August has four integers. When he adds them three at a time, the sums are 46,50,54 and 57. Compute the largest of the original four integers.

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- dtan5457

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- dtan5457

So what I tried so far was setting up
A+B+C=46
A+B+D=50
A+C+D=54
B+C+D=57
not sure how to solve this though

- dtan5457

- Jhannybean

@agent0smith or even @zepdrix ? A little help here :)

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## More answers

- dtan5457

I'm in a math research class so every question we get is supposed to be challenging lol

- campbell_st

well you can solve them as simultaneous equations using matrices
or just the hard slog of the elimination or substitution methods

- dtan5457

i have never done matrices before

- dtan5457

would substitution be a lot longer?

- anonymous

this should be linear algebra u have 4 equations to solve 4 unknowns, have u tried ?

- anonymous

what i'm not feeling okay about is this condition lol
"Compute the largest of the original four integers"

- dtan5457

i would rather just do this with substitution even if it took a while but im still a little stuck.

- anonymous

well good substitution is also away to solve i'll give you some hints

- mathmate

Using symmetry, you can solve it as follows.

- mathmate

subtract (4) from (3)
a-b=-3
subtract (3) from (2)
b-c=-4
subtract (2) from (1)
c-d=46-50=-4
observe the three equations, a

- mathmate

got to go, talkk later!

- dtan5457

finally solved it with subsitution and elimination, thanks anyways @mathmate

- dtan5457

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.

- anonymous

actually u can add them in 4! different way but it won't be a problem as there is only one set solution

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