Question

1. \\(\

$$\begin{cases}5x + 2y \\geq 4 \\\\ x + 4y < -8\\end{cases}$$

\\)

Explanation:

Step1: Analyze the first inequality \(5x + 2y\geq4\)

Rewrite it in slope - intercept form (\(y = mx + b\)):
\(2y\geq - 5x + 4\)
\(y\geq-\frac{5}{2}x + 2\)
The boundary line is \(y = -\frac{5}{2}x+2\) (solid line because of \(\geq\)). To graph this line, find two points. When \(x = 0\), \(y=2\); when \(y = 0\), \(0=-\frac{5}{2}x + 2\), \(\frac{5}{2}x=2\), \(x=\frac{4}{5}\). Shade the region above the line (since \(y\geq\)).

Step2: Analyze the second inequality \(x + 4y\lt - 8\)

Rewrite it in slope - intercept form:
\(4y\lt - x - 8\)
\(y\lt-\frac{1}{4}x - 2\)
The boundary line is \(y=-\frac{1}{4}x - 2\) (dashed line because of \(\lt\)). To graph this line, when \(x = 0\), \(y=-2\); when \(y = 0\), \(0=-\frac{1}{4}x-2\), \(\frac{1}{4}x=-2\), \(x=-8\). Shade the region below the line (since \(y\lt\)).

Step3: Find the intersection of the two regions

The solution of the system is the region that is shaded by both inequalities. We can also check for the existence of a solution by seeing if the two regions overlap. Let's check a point. For the first inequality, take a point above \(y = -\frac{5}{2}x + 2\), say \((0,3)\): \(5(0)+2(3)=6\geq4\) (satisfies first). For the second inequality, \(0 + 4(3)=12\lt - 8\)? No. Take a point below \(y=-\frac{1}{4}x - 2\), say \((0,-3)\): \(5(0)+2(-3)=-6\geq4\)? No. Let's solve the system of equations \(

$$\begin{cases}y = -\frac{5}{2}x + 2\\y=-\frac{1}{4}x - 2\end{cases}$$

\)
Set \(-\frac{5}{2}x + 2=-\frac{1}{4}x - 2\)
\(-\frac{5}{2}x+\frac{1}{4}x=-2 - 2\)
\(-\frac{10}{4}x+\frac{1}{4}x=-4\)
\(-\frac{9}{4}x=-4\)
\(x=\frac{16}{9}\approx1.78\)
\(y=-\frac{5}{2}\times\frac{16}{9}+2=-\frac{40}{9}+2=-\frac{40 + 18}{9}=-\frac{22}{9}\approx - 2.44\)
Now, check if this point satisfies both inequalities. For \(5x + 2y\): \(5\times\frac{16}{9}+2\times(-\frac{22}{9})=\frac{80-44}{9}=\frac{36}{9} = 4\) (satisfies \(5x + 2y\geq4\)). For \(x + 4y\): \(\frac{16}{9}+4\times(-\frac{22}{9})=\frac{16-88}{9}=\frac{-72}{9}=-8\). But the second inequality is \(x + 4y\lt - 8\), so the point of intersection of the boundary lines does not satisfy the second inequality. In fact, the two regions do not overlap. So the system of inequalities has no solution.

Answer:

The system of inequalities \(

$$\begin{cases}5x + 2y\geq4\\x + 4y\lt - 8\end{cases}$$

\) has no solution.