SOLUTION: I just can't figure this out no matter how I try. I am to find an intersection point(if there is one) and graph, shading the appropriate areas. y>=-1/2x + 1 and 2x-3y>=3. I ne

Question 13772: I just can't figure this out no matter how I try. I am to find an intersection point(if there is one) and graph, shading the appropriate areas.
y>=-1/2x + 1 and 2x-3y>=3. I need to know a step by step solution. I don't know what they are asking for. This is confusing to me. Any help would be appreciated.

Answer by longjonsilver(2297) (Show Source): You can put this solution on YOUR website!
right, this is a fairly straight forward question...biggest issue is visualising what is going on, so here goes.

First, you have 2 straight lines (if i have read your question correctly), namely
y >= (1/2)x + 1
2x-3y>=3.

Now, having just "=" means the points on the line, if you plot it. If you have ">" or "<", then these correspond to the region above and below the line.

Where do 2 lines cross? well first they cross at one point only (if at all).. this is the "root" or "solution". So lets find this point. Note the < has gone... i am interested to know where the 2 lines cross ie are equal.

y = (1/2)x + 1 --eqn1
2x-3y = 3 --eqn2

So, put eqn1 into eqn2 as:

--> x = 12

Now, if x=6, then from y = (1/2)x + 1, we get
y = (1/2)(6) + 1
y = 3 + 1
--> y = 4

Now, graph them. You mean plot or sketch? Plot is a proper graph on graph paper, a sketch is a rough drawing on normal paper, to highlight the rough shape.

To Sketch, you need to understand the "y=mx+c" version of any straight line...Look at my Lesson on this topic on this website... they will help.

jon.

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