# 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

Algebra ->  Algebra  -> Coordinate Systems and Linear Equations -> 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       Log On

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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(3464)   (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: