I don’t mean to be disrespectful. But there is only one correct answer here, m=6 and r=18. The solution set that has m=-12 and r=36 is incorrect. It doesn’t satisfy the terms of condition 1. Condition 1 states, “The distance between r and 0 is 3 times the distance between m and 0. Well, it turns out that when we write that out mathematically, we write, r=3m. Well, obviously 3(-12) is not equal to 36. We would have to use the absolute value when we set up to solve. Which is fine when using distance. But when we use absolute value we arrive at the same answer for both forms which is, m=6 and r=18. I repeat what I mentioned earlier in this note, the solution m=-12 and r=36 is incorrect. The following is the workout.
r = 3m
r - 3m = 0
| r - 3m | = 0
In order to solve the simultaneous equations, we use the forms,
r - 3m = 0 and
-r + 3m = 0
Now we can solve the simultaneous linear equations using both forms.
r + m = 24 (1)
r - 3m = 0 (2)
Subtract equation (2) from equation (1).
4m = 24
m = 6
Therefore, r = 18.
Now we use the second form of the equation, (-r + 3m = 0) and solve the simultaneous equations.
r + m = 24 (1)
-r + 3m = 0 (2)
Now, add equation (1) and equation (2) together in order to solve.
4m = 24
m = 6
Therefore, r = 18.
Thus, we see that even using absolute value, the solution is still m = 6 and r = 18. Any alternative solution does not meet the terms of the original equations in the setup.
Sean S.
05/21/18