SOLUTION: Solve the following system of linear inequalities by graphing. x – y < 3 x + 2y >6

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Question 83028: Solve the following system of linear inequalities by graphing.
x – y < 3
x + 2y >6
Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
Lets graph the equations first
x-y=3 Simply replace the inequality sign with an equal sign
x%2B2y=6

So lets graph x-y=3 first
Solved by pluggable solver: Graphing Linear Equations


1%2Ax-1%2Ay=3Start with the given equation



-1%2Ay=3-1%2Ax Subtract 1%2Ax from both sides

y=%28-1%29%283-1%2Ax%29 Multiply both sides by -1

y=%28-1%29%283%29%2B%281%29%281%29x%29 Distribute -1

y=-3%2B%281%29x Multiply

y=1%2Ax-3 Rearrange the terms

y=1%2Ax-3 Reduce any fractions

So the equation is now in slope-intercept form (y=mx%2Bb) where m=1 (the slope) and b=-3 (the y-intercept)

So to graph this equation lets plug in some points

Plug in x=-6

y=1%2A%28-6%29-3

y=-6-3 Multiply

y=-9 Add

So here's one point (-6,-9)





Now lets find another point

Plug in x=-5

y=1%2A%28-5%29-3

y=-5-3 Multiply

y=-8 Add

So here's another point (-5,-8). Add this to our graph





Now draw a line through these points

So this is the graph of y=1%2Ax-3 through the points (-6,-9) and (-5,-8)


So from the graph we can see that the slope is 1%2F1 (which tells us that in order to go from point to point we have to start at one point and go up 1 units and to the right 1 units to get to the next point) the y-intercept is (0,-3)and the x-intercept is (3,0) . So all of this information verifies our graph.


We could graph this equation another way. Since b=-3 this tells us that the y-intercept (the point where the graph intersects with the y-axis) is (0,-3).


So we have one point (0,-3)






Now since the slope is 1%2F1, this means that in order to go from point to point we can use the slope to do so. So starting at (0,-3), we can go up 1 units


and to the right 1 units to get to our next point



Now draw a line through those points to graph y=1%2Ax-3


So this is the graph of y=1%2Ax-3 through the points (0,-3) and (1,-2)



Now lets graph x%2B2y=6
Solved by pluggable solver: Graphing Linear Equations


1%2Ax%2B2%2Ay=6Start with the given equation



2%2Ay=6-1%2Ax Subtract 1%2Ax from both sides

y=%281%2F2%29%286-1%2Ax%29 Multiply both sides by 1%2F2

y=%281%2F2%29%286%29-%281%2F2%29%281%29x%29 Distribute 1%2F2

y=6%2F2-%281%2F2%29x Multiply

y=%28-1%2F2%29%2Ax%2B6%2F2 Rearrange the terms

y=%28-1%2F2%29%2Ax%2B3 Reduce any fractions

So the equation is now in slope-intercept form (y=mx%2Bb) where m=-1%2F2 (the slope) and b=3 (the y-intercept)

So to graph this equation lets plug in some points

Plug in x=-8

y=%28-1%2F2%29%2A%28-8%29%2B3

y=8%2F2%2B3 Multiply

y=14%2F2 Add

y=7 Reduce

So here's one point (-8,7)





Now lets find another point

Plug in x=-6

y=%28-1%2F2%29%2A%28-6%29%2B3

y=6%2F2%2B3 Multiply

y=12%2F2 Add

y=6 Reduce

So here's another point (-6,6). Add this to our graph





Now draw a line through these points

So this is the graph of y=%28-1%2F2%29%2Ax%2B3 through the points (-8,7) and (-6,6)


So from the graph we can see that the slope is -1%2F2 (which tells us that in order to go from point to point we have to start at one point and go down -1 units and to the right 2 units to get to the next point), the y-intercept is (0,3)and the x-intercept is (6,0) . So all of this information verifies our graph.


We could graph this equation another way. Since b=3 this tells us that the y-intercept (the point where the graph intersects with the y-axis) is (0,3).


So we have one point (0,3)






Now since the slope is -1%2F2, this means that in order to go from point to point we can use the slope to do so. So starting at (0,3), we can go down 1 units


and to the right 2 units to get to our next point



Now draw a line through those points to graph y=%28-1%2F2%29%2Ax%2B3


So this is the graph of y=%28-1%2F2%29%2Ax%2B3 through the points (0,3) and (2,2)



So lets graph y=x-3 and y=%28-1%2F2%29x%2B3 together



Now lets pick the test point (0,0) and evaluate x-y%3C3
%280%29-%280%29%3C3
0%3C3 true. Shade the region that does contain (0,0) for x-y=3. So this means you shade above the line y=x-3

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Now lets use the same test point (0,0) and evaluate x%2B2y%3E6
%280%29%2B2%280%29%3E6
0%3E6 false. Shade the region that doesn't contain (0,0) for x%2B2y=6. So this means you shade above the line y=%28-1%2F2%29x%2B3

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Now the overlapping regions is the complete shading. In other words, every point that satisfies

x-y%3C3
x%2B2y%3E6
is in this region. So this is what the overlapped shaded region looks like:

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