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You can put this solution on YOUR website!
Since I do not know how to use Algebra.com's software to actually draw the full, correct graph of the solution, I am going to describe the proper graph and provide a graph that is partially complete and partially incorrect.
\n" ); document.write( "To start with think of the related equations:
\n" ); document.write( "y = x + 1
\n" ); document.write( "y = (3/2)x + 3
\n" ); document.write( "I hope you would know how to graph these lines. They are both in Slope-intercept form so they should be easy. The first line would be the points where the y-coordinate equals (x+1). The second line would be the points where the y-coordinate equals ((3/2)x + 3). Since the both inequalities are not of the \"or equal to\" type, we do not want points where the y equals the right side. So the lines of the equations are NOT part of our solution. But, as we will soon see, they do serve as boundaries!
\n" ); document.write( "Now let's think about the inequality y < x + 1. This says the y is less than (x + 1). Where do we find lower y values? Up or down? Hopefully you understand that down is where we find lower y values. This means that the solution to y < x + 1 are the points below the line y = x + 1. If this was our only inequality then we would shade the area below the line y = x + 1 and we would be finished.
\n" ); document.write( "But we also have the second inequality y > (3/2)x + 3. This says y is greater than ((3/2)x + 3). Using logic like we did above we should know then that the solution to this equation is above the line y = (3/2)x + 3.
\n" ); document.write( "So our solution is the points in the area below y = x + 1 and above the line y = (3/2)x + 3.
\n" ); document.write( "Here's a step-by-step procedure:
- Graph, as a dotted (not solid), line the graph of y = x + 1. (The dotted line indicates that the points are not to be considered part of the solution. If the inequality has been an \"or equal to\" type, we would use a solid line indicating that the line, where y = ..., is part of the solution.)
- Graph, as a dotted (not solid), line the graph of y = (3/2)x + 3
- Now we have a graph of two dotted lines. These lines divide the graph into 4 regions. Find and shade the region that is both below the y = x + 1 and above the y = (3/2)x + 3 line. (Only 1 of the 4 regions fit this description.)
\n" ); document.write( "Below is a graph which
- Has no shading (which is the critical part of the graph),/li>
- Has solid lines where you need dotted lines
\n" ); document.write( "I'll leave it to you to figure out which line is which.
\n" ); document.write( "