Question 901078: find two consecutive odd integers such that three times the smaller exceeds tow time the larger by 7 Answer by JulietG(1812) (Show Source):
You can put this solution on YOUR website! A = B+2 (consecutive odd)
3B = 2A+7
Substitute the known value of A from the first equation into the second.
3B = 2A+7
3B = 2(B+2) + 7
Distribute
3B = 2B+4 + 7
3B = 2B + 11
Subtract 2B from each side
B = 11
If B is 11, then A is 13.
Prove: three times the smaller exceeds tow time the larger by 7
3 * 11 = (2*13)+7
33 = 26+7
33=33
