Systems of inequalities

[freeze44]

Need Help
I'm having problems graphing the inequalities and finding the corner points for the following:

x+y>=14
2x+y>=20
x+y>=24
x>=0
y>=0

Thankyou in advance.
freeze44

 

[Deleted member 4993]

Need Help
I'm having problems graphing the inequalities and finding the corner points for the following:

x+y>=14
2x+y>=20
x+y>=24
x>=0
y>=0

Thankyou in advance.
freeze44

Draw the lines assuming equality.

Shade the valid section - according to inequality.

Tell us exactly what part you are missing.

 

[imaginary # hater]

don't forget to draw dashed lines on signs that are not under lined and draw solid lines for the signs that are underlined

 

[soroban]

Hello, freeze44!

Exactly where is your difficulty?
. . You can't graph lines?
. . You can't shade the correct regions?

And are you sure you copied the problem correctly?
The inequalities are rather silly ones . . .

\(\displaystyle \begin{Bmatrix} x+y &\geq & 14 \\ 2x+y &\geq& 20 \\ x+y &\geq& 24 \\ x &\geq & 0 \\ y &\geq& 0 \end{Bmatrix}\)

From \(\displaystyle x \geq 0,\:y \geq 0\), we are limited to Quadrant 1.

The line, \(\displaystyle x+y\:=\:14\), has intercepts (14,0) and (0,14).
. . Shade the region above the line.

The line \(\displaystyle 2x+y \:=\:20\) has intercepts (10,0) and (0,20).
. . Shade the region above the line.

The line \(\displaystyle x+y \:=\:24\) has intercepts (24,0) and (0,24).
. . Shade the region above the line.


The region that satisfies all three inequalities (shaded three times) looks like this:

        |::::::::::::
     24 *:::::::::::::
        | *::::::::::::
        |   *:::::::::::
        |     *::::::::::
        |       *:::::::::
        |         * :::::::
    - - + - - - - - * - - - -
        |          24