Question 87161
Start with the given system of inequalities

{{{y>x+1}}}

{{{y<(3/2)x+3}}}


In order to graph this system of inequalities, we need to graph each inequality one at a time.



First lets graph the first inequality {{{y>x+1}}}

In order to graph {{{y>x+1}}}, we need to graph the <b>equation</b> {{{y=x+1}}} (just replace the inequality sign with an equal sign).
So lets graph the line {{{y=x+1}}} (note: if you need help with graphing, check out this <a href=http://www.algebra.com/algebra/homework/Linear-equations/graphing-linear-equations.solver>solver<a>)


{{{ graph( 500, 500, -5, 5, -5, 5, x+1) }}} graph of {{{y=x+1}}} 

Now lets pick a test point, say (0,0) (any point will work, but this point is the easiest to work with), and evaluate the inequality {{{y>x+1}}}


Substitute (0,0) into the inequality

{{{(0)>(0)+1}}} Plug in {{{x=0}}} and {{{y=0}}}

{{{0>1}}} Simplify

Since this inequality is <b>not</b> true, we do <b>not</b> shade the entire region that contains (0,0). So this means we shade the region that is on the opposite side of the line

{{{drawing( 500, 500, -5, 5, -5, 5,
graph(  500, 500, -5, 5, -5, 5,x+1,x+1+0.5),
graph(  500, 500, -5, 5, -5, 5,x+1,x+1+1),
graph(  500, 500, -5, 5, -5, 5,x+1,x+1+1.5),
graph(  500, 500, -5, 5, -5, 5,x+1,x+1+2),
graph(  500, 500, -5, 5, -5, 5,x+1,x+1+2.5),
graph(  500, 500, -5, 5, -5, 5,x+1,x+1+3),
graph(  500, 500, -5, 5, -5, 5,x+1,x+1+3.5),
graph(  500, 500, -5, 5, -5, 5,x+1,x+1+4),
graph(  500, 500, -5, 5, -5, 5,x+1,x+1+4.5),
graph(  500, 500, -5, 5, -5, 5,x+1,x+1+5),
graph(  500, 500, -5, 5, -5, 5,x+1,x+1+5.5),
graph(  500, 500, -5, 5, -5, 5,x+1,x+1+6),
graph(  500, 500, -5, 5, -5, 5,x+1,x+1+6.5),
graph(  500, 500, -5, 5, -5, 5,x+1,x+1+7),
graph(  500, 500, -5, 5, -5, 5,x+1,x+1+7.5),
graph(  500, 500, -5, 5, -5, 5,x+1,x+1+8),
graph(  500, 500, -5, 5, -5, 5,x+1,x+1+8.5),
graph(  500, 500, -5, 5, -5, 5,x+1,x+1+9),
graph(  500, 500, -5, 5, -5, 5,x+1,x+1+9.5))}}} Graph of {{{y>x+1}}} with the boundary (which is the line {{{y=x+1}}} in red) and the shaded region (in green)(note: since the inequality has a greater-than sign, this means the boundary is excluded. This means the solid line is really a dashed line)




---------------------------------------------------------------



Now lets graph the second inequality {{{y<(3/2)x+3}}}

In order to graph {{{y<(3/2)x+3}}}, we need to graph the <b>equation</b> {{{y=(3/2)x+3}}} (just replace the inequality sign with an equal sign).
So lets graph the line {{{y=(3/2)x+3}}} (note: if you need help with graphing, check out this <a href=http://www.algebra.com/algebra/homework/Linear-equations/graphing-linear-equations.solver>solver<a>)


{{{ graph( 500, 500, -5, 5, -5, 5, (3/2)x+3) }}} graph of {{{y=(3/2)x+3}}} 

Now lets pick a test point, say (0,0) (any point will work, but this point is the easiest to work with), and evaluate the inequality {{{y<(3/2)x+3}}}


Substitute (0,0) into the inequality

{{{(0)<(3/2)(0)+3}}} Plug in {{{x=0}}} and {{{y=0}}}

{{{0<3}}} Simplify

Since this inequality is true, we simply shade the entire region that contains (0,0)

