this is what i wrote on paper btw
4 integers=A,B,C,D
"When he adds them three at a time"
A+B+C=46
A+B+D=50
A+C+D=54
B+C+D=57
You can add them 3 at a time in 4 different ways, which is correctly shown in the system above.
Step Two: Now, solving for just 1 of the 4 variables would easily lead the answers of the other three. For example, we will eliminate the variable A.
1. We can take the first two equations, and subtract them, to eliminate A.
(We would get C-D=-4)
2. Next, we want to get another equation, without the variable A in it. We can use elimination for the 2nd and 3rd equation. We could get (B-C=-4)
3. Finally, we can use the last equation (also without the variable A). (B+C+D)=57
4. Now, we have the 3 equations:
B+C+D=57. C-D=-4, B-C=-4
Beginning the elimination process, we can add the first 2 equations together and get B+2C=53. Now, we need either variable B or C in terms of the other, such as B-C=-4. We can say that B=-4+C, which we will then plug back into -4+2C+2C=53. -4+4C=53. C=19. We have now solved one of the 4 variables.
Step Three: Now that we know C=19, here comes the easy part, to solve for the other 3 variables, we simply plug into equation that contains the variable C. How about that equation C-D=-4 we received from the first elimination? (19)-D=-4, D=23.
Now, we know that C=19 and D=23. We now have a even easier time solving for the last 2 variables.
B+C+D=57, B+19+23=57, B=15
A+B+C=46, A+15+19=46, A=12
Step Four: Now, we have solved for all 4 variables. A=12, B=15, C=19, D=23. The question asks for the largest integer which is D=23, which is our final answer.