SOLUTION: Find two numbers where three times the smaller number exceeds the larger by 5 and the sum of the numbers is 11.

Question 1051836: Find two numbers where three times the smaller number exceeds the larger by 5 and the sum of the numbers is 11.
Answer by addingup(3677) (Show Source): You can put this solution on YOUR website!
x+y = 11 and if we subtract y on both sides:
x = 11-y we'll use this value for x below.
Then the problem says:
3x = y+5 in this equation, substitute for x:
3(11-y) = y+5
33-3y = y+5 subtract 5 and add 3y on both sides:
28 = 4y
or
4y = 28 divide both sides by 4
y = 7
and we said that x = 11-y = 11-7 = 4
Your answer:
x = 4
y = 7
-------------------
Check:
3*the smaller exceeds the larger by 5:
3*4 = 12 and 12-7 = 5 Correct
:
John
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