Q28. - Algebra 2

Question

Find the coordinates of the vertices of the figure formed by each system of inequalities. 

 

$\begin{array}{c}28.y\ge -3\\ x\le 6\\ y\ge x-2\\ 2y\le x+5\end{array}$

Step-by-Step Solution

Answer

The figure formed by system of inequalities is a quadrilateral.

The coordinates of the vertices of quadrilateral are $\left(-11,-3\right)$, $\left(-1,-3\right)$, $\left(6,4\right)$ and $\left(6,5.5\right)$.

Step-1 – Apply the concept of graphing the inequality

The steps to graph the inequality are provided below.

1. If the inequality contains greater than or less than sign then the boundary of the line will be dashed. If the inequality contains signs of greater than or equal to or less than or equal to then the boundary of the line will be solid. 

2. Select a point (known as test point) from the plane that does not lie on the boundary on the line and substitute it in the inequality. 

3. If the inequality is true then shade the region that contains the test point otherwise shade the other region when inequality is false.

Step-2 – Interpret the sign of the inequality

Consider the inequality provided below.

$y\ge -3$

The inequality contains the sign of greater than or equal to.

Therefore, the boundary line will be solid.

Next, consider the inequality $x\le 6$.

The inequality contains the sign of less than or equal to.

Therefore, the boundary line will be solid.

Next, consider the inequality $y\ge x-2$.

The inequality contains the sign of greater than or equal to.

Therefore, the boundary line will be solid.

Next, consider the inequality $2y\le x+5$.

The inequality contains the sign of less than or equal to.

Therefore, the boundary line will be solid.

Step-3 – Graph the inequalities

Graph the inequalities $y\ge -3$, $x\le 6$, $y\ge x-2$ and $2y\le x+5$ on same plane and shade the region.

Draw the line.

$y=-3$

Take a test point that does not lie on the boundary of the line, say $\left(0,0\right)$. 

Substitute the point $\left(0,0\right)$ in the inequality and check whether it’s true or not.

$0\ge -3$

This is true.

Therefore, shade the region containing the point $\left(0,0\right)$.

Draw the line.

$x=6$

Take a test point that does not lie on the boundary of the line, say $\left(0,0\right)$. 

Substitute the point $\left(0,0\right)$ in the inequality and check whether it’s true or not.

$0\le 6$

This is true.

Therefore, shade the region containing the point $\left(0,0\right)$.

Draw the line $y=x-2$.

Take a test point that does not lie on the boundary of the line, say $\left(0,0\right)$. 

Substitute the point $\left(0,0\right)$ in the inequality and check whether it’s true or not.

 $\begin{array}{c}0\ge 0-2\\ 0\ge -2\end{array}$

This is true.

Therefore, shade the region containing the point $\left(0,0\right)$.

Draw the line $2y=x+5$.

Take a test point that does not lie on the boundary of the line, say $\left(0,0\right)$. 

Substitute the point $\left(0,0\right)$ in the inequality and check whether it’s true or not.

 $\begin{array}{c}2\left(0\right)\le 0+5\\ 0\le 5\end{array}$

This is true.

Therefore, shade the region containing the point $\left(0,0\right)$.

Thus, the common shaded region is provided below.

The green line denotes the equation $y=-3$, purple line denotes the equation $x=6$, black line denotes the equation $y=x-2$ and red line denotes the equation $2y=x+5$.

Step-4 – Find the coordinates

The point of intersection of inequalities are the required coordinates of the figure.

The figure obtained by system of inequalities is a quadrilateral.

The coordinates of three vertices of the quadrilateral are $\left(-11,-3\right)$, $\left(-1,-3\right)$ and $\left(6,4\right)$.

To find the fourth coordinate compute the intersection point of the lines $x=6$ and $2y=x+5$.

Apply the method of substitution to find the coordinates.

Substitute $x=6$ in the equation $2y=x+5$ to solve for y.

$\begin{array}{c}2y=6+5\\ y=\frac{11}{2}\\ y=5.5\end{array}$

Therefore, the point of intersection is $\left(6,5.5\right)$.

Thus, the coordinates of the vertices of quadrilateral are $\left(-11,-3\right)$, $\left(-1,-3\right)$, $\left(6,4\right)$ and $\left(6,5.5\right)$.