Question

question 9 6-5 graded practice solve the system of inequalities by graphing. if there is no solut 2x + 5y ≥ 5 -3x + 4y ≤ 4

Explanation:

Step1: Rewrite the first inequality

For the inequality \(2x + 5y\geq5\), we solve for \(y\):
\[

$$\begin{align*} 5y&\geq - 2x + 5\\ y&\geq-\frac{2}{5}x + 1 \end{align*}$$

\]
This is a linear inequality. The boundary line is \(y = -\frac{2}{5}x+1\) (a straight line with slope \(-\frac{2}{5}\) and \(y\)-intercept \(1\)). Since the inequality is \(\geq\), we draw a solid line and shade the region above the line.

Step2: Rewrite the second inequality

For the inequality \(-3x + 4y\leq4\), we solve for \(y\):
\[

$$\begin{align*} 4y&\leq3x + 4\\ y&\leq\frac{3}{4}x + 1 \end{align*}$$

\]
The boundary line is \(y=\frac{3}{4}x + 1\) (a straight line with slope \(\frac{3}{4}\) and \(y\)-intercept \(1\)). Since the inequality is \(\leq\), we draw a solid line and shade the region below the line.

Step3: Find the intersection region

To solve the system, we find the region that is shaded by both inequalities.

• First, plot the line \(y = -\frac{2}{5}x + 1\): when \(x = 0\), \(y=1\); when \(y = 0\), \(2x=5\) so \(x=\frac{5}{2}=2.5\).
• Then, plot the line \(y=\frac{3}{4}x + 1\): when \(x = 0\), \(y = 1\); when \(y=0\), \(- 3x=4\) so \(x=-\frac{4}{3}\approx - 1.33\).

The two lines intersect at \((0,1)\) (by setting \(-\frac{2}{5}x + 1=\frac{3}{4}x + 1\), we get \(-\frac{2}{5}x=\frac{3}{4}x\), \(-\frac{8}{20}x-\frac{15}{20}x = 0\), \(-\frac{23}{20}x=0\) so \(x = 0\) and then \(y = 1\)).

The solution region is the set of all points \((x,y)\) that are above (or on) \(y=-\frac{2}{5}x + 1\) and below (or on) \(y=\frac{3}{4}x + 1\). To describe the solution, we can also find the intersection of the two half - planes.

Answer:

The solution to the system of inequalities is the region that lies above (or on) the line \(y = -\frac{2}{5}x + 1\) and below (or on) the line \(y=\frac{3}{4}x + 1\). In set - builder notation, it is \(\{(x,y):y\geq-\frac{2}{5}x + 1\text{ and }y\leq\frac{3}{4}x + 1\}\). If we want to find the vertices of the solution region (if it is a polygon), we can also find the intersection of the boundary lines with the axes and with each other. The intersection of the two boundary lines is \((0,1)\), the intersection of \(y = -\frac{2}{5}x + 1\) with the \(x\) - axis is \((2.5,0)\), and the intersection of \(y=\frac{3}{4}x + 1\) with the \(x\) - axis is \((-\frac{4}{3},0)\). But the main description of the solution is the region defined by the two inequalities.