5th Ed. Word Problems, p. 35
I understand how to solve this problem using a combined equation (.5r x 8/3 = 40) but wanted to see if solving in a more traditional way would work too. I tried solving one of the two equations for d and plugging it into the other and kept coming up with a negative number.
1st two equations:
3/2r x 8/3 = d
r x 8/3 = d + 40
Solve for d:
d = 8/3r - 40
Plug into the other one:
8/3r - 40 = 3/2r x 8/3
8/3r - 40 = 4r
40 = -4/3r
-30 = r
Should solving for d and plugging it into the other equation work (i.e. did i just make a misstep in my algebra to end up with a negative answer) or is a singular equation the only way to solve for this?