1. Preeta watches an ant and a beetle crawl in a hole in the ground. The beetle is inch below ground level. The beetle climbs the distance the ant is below ground level. The beetle is now inches below ground level.
(a) Let x = the position of the ant relative to ground level. What equation can be written to solve for x?
(b) Solve the equation from Part (a). Show your work.
(c) What is the distance between where the ant is and where the beetle is? Answer the question in a complete sentence. Show your work.

1a. To build the equation, we need to see that the beetle started at 3/4th inches below ground level, then "climbed" (which is poor English, it should be descended) a certain distance, and then it ended up at 2 1/2 inches below ground. The distance "climbed" can be represented as 1/3X, where "X" is the ant's current distance. We would do 1/3rd times "X" to figure out how much the beetle climbed down.

So that means, if we added together the starting position of 3/4, plus the distance the beetle descended (1/3X), that would get us TO 2 1/2 inches below ground (the beetle's new current position). So therefore our equation is:

[tex]\frac{3}{4}[/tex] + [tex]\frac{1}{3}x[/tex] = [tex]2\frac{1}{2}[/tex]

1b. Solve the equation written above from 1a.

[tex]\frac{3}{4}[/tex] + [tex]\frac{1}{3}x[/tex] = [tex]2\frac{1}{2}[/tex]

-3/4 -3/4 [subtract 3/4 to start the process of getting x by itself]

[tex]\frac{1}{3}x[/tex] = 1[tex]\frac{3}{4}[/tex]

*3/1 *3/1 [Multiply Reciprocal to get X by itself]

x= [tex]5\frac{1}{4}[/tex]

1c. So we take the position of the ant figured out on 1b (5 1/4 inches), and the new current position of the beetle (2 1/2 inches), and subtract them

5 1/4 - 2 1/2 = [tex]2\frac{3}{4}[/tex]

These answers were found by using this info:

the beetle is 3/4 inch below the ground level. The beetle climbs 1/3 the distance. The beetle is now 2 1/2 inches below ground.