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

Dec 29, 2017

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 : the X-axis
In this case we have a triangle with a base of $4.5$ units (from $\left(1.5 , 0\right) \text{ to } \left(6 , 0\right)$) and a height of $9$ units (to the point $\left(- 3 , 9\right)$)
$\text{Area"_triangle = (4.5xx9)/2=20.25" sq.units}$

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

Possible intended third and fourth boundaries : 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.
$\text{Area"=20.25-4.5=15.75" sq.units}$