be dashed and we shade the region above the line.
Step3: Graph the boundary lines and find the intersection
- For \(y=x + 4\): When \(x = 0\), \(y = 4\); when \(y=0\), \(x=-4\). Plot the points \((0,4)\) and \((-4,0)\) and draw a dashed line. Shade above the line.
- For \(y=-x + 2\): When \(x = 0\), \(y = 2\); when \(y = 0\), \(x = 2\). Plot the points \((0,2)\) and \((2,0)\) and draw a dashed line. Shade above the line.
The solution to the system of inequalities is the region where the two shaded regions overlap.
(If we were to graph it on the given coordinate plane with the \(x\) - axis and \(y\) - axis, we would first draw the two dashed lines and then identify the overlapping region. For example, we can test a point in the overlapping region. Let's take the point \((0,5)\). For \(y - x>4\): \(5-0 = 5>4\), which is true. For \(x + y>2\): \(0 + 5=5>2\), which is true. Another point, say \((1,6)\): \(6 - 1=5>4\) and \(1 + 6 = 7>2\), also true. A point outside the overlapping region, like \((0,3)\): For \(y - x>4\), \(3-0=3
ot>4\), so it's not in the solution region.)
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To graph the system \(
\), we:
- Graph \(y=x + 4\) as a dashed line and shade above it.
- Graph \(y=-x + 2\) as a dashed line and shade above it.
- The solution is the region where the two shaded regions overlap.