QUESTION IMAGE
Question
linear systems
graphing a system of two linear inequalities: advanced
(a) graph the solution to the following system of inequalities.
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graph grid
(b) give the coordinates of one point in the solution set.
point in the solution set: (\square, \square)
Part (a) Graphing the System of Inequalities
Step 1: Analyze the first inequality \(7x + 4y \leq -4\)
- Rewrite it in slope - intercept form (\(y=mx + b\)):
- Subtract \(7x\) from both sides: \(4y\leq - 7x-4\).
- Divide both sides by 4: \(y\leq-\frac{7}{4}x - 1\).
- The boundary line is \(y =-\frac{7}{4}x - 1\) (a straight line with slope \(m =-\frac{7}{4}\) and \(y\) - intercept \(b=-1\)). Since the inequality is \(\leq\), the line is solid, and we shade the region below the line.
Step 2: Analyze the second inequality \(y>-\frac{7}{4}x - 7\)
- The boundary line is \(y =-\frac{7}{4}x - 7\) (a straight line with slope \(m =-\frac{7}{4}\) and \(y\) - intercept \(b = - 7\)). Since the inequality is \(>\), the line is dashed, and we shade the region above the line.
Step 3: Find the overlapping region
- The two lines \(y =-\frac{7}{4}x - 1\) and \(y=-\frac{7}{4}x - 7\) are parallel (same slope). The solution to the system is the region that is below the solid line \(y =-\frac{7}{4}x - 1\) and above the dashed line \(y=-\frac{7}{4}x - 7\).
Part (b) Finding a Point in the Solution Set
Step 1: Choose a value for \(x\)
Let's choose \(x = 0\).
Step 2: Check the first inequality \(7x + 4y\leq - 4\)
Substitute \(x = 0\) into \(7x+4y\leq - 4\):
- \(7(0)+4y\leq - 4\), which simplifies to \(4y\leq - 4\), or \(y\leq - 1\).
Step 3: Check the second inequality \(y>-\frac{7}{4}x - 7\)
Substitute \(x = 0\) into \(y>-\frac{7}{4}x - 7\):
- \(y> - 7\).
Step 4: Find a \(y\) - value that satisfies both
We need \(y\) such that \(-7
- For \(7x + 4y\leq - 4\): \(7(0)+4(-2)=-8\leq - 4\) (true).
- For \(y>-\frac{7}{4}x - 7\): \(-2>-\frac{7}{4}(0)-7=-7\) (true).
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(a) The graph has a solid line \(y =-\frac{7}{4}x - 1\) (shade below) and a dashed line \(y=-\frac{7}{4}x - 7\) (shade above), with the overlapping region as the solution.
(b) One point in the solution set is \((0, - 2)\) (other valid points like \((-4, - 5)\) etc. are also correct).