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Question 350815: problem 1. Twice the sum of two numbers exceeds three times their difference by 8, while half the sum is one more than the difference. what are the numbers?
Answer by sudhanshu_kmr(1152)   (Show Source):
You can put this solution on YOUR website! Let the numbers are x and y, their sum =(x+y) , difference =(x -y)
according to question.
2(x+y) = 3(x-y) + 8 .....(1)
and (x+y)/2 = (x-y) + 1 ....... (2)
using (1) we got -x +5y =8
using (2) we got -x +3y =2
now, we solve these two equations...
subtract second equation from first....
2y =6, y =3
using first -x + 5*3 = 8
=> -x = 8-15
=> -x = -7
=> x = 7
so, numbers are 7 and 8
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