Solution

1. The intersection of the two lines can be found by algebraic substitution. There are multiple ways the substitution can be done but here's one approach. Start by multiplying both sides of equation 2 by 1/0.05 to obtain:
   
   `1/0.05*(x+1) = y`
   
   Since 1/0.05 = 20, this can be written more compactly as:
   
   `20 x + 20 = y`
   
   Use the equation above to eliminate `y` from the first equation in the problem:
   
   `20 x + 20 = 200 - 10 x`
   
   Collecting terms and simplifying:
   
   `30 x = 180`
   
   `x = 6`
   
   To obtain `y`, substitute `x` back into the first equation:
   
   `y = 200 - 10 \cdot 6 = 140`
   
   To check for errors, also plug `x` into the second equation to make sure it gives the same value for `y`:
   
   `6 + 1 = 0.05 y`
   
   `7/0.05 = y`
   
   `y = 140`
   
   The intersection is thus at the point `x`=6, `y`=140. The finished diagram is shown below. For line 2, the x intercept is actually at `x`=-1, but that's not shown in the diagram. Also, please note that the diagram is not drawn to scale.
   
   
2. The triangle is shown below. Its area is one half of its base times its height. Since it's a right triangle, either side can be taken to be the base, and the other will be the height. Taking the base to be 6 and the height to be 200-140=60 shows that the area is 0.5*6*60 = 180.
   
   
3. Finding the intersection of lines 1 and 2:
   
   `1000 - 10 x = 200 + 10 x`
   
   `800 = 20 x`
   
   `x = 40`
   
   `y = 1000 - 10 \cdot40 = 600`
   
   Checking:
   
   `y = 200 + 10 \cdot 40 = 600`
   
   Lines 1 and 2 thus intersect at `x`=40, `y`=600. Finding the intersection of lines 1 and 3 is easy because line 3 gives the value of `x` immediately: `x`=20. To find `y`:
   
   `y = 1000 - 10 \cdot 20 = 800`
   
   Lines 1 and 3 therefore intersect at `x`=20, `y`=800. Finding the intersection of lines 2 and 3 is similar:
   
   `y = 200 + 10 \cdot 20 = 400`
   
   Lines 2 and 3 intersect at `x`=20, `y`=400.
   
   The finished diagram is shown below, where the coordinates of point A are `x`=40, `y`=600, point B are `x`=20, `y`=800, and point C are `x`=20, `y`=400.
   
   

   The easiest way to figure out the area of the triangle is to take the base to be the length of the vertical line at the left: the difference between the y values of points B and C. That's 800-400=400. The height of the triangle is then the difference between the `x`=20 line and the x coordinate of point A, which is 40. The height is thus 40-20=20. The area is 0.5*400*20=4,000. (Remember: the 0.5*b*h formula requires that there be a right angle between b and h.)

4. Since `x_3` is the sum of `x_1` and `x_2`, it can be found as follows:
   
   `x_3 = x_1 + x_2`
   
   `x_3 =(10-y) + (20-2 y)`
   
   `x_3 = 30 - 3 y`
   
   The graphs are shown under part 5.
5. The first step in finding the areas of the three triangles is to calculate the `x` value of each curve at `y`=5:
   
   `x_1 = 10 - 5 = 5`
   
   `x_2 = 20 - 2 \cdot 5 = 10`
   
   `x_3 = 30 - 3 \cdot 5 = 15`
   
   The other piece of information needed is the y intercept for each curve. For all three lines, that can be shown to be 10. Thus the graphs look as follows (not to scale):
   
   
   
   The height of all three triangles is 10-5 = 5. Now the three areas can be calculated using the triangle formula:
   
   Area 1: 0.5*5*5 = 12.5
   
   Area 2: 0.5*5*10 = 25
   
   Area 3: 0.5*5*15 = 37.5
   
   Area 3 is exactly equal to the sum of areas 1 and 2.
   
   This sort of relationship turns up a lot in economics. For example, when there are two buyers in a market, it's possible to calculate total consumer surplus by calculating each buyer's surplus and adding them together (similar to calculating areas 1 and 2 and then adding them), or it can be calculated by computing the area under the market demand (similar to calculating area 3).