Question 834478
Here's what we know:
B = A+1
C = A+2 (consecutive)
A + 2B = 98 - 3C
Substitute the known values
A + 2(A+1) = 98 - 3(A+2)
Distribute
A + 2A + 2 = 98 - 3A - 6
Add the As
3A + 2 = 98 - 3A - 6
Add 3A to each side
6A + 2 = 98 - 6
Subtract 2 from each side
6A = 98-6-2
Do the math
6A = 90
Divide each side by 6
A = 15
.
If A is 15, then B is 16, and C is 17
.
Let's see if it works.
(...the sum of the first and twice the second is 98 minus three times the third.)
15 + 32 = 98 - 51
47 = 47
Success!