SOLUTION: x>=2 y>=5 y<=3x-12 y<=-x+15 Solve the given inqualities for y if possible. I began by pluggin y=5 into the 3rd equation. 5=3x-12 17/3=x (17/3,5) but now I am stuck...

Click here to see ALL problems on Inequalities

Question 220656: x>=2
y>=5
y<=3x-12
y<=-x+15
Solve the given inqualities for y if possible.
I began by pluggin y=5 into the 3rd equation. 5=3x-12 17/3=x (17/3,5)
but now I am stuck...
Answer by Theo(13342) (Show Source):
You can put this solution on YOUR website!

looking at the graph, it looks like you have potential solutions when x >= 5+(2/3) and when x >= 13+(1/2)
-----
between x = 5+(2/3) and x = 13+(1/2) your y value would have to be >= 5 and <= 3x-12.
-----
from x = 13+(1/2) to infinity your y value would have to be >= 5 and <= x+15.
-----
hopefully that's the answer you are looking for.
solving this without looking at a graph is not something I would look forward to doing.
-----
It would be best to solve the equalities first and then get the intersection points.
-----
you have 3 equations.
solving for equalities, you would get:
y = 5
y = 3x-12
y = x+15
-----
solving for the intersection points you would get:
-----
y = 5 intersect with y = 3x-12 at x = 5+(2/3)
-----
y = 5 intersects with y = x+15 at x = -10
-----
y = 3x-12 intersects with y = x+15 at x = 13+(1/2)
-----
now that you have the intersection points, you can try to find where y has to fit in those intervals.
-----
the intervals are:
x = -10
x = 5+(2/3)
x = 13+(1/2)
-----
you can forget x = -10 because x has to be > 2 (one of the equations shown)
-----
when x = 2, you have 3 solutions of equality.
y = 5
y = 3x-12 = 3*2-12 = 6-12 = -6
y = x+15 = 3+15 = 18
-----
y can be greater than 5 and smaller than 18 at the same time, but it cannot be smaller than -6 and greater than 5 at the same time so it looks like the interval between 2 and 5+(2/3) is not going to be good. The graph shows that easily, but without the graph, you might need to plot more points to see it. Logiclly it would make sense since 3x-12 is below y = 5 and intersect with y = 5 at x = 5+(2/3) which means it must have been below it up to that point if it started below it at x = 2.
-----
you would do the same thing with the other intervals between the intersection points.
-----
as I said before, looking at the graph make the job so much easier. you still have to logically deduce what's going on but seeing what it looks like on the graph is much better than not.
-----
you can see a picture of the graph with comments by clicking on the following hyperlink.
-----
Picture of Graph with Comments