Question

sketch the solution to each system of ineqs 27) $ygeq-\frac{5}{3}x + 2$ $yleq-\frac{5}{3}x - 1$

Explanation:

Step1: Identify the boundary - lines

The boundary - lines for the inequalities $y\geq-\frac{5}{3}x + 2$ and $y\leq-\frac{5}{3}x-1$ are $y =-\frac{5}{3}x + 2$ and $y=-\frac{5}{3}x - 1$ respectively. These are two parallel lines with slope $m =-\frac{5}{3}$. The $y$ - intercept of $y =-\frac{5}{3}x + 2$ is $b_1 = 2$ and the $y$ - intercept of $y=-\frac{5}{3}x - 1$ is $b_2=-1$.

Step2: Determine the shading for $y\geq-\frac{5}{3}x + 2$

For the inequality $y\geq-\frac{5}{3}x + 2$, we test a point not on the line, say $(0,0)$. Substitute $x = 0$ and $y = 0$ into the inequality: $0\geq-\frac{5}{3}(0)+2$, or $0\geq2$, which is false. So, we shade the region above the line $y =-\frac{5}{3}x + 2$.

Step3: Determine the shading for $y\leq-\frac{5}{3}x - 1$

For the inequality $y\leq-\frac{5}{3}x - 1$, we test the point $(0,0)$. Substitute $x = 0$ and $y = 0$ into the inequality: $0\leq-\frac{5}{3}(0)-1$, or $0\leq - 1$, which is false. So, we shade the region below the line $y=-\frac{5}{3}x - 1$.

Step4: Analyze the solution set

Since we have $y\geq-\frac{5}{3}x + 2$ and $y\leq-\frac{5}{3}x - 1$, and the two lines are parallel with the upper - bound line $y =-\frac{5}{3}x + 2$ above the lower - bound line $y=-\frac{5}{3}x - 1$, there is no intersection between the two shaded regions. So the solution set of the system of inequalities is the empty set.

Answer:

The solution set of the system of inequalities is the empty set, $\varnothing$.