dtan5457
  • dtan5457
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.
Mathematics
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dtan5457
  • dtan5457
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.
Mathematics
chestercat
  • chestercat
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dtan5457
  • 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
  • dtan5457
Jhannybean
  • Jhannybean
@agent0smith or even @zepdrix ? A little help here :)

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dtan5457
  • dtan5457
I'm in a math research class so every question we get is supposed to be challenging lol
campbell_st
  • campbell_st
well you can solve them as simultaneous equations using matrices or just the hard slog of the elimination or substitution methods
dtan5457
  • dtan5457
i have never done matrices before
dtan5457
  • dtan5457
would substitution be a lot longer?
anonymous
  • anonymous
this should be linear algebra u have 4 equations to solve 4 unknowns, have u tried ?
anonymous
  • anonymous
what i'm not feeling okay about is this condition lol "Compute the largest of the original four integers"
dtan5457
  • dtan5457
i would rather just do this with substitution even if it took a while but im still a little stuck.
anonymous
  • anonymous
well good substitution is also away to solve i'll give you some hints
mathmate
  • mathmate
Using symmetry, you can solve it as follows.
mathmate
  • 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
  • mathmate
got to go, talkk later!
dtan5457
  • dtan5457
finally solved it with subsitution and elimination, thanks anyways @mathmate
dtan5457
  • 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
  • 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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