{{{drawing( 500, 500, -5, 5, -5, 5,
graph(  500, 500, -5, 5, -5, 5,(3/2)x+3,(3/2)x+3+-0.5),
graph(  500, 500, -5, 5, -5, 5,(3/2)x+3,(3/2)x+3+-1),
graph(  500, 500, -5, 5, -5, 5,(3/2)x+3,(3/2)x+3+-1.5),
graph(  500, 500, -5, 5, -5, 5,(3/2)x+3,(3/2)x+3+-2),
graph(  500, 500, -5, 5, -5, 5,(3/2)x+3,(3/2)x+3+-2.5),
graph(  500, 500, -5, 5, -5, 5,(3/2)x+3,(3/2)x+3+-3),
graph(  500, 500, -5, 5, -5, 5,(3/2)x+3,(3/2)x+3+-3.5),
graph(  500, 500, -5, 5, -5, 5,(3/2)x+3,(3/2)x+3+-4),
graph(  500, 500, -5, 5, -5, 5,(3/2)x+3,(3/2)x+3+-4.5),
graph(  500, 500, -5, 5, -5, 5,(3/2)x+3,(3/2)x+3+-5),
graph(  500, 500, -5, 5, -5, 5,(3/2)x+3,(3/2)x+3+-5.5),
graph(  500, 500, -5, 5, -5, 5,(3/2)x+3,(3/2)x+3+-6),
graph(  500, 500, -5, 5, -5, 5,(3/2)x+3,(3/2)x+3+-6.5),
graph(  500, 500, -5, 5, -5, 5,(3/2)x+3,(3/2)x+3+-7),
graph(  500, 500, -5, 5, -5, 5,(3/2)x+3,(3/2)x+3+-7.5),
graph(  500, 500, -5, 5, -5, 5,(3/2)x+3,(3/2)x+3+-8),
graph(  500, 500, -5, 5, -5, 5,(3/2)x+3,(3/2)x+3+-8.5),
graph(  500, 500, -5, 5, -5, 5,(3/2)x+3,(3/2)x+3+-9),
graph(  500, 500, -5, 5, -5, 5,(3/2)x+3,(3/2)x+3+-9.5))}}} Graph of {{{y<(3/2)x+3}}} with the boundary (which is the line {{{y=(3/2)x+3}}} in red) and the shaded region (in green) (note: since the inequality has a less-than sign, this means the boundary is excluded. This means the solid line is really a dashed line)




---------------------------------------------------------------



So we essentially have these 2 regions:


Region #1
{{{drawing( 500, 500, -5, 5, -5, 5,
graph( 500, 500, -5, 5, -5, 5,x+1),
graph( 500, 500, -5, 5, -5, 5,x+1,x+1+0.5),
graph( 500, 500, -5, 5, -5, 5,x+1,x+1+1),
graph( 500, 500, -5, 5, -5, 5,x+1,x+1+1.5),
graph( 500, 500, -5, 5, -5, 5,x+1,x+1+2),
graph( 500, 500, -5, 5, -5, 5,x+1,x+1+2.5),
graph( 500, 500, -5, 5, -5, 5,x+1,x+1+3),
graph( 500, 500, -5, 5, -5, 5,x+1,x+1+3.5),
graph( 500, 500, -5, 5, -5, 5,x+1,x+1+4),
graph( 500, 500, -5, 5, -5, 5,x+1,x+1+4.5),
graph( 500, 500, -5, 5, -5, 5,x+1,x+1+5),
graph( 500, 500, -5, 5, -5, 5,x+1,x+1+5.5),
graph( 500, 500, -5, 5, -5, 5,x+1,x+1+6),
graph( 500, 500, -5, 5, -5, 5,x+1,x+1+6.5),
graph( 500, 500, -5, 5, -5, 5,x+1,x+1+7),
graph( 500, 500, -5, 5, -5, 5,x+1,x+1+7.5),
graph( 500, 500, -5, 5, -5, 5,x+1,x+1+8),
graph( 500, 500, -5, 5, -5, 5,x+1,x+1+8.5),
graph( 500, 500, -5, 5, -5, 5,x+1,x+1+9),
graph( 500, 500, -5, 5, -5, 5,x+1,x+1+9.5))}}} Graph of {{{y>x+1}}}



