Question

question 10
$5x + 4y \geq 4$
$3x - y \leq 1$

Explanation:

Step1: Rewrite inequalities as slope - intercept form

For the first inequality \(5x + 4y\geq4\), solve for \(y\):
Subtract \(5x\) from both sides: \(4y\geq - 5x + 4\)
Divide both sides by \(4\): \(y\geq-\frac{5}{4}x + 1\)
The boundary line is \(y = -\frac{5}{4}x+1\), with a slope of \(-\frac{5}{4}\) and a \(y\) - intercept of \(1\). Since the inequality is \(\geq\), the line is solid and we shade above the line.

For the second inequality \(3x - y\leq1\), solve for \(y\):
Subtract \(3x\) from both sides: \(-y\leq - 3x + 1\)
Multiply both sides by \(- 1\) (and reverse the inequality sign): \(y\geq3x - 1\)
The boundary line is \(y = 3x - 1\), with a slope of \(3\) and a \(y\) - intercept of \(-1\). Since the inequality is \(\geq\) (after multiplying by \(-1\)), the line is solid and we shade above the line.

Step2: Graph the boundary lines

• For \(y = -\frac{5}{4}x + 1\):
• The \(y\) - intercept is \((0,1)\). To find another point, use the slope. The slope \(-\frac{5}{4}\) means from \((0,1)\), we move down \(5\) units and right \(4\) units to get the point \((4,1 - 5)=(4,-4)\). Draw a solid line through \((0,1)\) and \((4,-4)\).
• For \(y = 3x-1\):
• The \(y\) - intercept is \((0, - 1)\). Using the slope \(3\) (which is \(\frac{3}{1}\)), from \((0,-1)\), we move up \(3\) units and right \(1\) unit to get the point \((1,2)\). Draw a solid line through \((0,-1)\) and \((1,2)\).

Step3: Shade the solution region

The solution to the system of inequalities is the region that is shaded above both \(y = -\frac{5}{4}x + 1\) and \(y = 3x - 1\). To find the intersection point of the two lines, set \(-\frac{5}{4}x + 1=3x - 1\)
Multiply through by \(4\) to clear the fraction: \(-5x + 4 = 12x-4\)
Add \(5x\) to both sides: \(4=17x - 4\)
Add \(4\) to both sides: \(8 = 17x\)
So \(x=\frac{8}{17}\approx0.47\)
Substitute \(x = \frac{8}{17}\) into \(y = 3x - 1\): \(y=3\times\frac{8}{17}-1=\frac{24}{17}- \frac{17}{17}=\frac{7}{17}\approx0.41\)
The intersection point is \((\frac{8}{17},\frac{7}{17})\). The solution region is the area that is above both lines (including the lines themselves).

Answer:

To graph the system \(

$$\begin{cases}5x + 4y\geq4\\3x - y\leq1\end{cases}$$

\), we first rewrite the inequalities in slope - intercept form as \(y\geq-\frac{5}{4}x + 1\) and \(y\geq3x - 1\). Then we graph the solid lines \(y = -\frac{5}{4}x + 1\) (with slope \(-\frac{5}{4}\) and \(y\) - intercept \(1\)) and \(y = 3x - 1\) (with slope \(3\) and \(y\) - intercept \(-1\)). The solution region is the area above both lines (including the lines), with the intersection point of the two lines being \((\frac{8}{17},\frac{7}{17})\).