# Math Help - Cornerpoints of solution region of linear system.

1. ## Cornerpoints of solution region of linear system.

I am having trouble finding the cornerpoints of the solution region of a linear system. There seems to be discrepancy between the graph of the solution region and the cornerpoints I can find algebraically.

Here is the linear system:
x, y ≥0
1x + 11y ≤ 1
7x + 2y ≤1

This is a picture of the linear system plotted on a graph and of the system's small bounded solution region:

The cornerpoints that I have obtained algebraically are:

(i) for the intersection of y = 0 and x = 0, (0, 0)
(ii) for the intersection of y = 0 and 7x + 2y = 1, (1/7, -1/2)
(iii) for the intersection of x = 0 and 1x + 11y = 1, (1,0)
(iv) for the intersection of 1x + 11y = 1 and 7x + 2y = 1, (8.3, -0.7)

Those cornerpoints don't seem to be right according to the bounded region in the graph.
Here is an example of how I algebraically came to a cornerpoint:

Intersection of 1x + 11y = 1 and 7x + 2y = 1

(1-7x)/2 = (1-1x)/11
11(1-7x) = 2(1-1x)
11 - 77x = 2 - 2x
11 - 77x - 2 + 2x = 0
9 - 75x = 0
-75x = -9
x = 8.333333333

1(8.333333333) + 11y = 1
8.333333333 + 11y = 1
11y = -7.333333333
y = -0.666666666 or -0.7

2. ## Re: Cornerpoints of solution region of linear system.

Hi Yoodle15!

Originally Posted by Yoodle15
-75x = -9
x = 8.333333333
Could you calculate that again?

3. ## Re: Cornerpoints of solution region of linear system.

$\frac{9}{75}= \frac{3(3)}{3(25)}= \frac{3}{25}= \frac{3(4)}{25(4)}= \frac{12}{100}$

4. ## Re: Cornerpoints of solution region of linear system.

Such a silly mistake on my part! Thank you for pointing it out to me! *feels silly*

5. ## Re: Cornerpoints of solution region of linear system.

No problem. Cheers!