Find the coordinates of the vertices of the triangle formed by each system of inequalities.
[tex]6y - 24x \geqslant - 168 \\ 8y + 7x > 10 \\ 20y - 2x \leqslant 64[/tex]
I've solved the other previous questions and I'm just really stuck on this one for some reason. I've given an example of how it should be solved based on how a previous question I solved.

Knowing that the inequalities form a triangle, we know that we need to find three distinct solutions when two of the given inequalities intersect.

Since the intersection of two inequalities occur when their lines do, we can treat all of the inequalities as a system of equations or as a linear functions to find where they intersect.

What differentiates a line and an inequality is that one has a shaded region and the possibility of being a dotted line while the other does not. Knowing that the problem is concerned with the inequalities intersections rather than their overlapping regions, we don't need to bother with the <, >, ≥, or ≤ signs.

So we need to solve three systems of equations,

(1) 6y - 24x = -168

8y + 7x = 10,

(2) 8y + 7x = 10

20y - 2x = 64

and,

(3) 6y - 24x = -168

20y - 2x = 64.

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Solving the Problem

Let's start with (1).

Identifying that the Least Common Multiple (LCM) of 6 and 8 is 24, we can multiply each equation by a factor to make the y terms have the same value.

4(6y - 24x = -168) --> 24y - 96x = -672

3(8y +7x = 10) --> 24y + 21x = 30

Now that they both have the same y term, we can subtract them from each other to solve for x in which we can plug that value in to find y!

24y - 96x = -672

-(24y + 21x = 30)

-117x = -702

x = 6

Plug x = 6 into the second equation:

8y + 7(6) = 10

8y + 42 = 10

8y = -32

y = -4.

So, one of the intersections is at (6, -4).

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Next, let's do (2).

Using the LCM of 8 and 20, we multiply each of the equations by an appropriate factor to get y terms that have the same coefficient.

5(8y + 7x = 10) --> 40y + 35x = 50

2(20y - 2x = 64) --> 40y - 4x = 128

Subtracting the equations, the value of x is,

40y + 35x = 50

-(40y - 4x = 128)

39x = -78

x = -2.

Plugging x = -2 into the first equation:

8y + 7(-2) = 10

8y - 14 = 10

8y = 24

y = 3.

Another intersection occurs at (-2, 3).

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Lastly, let's solve (3).

Multiplying both equation with an appropriate factor that lets y terms have the same coefficient, we have,

10(6y - 24x = -168) --> 60y - 240x = -1680

3(20y - 2x = 64) --> 60y - 6x = 192.

Subtracting those equations,

60y - 240x = -1680

-(60y - 6x = 192)

-234x = -1872

x = 8.

Plugging x = 8 into the second equation:

20y - 2(8) = 64

20y - 16 = 64

20y = 80

y = 4.

The last intersection is at (8, 4).