Question

Given below are the graphs of two lines, y=-0.5 + 5 and y=-1.25x + 8 and several regions and points are shown. Note that C is the region that appears completely white in the graph.

CAN SOMEONE HELP ME WITH BOTH A AND B ASAP

Answer 1

We have the following equations:


`(1) \ y=-0.5x+5 \\ (2) \ y=-1.25x+8`

So we are asked to write a system of equations or inequalities for each region and each point.

Part a)

Region Example A


`y \leq -0.5x+5 \\ y \leq -1.25x+8`

Region B.

Let's take a point that is in this region, that is:


`P(0,6)`

So let's find out the signs of each inequality by substituting this point in them:


`y \ (?)-0.5x+5 \\ 6 \ (?) -0.5(0)+5 \\ 6 \ (?) \ 5 \\ 6\ \textgreater \ 5 \\ \\ y \ (?) \ -1.25x+8 \\ 6 \ (?) -1.25(0)+8 \\ 6 \ (?) \ 8 \\ 6\ \textless \ 8`

So the inequalities are:


`(1) \ y \geq -0.5x+5 \\ (2) \ y \leq -1.25x+8`

Region C.

A point in this region is:


`P(0,10)`

So let's find out the signs of each inequality by substituting this point in them:


`y \ (?)-0.5x+5 \\ 10 \ (?) -0.5(0)+5 \\ 10 \ (?) \ 5 \\ 10\ \textgreater \ 5 \\ \\ y \ (?) \ -1.25x+8 \\ 10 \ (?) -1.25(0)+8 \\ 10 \ (?) \ 8 \\ 10 \ \ \textgreater \ \ 8`

So the inequalities are:


`(1) \ y \geq -0.5x+5 \\ (2) \ y \geq -1.25x+8`

Region D.

A point in this region is:


`P(8,0)`

So let's find out the signs of each inequality by substituting this point in them:


`y \ (?)-0.5x+5 \\ 0 \ (?) -0.5(8)+5 \\ 0 \ (?) \ 1 \\ 0 \ \ \textless \ \ 1 \\ \\ y \ (?) \ -1.25x+8 \\ 0 \ (?) -1.25(8)+8 \\ 0 \ (?) \ -2 \\ 0 \ \ \textgreater \ \ -2`

So the inequalities are:


`(1) \ y \leq -0.5x+5 \\ (2) \ y \geq -1.25x+8`

Point P:

This point is the intersection of the two lines. So let's solve the system of equations:


`(1) \ y=-0.5x+5 \\ (2) \ y=-1.25x+8 \\ \\ Subtracting \ these \ equations: \\ 0=0.75x-3 \\ \\ Solving \ for \ x: \\ x=4 \\ \\ Solving \ for \ y: \\ y=-0.5(4)+5=3`

Accordingly, the point is:


`\boxed{p(4,3)}`

Point q:

This point is the
`x-intercept` of the line:


`y=-0.5x+5`

So let:


`y=0`

Then


`x=(5)/(0.5)=10`

Therefore, the point is:


`\boxed{q(10,0)}`

Part b)

The coordinate of a point within a region must satisfy the corresponding system of inequalities. For each region we have taken a point to build up our inequalities. Now we will take other points and prove that these are the correct regions.

Region Example A

The origin is part of this region, therefore let's take the point:


`O(0,0)`

Substituting in the inequalities:


`y \leq -0.5x+5 \\ 0 \leq -0.5(0)+5 \\ \boxed{0 \leq 5} \\ \\ y \leq -1.25x+8 \\ 0 \leq -1.25(0)+8 \\ \boxed{0 \leq 8}`

It is true.

Region B.

Let's take a point that is in this region, that is:


`P(0,7)`

Substituting in the inequalities:


`y \geq -0.5x+5 \\ 7 \geq -0.5(0)+5 \\ \boxed{7 \geq \ 5} \\ \\ y \leq \ -1.25x+8 \\ 7 \ \leq -1.25(0)+8 \\ \boxed{7 \leq \ 8}`

It is true

Region C.

Let's take a point that is in this region, that is:


`P(0,11)`

Substituting in the inequalities:


`y \geq -0.5x+5 \\ 11 \geq -0.5(0)+5 \\ \boxed{11 \geq \ 5} \\ \\ y \geq \ -1.25x+8 \\ 11 \ \geq -1.25(0)+8 \\ \boxed{11 \geq \ 8}`

It is true

Region D.

Let's take a point that is in this region, that is:


`P(9,0)`

Substituting in the inequalities:


`y \leq -0.5x+5 \\ 0 \leq -0.5(9)+5 \\ \boxed{0 \leq \ 0.5} \\ \\ y \geq \ -1.25x+8 \\ 0 \geq -1.25(9)+8 \\ \boxed{0 \geq \ -3.25}`

It is true