Unless this is some perplexed kind of math, how can you solve the variables when there is only one number to go from?
Is there a 'not enough information' answer choice
Answer: It would be d,15m
Answer:
1. 
2. 
3. 
Step-by-step explanation:
1. 
I like to make it an improper fraction first.

Now take the LCM. The LCM is 20.

Now add them.

Make this into a mixed number.

That's your answer for number 1.
2. 
Do the same thing as the previous addition problem but subtract instead.



is your answer for number 2.
3. 



is your answer for number 3.
I hope this helps! Let me know if you need help or if I got anything wrong :)
So... let's say, we have two numbers "a" and "b",
"a" is the smaller one, "b" is the larger one
so, their sum is 6
a + b = 6
3 times the smaller => 3 * a or 3a
now, two less than that
3a - 2
that equals the larger "b"
3a - 2 = b
solve for "a",
now, what's "b"? well, 3a - 2 = b :)