Region #2
{{{drawing( 500, 500, -5, 5, -5, 5,
graph( 500, 500, -5, 5, -5, 5,(3/2)x+3),
graph( 500, 500, -5, 5, -5, 5,(3/2)x+3,(3/2)x+3+-0.5),
graph( 500, 500, -5, 5, -5, 5,(3/2)x+3,(3/2)x+3+-1),
graph( 500, 500, -5, 5, -5, 5,(3/2)x+3,(3/2)x+3+-1.5),
graph( 500, 500, -5, 5, -5, 5,(3/2)x+3,(3/2)x+3+-2),
graph( 500, 500, -5, 5, -5, 5,(3/2)x+3,(3/2)x+3+-2.5),
graph( 500, 500, -5, 5, -5, 5,(3/2)x+3,(3/2)x+3+-3),
graph( 500, 500, -5, 5, -5, 5,(3/2)x+3,(3/2)x+3+-3.5),
graph( 500, 500, -5, 5, -5, 5,(3/2)x+3,(3/2)x+3+-4),
graph( 500, 500, -5, 5, -5, 5,(3/2)x+3,(3/2)x+3+-4.5),
graph( 500, 500, -5, 5, -5, 5,(3/2)x+3,(3/2)x+3+-5),
graph( 500, 500, -5, 5, -5, 5,(3/2)x+3,(3/2)x+3+-5.5),
graph( 500, 500, -5, 5, -5, 5,(3/2)x+3,(3/2)x+3+-6),
graph( 500, 500, -5, 5, -5, 5,(3/2)x+3,(3/2)x+3+-6.5),
graph( 500, 500, -5, 5, -5, 5,(3/2)x+3,(3/2)x+3+-7),
graph( 500, 500, -5, 5, -5, 5,(3/2)x+3,(3/2)x+3+-7.5),
graph( 500, 500, -5, 5, -5, 5,(3/2)x+3,(3/2)x+3+-8),
graph( 500, 500, -5, 5, -5, 5,(3/2)x+3,(3/2)x+3+-8.5),
graph( 500, 500, -5, 5, -5, 5,(3/2)x+3,(3/2)x+3+-9),
graph( 500, 500, -5, 5, -5, 5,(3/2)x+3,(3/2)x+3+-9.5))}}} Graph of {{{y<(3/2)x+3}}}





When these inequalities are graphed on the same coordinate system, the regions overlap to produce this region. It's a little hard to see, but after evenly shading each region, the intersecting region will be the most shaded in.




{{{drawing( 500, 500, -5, 5, -5, 5,
graph( 500, 500, -5, 5, -5, 5,x+1,x+1+0.5),
graph( 500, 500, -5, 5, -5, 5,x+1,x+1+1),
graph( 500, 500, -5, 5, -5, 5,x+1,x+1+1.5),
graph( 500, 500, -5, 5, -5, 5,x+1,x+1+2),
graph( 500, 500, -5, 5, -5, 5,x+1,x+1+2.5),
graph( 500, 500, -5, 5, -5, 5,x+1,x+1+3),
graph( 500, 500, -5, 5, -5, 5,x+1,x+1+3.5),
graph( 500, 500, -5, 5, -5, 5,x+1,x+1+4),
graph( 500, 500, -5, 5, -5, 5,x+1,x+1+4.5),
graph( 500, 500, -5, 5, -5, 5,x+1,x+1+5),
graph( 500, 500, -5, 5, -5, 5,x+1,x+1+5.5),
graph( 500, 500, -5, 5, -5, 5,x+1,x+1+6),
graph( 500, 500, -5, 5, -5, 5,x+1,x+1+6.5),
graph( 500, 500, -5, 5, -5, 5,x+1,x+1+7),
graph( 500, 500, -5, 5, -5, 5,x+1,x+1+7.5),
graph( 500, 500, -5, 5, -5, 5,x+1,x+1+8),
graph( 500, 500, -5, 5, -5, 5,x+1,x+1+8.5),
graph( 500, 500, -5, 5, -5, 5,x+1,x+1+9),
graph( 500, 500, -5, 5, -5, 5,x+1,x+1+9.5),
graph( 500, 500, -5, 5, -5, 5,(3/2)x+3,(3/2)x+3+-0.5),
graph( 500, 500, -5, 5, -5, 5,(3/2)x+3,(3/2)x+3+-1),
graph( 500, 500, -5, 5, -5, 5,(3/2)x+3,(3/2)x+3+-1.5),
graph( 500, 500, -5, 5, -5, 5,(3/2)x+3,(3/2)x+3+-2),
graph( 500, 500, -5, 5, -5, 5,(3/2)x+3,(3/2)x+3+-2.5),
graph( 500, 500, -5, 5, -5, 5,(3/2)x+3,(3/2)x+3+-3),
graph( 500, 500, -5, 5, -5, 5,(3/2)x+3,(3/2)x+3+-3.5),
graph( 500, 500, -5, 5, -5, 5,(3/2)x+3,(3/2)x+3+-4),
graph( 500, 500, -5, 5, -5, 5,(3/2)x+3,(3/2)x+3+-4.5),
graph( 500, 500, -5, 5, -5, 5,(3/2)x+3,(3/2)x+3+-5),
graph( 500, 500, -5, 5, -5, 5,(3/2)x+3,(3/2)x+3+-5.5),
graph( 500, 500, -5, 5, -5, 5,(3/2)x+3,(3/2)x+3+-6),
graph( 500, 500, -5, 5, -5, 5,(3/2)x+3,(3/2)x+3+-6.5),
graph( 500, 500, -5, 5, -5, 5,(3/2)x+3,(3/2)x+3+-7),
graph( 500, 500, -5, 5, -5, 5,(3/2)x+3,(3/2)x+3+-7.5),
graph( 500, 500, -5, 5, -5, 5,(3/2)x+3,(3/2)x+3+-8),
graph( 500, 500, -5, 5, -5, 5,(3/2)x+3,(3/2)x+3+-8.5),
graph( 500, 500, -5, 5, -5, 5,(3/2)x+3,(3/2)x+3+-9),
graph( 500, 500, -5, 5, -5, 5,(3/2)x+3,(3/2)x+3+-9.5))}}}




