Question 13772
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: {{{2x-3((1/2)x + 1) = 3}}}
{{{2x - (3/2)x - 3 = 3}}}
{{{2x - (3/2)x = 6}}}
{{{(1/2)x = 6}}}
--> 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.