How do you find the area of region bounded by the graphs of y +x= 6 and y +2x-3=0?

1 Answer
Dec 29, 2017

Answer:

You need at least one more line to bound the area.
See below for possibilities

Explanation:

The two given equations form the graph below:
graph{(y+2x-3)(y+x-6)=0 [-13.19, 12.13, -1.93, 10.73]}
These intersecting lines divide the plane into four (infinite) regions.

Possible intended third boundary [1]: the X-axis
In this case we have a triangle with a base of #4.5# units (from #(1.5,0) " to " (6,0)#) and a height of #9# units (to the point #(-3,9)#)
#"Area"_triangle = (4.5xx9)/2=20.25" sq.units"#

Possible intended third boundary [2]: the Y-axis
In this case we hav a triangle with a base of #3# units (from #(0,3)" to "(0,6)#) and a height of #3# units (to the point #(-3,9)# from the Y-axis)
#"Area"_triangle=(3xx3)/2=4.5" sq.units"#

Possible intended third and fourth boundaries [3]: both the X and Y-axes
This would give us a quadrilateral region with an area equal to the difference between the two triangular areas calculated above.
#"Area"=20.25-4.5=15.75" sq.units"#