SOLUTION: My homework is to graph systems of inequalities. The problem I'm having trouble with is: x-3y>(or equal to)-7 5x+y<(or equal to)13 x+6y>(or equal to)-9 3x-2y>(or equal to)-7

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Question 279605: My homework is to graph systems of inequalities. The problem I'm having trouble with is:
x-3y>(or equal to)-7
5x+y<(or equal to)13
x+6y>(or equal to)-9
3x-2y>(or equal to)-7
I have successfully graphed the first two, but I cannot figure out the third or last one yet. I graphed what I thought the third one looked like, but it didn't make any sense to the problem. I set the y-intercept at -1 1/3 with a slope of -9/6 which reduces to -3/2.
Answer by Theo(13342) (Show Source):
You can put this solution on YOUR website!
third equation is x + 6y >= -9

subtract x from both sides of this equation to get:

6y >= -x - 9

divide both sides of this equation by 6 to get:

y = (-1/6)x - (9/6)

graph the equation of y = -(1/6)x - (9/6)

the shaded area will be everything above or equal to that line.

any value of y above or on that line will be valid.

the graph will look like this:

your slope should have been -(1/6), and not -(9/6).

you solve your fourth equation in a similar manner.

your fourth equation is 3x-2y >= -7.

subtract 3x from both sides of this equation to get:

-2y >= -3x - 7

multiply both sides of this equation by (-1) to get:

2y <= 3x + 7

note that multiplying both sides of an inequality reverses the inequality.

if 5 > 3, then -5 < -3 for example.

your equation is now:

2y <= 3x + 7

divide both sides of this equation by 2 to get:

y <= (3/2)*x + (7/2)

graph the equation of y = (3/2)*x + (7/2)

your shaded area will be everything below or on that line.

any value of y below or on that line will be valid.

a graph of that equation looks like this: