SOLUTION: One number exceeds three times a second number by 7. The sum of twice the larger number and four times the smaller nu mber is 44. What are the numbers?

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Question 124940: One number exceeds three times a second number by 7. The sum of twice the larger number and four times the smaller nu
mber is 44. What are the numbers?
Found 2 solutions by stanbon, chitra:
Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
One number exceeds three times a second number by 7. The sum of twice the larger number and four times the smaller nu
mber is 44. What are the numbers?
------------------------------------
x = 3y +7
2x+4y = 44
-------------
Rearrange:
x = 3y+7
x+2y = 22
--------------
Substitute to solve for "Y":
3y+7 +2y=22
5y = 155
y = 3
--------
Substitute for "x"
x = 3*3+7
x = 16
============
Cheers,
Stan H.

Answer by chitra(359) About Me  (Show Source):
You can put this solution on YOUR website!
Let the numbers be x & y.

According to the data given we write "x" in terms of "y".

x = 3y + 7

Given that:
2(3y + 7) + 4y = 44

6y + 14 + 4y = 44

10y = 44 - 14

10y = 30

y = 30%2F10

y = 3

By back substitution, we get:

x = 3(y) + 7

x = 3(3) + 7

x = 9 + 7

x = 16

Thus the values are x = 16 & y = 3

Hence, the solution.