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dtan5457

  • one year ago

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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  1. dtan5457
    • one year ago
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    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

  2. dtan5457
    • one year ago
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    @Jhannybean @Nnesha @Loser66

  3. Jhannybean
    • one year ago
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    @agent0smith or even @zepdrix ? A little help here :)

  4. dtan5457
    • one year ago
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    I'm in a math research class so every question we get is supposed to be challenging lol

  5. campbell_st
    • one year ago
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    well you can solve them as simultaneous equations using matrices or just the hard slog of the elimination or substitution methods

  6. dtan5457
    • one year ago
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    i have never done matrices before

  7. dtan5457
    • one year ago
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    would substitution be a lot longer?

  8. anonymous
    • one year ago
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    this should be linear algebra u have 4 equations to solve 4 unknowns, have u tried ?

  9. anonymous
    • one year ago
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    what i'm not feeling okay about is this condition lol "Compute the largest of the original four integers"

  10. dtan5457
    • one year ago
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    i would rather just do this with substitution even if it took a while but im still a little stuck.

  11. anonymous
    • one year ago
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    well good substitution is also away to solve i'll give you some hints

  12. mathmate
    • one year ago
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    Using symmetry, you can solve it as follows.

  13. mathmate
    • one year ago
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    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<b<c<d, so d is the largest of the 4. Now we know that adding all four equations together gives by symmetry 3(a+b+c+d)=207 a+b+c+d=69 substitute c=d-4, b=d-4-4=d-8, a=d-4-4-3=d-11 we get 4d-23=69 4d=92 d=23 voila

  14. mathmate
    • one year ago
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    got to go, talkk later!

  15. dtan5457
    • one year ago
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    finally solved it with subsitution and elimination, thanks anyways @mathmate

  16. dtan5457
    • one year ago
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    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.

  17. anonymous
    • one year ago
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    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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