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?
------------------------------------
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.