Here is a cleaner look at the intersection of regions





{{{drawing( 500, 500, -5, 5, -5, 5,
          graph( 500, 500, -5, 5, -5, 5,x+1,(3/2)x+3),circle(-2.5,-1,0.05),
circle(-2,-0.5,0.05),
circle(-1.5,0,0.05),
circle(-1.5,0.5,0.05),
circle(-1,0.5,0.05),
circle(-1,1,0.05),
circle(-0.5,1,0.05),
circle(-0.5,1.5,0.05),
circle(-0.5,2,0.05),
circle(0,1.5,0.05),
circle(0,2,0.05),
circle(0,2.5,0.05),
circle(0.5,2,0.05),
circle(0.5,2.5,0.05),
circle(0.5,3,0.05),
circle(0.5,3.5,0.05),
circle(1,2.5,0.05),
circle(1,3,0.05),
circle(1,3.5,0.05),
circle(1,4,0.05),
circle(1.5,3,0.05),
circle(1.5,3.5,0.05),
circle(1.5,4,0.05),
circle(1.5,4.5,0.05),
circle(2,3.5,0.05),
circle(2,4,0.05),
circle(2,4.5,0.05),
circle(2.5,4,0.05),
circle(2.5,4.5,0.05),
circle(3,4.5,0.05),
circle(-2.5,-1,0.05),
circle(-2,-0.5,0.05),
circle(-1.5,0,0.05),
circle(-1.5,0.5,0.05),
circle(-1,0.5,0.05),
circle(-1,1,0.05),
circle(-0.5,1,0.05),
circle(-0.5,1.5,0.05),
circle(-0.5,2,0.05),
circle(0,1.5,0.05),
circle(0,2,0.05),
circle(0,2.5,0.05),
circle(0.5,2,0.05),
circle(0.5,2.5,0.05),
circle(0.5,3,0.05),
circle(0.5,3.5,0.05),
circle(1,2.5,0.05),
circle(1,3,0.05),
circle(1,3.5,0.05),
circle(1,4,0.05),
circle(1.5,3,0.05),
circle(1.5,3.5,0.05),
circle(1.5,4,0.05),
circle(1.5,4.5,0.05),
circle(2,3.5,0.05),
circle(2,4,0.05),
circle(2,4.5,0.05),
circle(2.5,4,0.05),
circle(2.5,4.5,0.05),
circle(3,4.5,0.05))}}} Here is the intersection of the 2 regions represented by the series of dots