Question 909336
A + B + C = 77
A + B = C + 99
C = A - 65
Substitute the known values
A + B + C = 77
(C + 99) + C = 77
2C + 99 = 77
2C = -22
C = -11
Now substitute the value of C into the 3rd equation
C = A - 65
-11 = A - 65
54 = A
Back the the first equation
A + B + C = 77
54 + B - 11 = 77
Subtract 54 and add 11 to each side
B = 34
A=54, B=34, C=-11<br>
Check:
The sum of the first and second integers exceeds the third by 99.-- 54+34=99-11
88=88
The third integer is 65 less than the first. 54-65 = -11
the sum of 3 integers is 77. 54 + 34 - 11 = 77