SOLUTION: Find the area bounded by the equation x/- 5 + y/2 = 1 on the first quadrant and the line x = 10

Question 1185925: Find the area bounded by the equation x/- 5 + y/2 = 1 on the first quadrant and the line x = 10
Answer by CPhill(1987) (Show Source): You can put this solution on YOUR website!
Here's how to find the area:
1. **Rewrite the equation:**
The equation x/-5 + y/2 = 1 can be rewritten as:
y/2 = 1 + x/5
y = 2 + (2/5)x
2. **Find the intersection points:**
* **Intersection with the x-axis:** Set y = 0 in the original equation:
x/-5 + 0/2 = 1
x/-5 = 1
x = -5
This point is (-5, 0). However, we are only interested in the first quadrant, where x and y are both positive.
* **Intersection with the y-axis:** Set x = 0 in the original equation:
0/-5 + y/2 = 1
y/2 = 1
y = 2
This point is (0, 2).
* **Intersection with the line x = 10:** Substitute x = 10 into the equation:
10/-5 + y/2 = 1
-2 + y/2 = 1
y/2 = 3
y = 6
This point is (10, 6).
3. **Visualize the area:** The area we're looking for is a trapezoid. The vertices of the trapezoid are (0, 2), (10, 6), and the points where x=10 intersects the x-axis (10,0).
4. **Calculate the area of the trapezoid:** The formula for the area of a trapezoid is:
Area = (1/2) * (base1 + base2) * height
In our case:
* base1 = the length of the vertical line at x=0, which is 2
* base2 = the length of the vertical line at x=10, which is 6
* height = the horizontal distance between the two vertical lines, which is 10
Area = (1/2) * (2 + 6) * 10
Area = (1/2) * 8 * 10
Area = 40
Therefore, the area bounded by the given equation, the x and y axes in the first quadrant, and the line x = 10 is $\boxed{40}$ square units.

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