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

Question

Paisley Johnson · Accepted Answer

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

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"#

Emilia Aguilar · Answer

undefined To find the area of the region bounded by the graphs of $y + x = 6$ and $y + 2x - 3 = 0$, follow these steps:

1. Solve each equation for y to express them in terms of x.
2. Determine the points where the two lines intersect by solving the system of equations.
3. Find the x-values of the intersection points.
4. Calculate the area between the curves by integrating the absolute difference of the y-values with respect to x over the interval between the x-values of the intersection points.

Let's solve it step by step:

1. Equation 1: $y + x = 6$
   Solve for y: $y = 6 - x$

2. Equation 2: $y + 2x - 3 = 0$
   Solve for y: $y = -2x + 3$

3. Set the expressions for y equal to each other and solve for x:
   $6 - x = -2x + 3$
   $3 = -3x$
   $x = -1$

4. Now that we have the x-value of the intersection point, substitute it into either equation to find the corresponding y-value. Let's use $y = 6 - x$:
   $y = 6 - (-1)$
   $y = 7$

5. Therefore, the intersection point is $(-1, 7)$.

6. Now, integrate the absolute difference of the y-values between the curves with respect to x from the x-coordinate of the first intersection point to the x-coordinate of the second intersection point:
   $$\text{Area} = \int_{-1}^{3} |(6 - x) - (-2x + 3)| \, dx$$

7. Evaluate the integral:
   $$\text{Area} = \int_{-1}^{3} |6 - x + 2x - 3| \, dx$$
   $$\text{Area} = \int_{-1}^{3} |6 + x - 3| \, dx$$
   $$\text{Area} = \int_{-1}^{3} |3 + x| \, dx$$
   $$\text{Area} = \int_{-1}^{3} (3 + x) \, dx$$

8. Integrate the function:
   $$\text{Area} = \left[\frac{1}{2}x^2 + 3x\right]_{-1}^{3}$$
   $$\text{Area} = \left[\frac{1}{2}(3)^2 + 3(3)\right] - \left[\frac{1}{2}(-1)^2 + 3(-1)\right]$$
   $$\text{Area} = \left[\frac{9}{2} + 9\right] - \left[\frac{1}{2} - 3\right]$$
   $$\text{Area} = \left[\frac{27}{2}\right] - \left[\frac{5}{2}\right]$$
   $$\text{Area} = \frac{22}{2}$$
   $$\text{Area} = 11$$

So, the area of the region bounded by the graphs of $y + x = 6$ and $y + 2x - 3 = 0$ is $11$ square units.