Which system of linear inequalities has the point (3,2) in its solution set?

Hello there!

we have the point (3,2) which means that x = 3 and y = 2

The systems of equations are:

Y <  - 2

Y ≥ (2/3)x - 4

and

Y > - 2

Y ≤ (2/3)x - 4

The points were the systems are true in the white areas, the colored areas are the excluded ones.

Now we need to put our point in both systems and see if the point is a solution or not,

In the first one, you can see that y needs to be less than -2, and in out point y is equal to 2, then the point (3,2) cant is a solution of the first system.

let's see the second system:

Y > - 2

Y ≤ (2/3)x - 4

valuate it in the point (3,2)

2 > -2

2 ≤ (2/3)*3 - 4 = - 2

this is also false.

Then the point (3,2) is not a solution for neither system, and you can see it in the graphs, in the first graph the point (3,2) is in the black area, and in the second one is in the red area.

step-by-step explanation:

given perimeter: 28cm

length: 9cm

atp,

2(l+w)=p of the reactangle

2(9+x)=28

18+2x=28

2x=28-18

x=10/2

x=5

therefore width is 5 cm