Please help

Jomo

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E651318C-2F91-48DF-9DAE-1E1945C3980C.jpeg

Some subtraction on the left side are wrong. Go back and fix them.
 

Otis

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… I don’t think I've taken the right approach …
Hi Mohammed. Your approach is okay. Using algebra to eliminate variables or to solve pairs of equations for the same expression is a good start. When you find a certain variable or expression repeated, you can proceed with substitutions, to make progress.

However, some of your work shows that you're assuming {a,b,c,d} are Integers. They are not Integers. (If you were expecting Integer solutions, then please check to be sure you copied the exercise correctly.)

😎
 

Jomo

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By any chance did the problem say that a, b, c and d are integers. If yes, then why not tell us? In no, then why when a product of two factors equal 13 must the factors be 1 and 13. Why not 11sqrt3 and 13/911sqrt3 ?
 

Mohammed 23

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By any chance did the problem say that a, b, c and d are integers. If yes, then why not tell us? In no, then why when a product of two factors equal 13 must the factors be 1 and 13. Why not 11sqrt3 and 13/911sqrt3 ?
Well I don’t know if it’s necessary that a b c and d need to be integers , it’s not mentioned in the question and was just an assumption I made . But , the final answer of a+b+c+d needs to be an integer for sure
I don’t think my approach will give me the answer because the question hasn’t asked us to find out the individual values but only the sum total of values
 

Mohammed 23

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Hi Mohammed. Your approach is okay. Using algebra to eliminate variables or to solve pairs of equations for the same expression is a good start. When you find a certain variable or expression repeated, you can proceed with substitutions, to make progress.

However, some of your work shows that you're assuming {a,b,c,d} are Integers. They are not Integers. (If you were expecting Integer solutions, then please check to be sure you copied the exercise correctly.)

😎
Yes I know they need not be integers , but I assumed so that it’s easier to find out the individual values if they are taken as integers . The final sum however needs to be an integer . I haven’t been able to get any equation which helps me eliminate a variable yet .
 

Otis

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Hi Mohammed. I played around with the system as posted, using paper & pencil. First, I solved for {b,c,d} each in terms of a. Substitution then yielded:

(679/36)a + (14√142/9)a + 13√142/9 = -149/36

I obtained two solutions, for a. (One of them turned out to be extraneous, but I didn't realize it until later.)

At that point, I'd spent nearly two hours, so I finished the back-substitution using a computer.

I would hope there's a shortcut that I missed, if that system is supposed to be solved by hand.

😎
 

Mohammed 23

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Some subtraction on the left side are wrong. Go back and fix them.
Actually, I don’t think all this is needed .I don’t feel we need to find out individual values to find the sum of all the values
 

Mohammed 23

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Hi Mohammed. I played around with the system as posted, using paper & pencil. First, I solved for {b,c,d} each in terms of a. Substitution then yielded:

(679/36)a + (14√142/9)a + 13√142/9 = -149/36

I obtained two solutions, for a. (One of them turned out to be extraneous, but I didn't realize it until later.)

At that point, I'd spent nearly two hours, so I finished the back-substitution using a computer.

I would hope there's a shortcut that I missed, if that system is supposed to be solved by hand.

😎
Oh thanks , I think substitution is the only way then . Sorry for taking so much of your time actually, it is to be done by hand , without use of calculator
 

Otis

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Oops! I'd completely forgotten the goal (to find the sum of a+b+c+d).

Yes, I'm sure there must be an easier way than solving for {a,b,c,d}.

… Sorry for taking so much of your time
No need to apologize. The work that I did was for fun.

😎
 

Otis

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Well, darn it all. I've got more than a dozen sheets of paper with over 100 equations, but I can't figure out why I'd scribbled:

ab + cd = -27

If we can justify that, then here's how it goes from there.

We add bc to each side:

ab + bc + cd = bc - 27

Now, we know that bc - 27 = -a - d, so we have:

ab + bc + cd = -a - d

And we know that da = 17 - b - c, so adding the respective sides gives:

ab + bc + cd + da = -a - d - b - c + 17

You'd written:

ab + bc + cd + da + 2(a + b + c + d) = 87

Therefore:

-a - b - c - d + 17 + 2(a + b + c + d) = 87

or

2(a + b + c + d) - (a + b + c + d) = 70

My brain is a bit fried; I'll try to revisit my papers later.

o_O
 

Mohammed 23

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Nov 5, 2020
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Well, darn it all. I've got more than a dozen sheets of paper with over 100 equations, but I can't figure out why I'd scribbled:

ab + cd = -27

If we can justify that, then here's how it goes from there.

We add bc to each side:

ab + bc + cd = bc - 27

Now, we know that bc - 27 = -a - d, so we have:

ab + bc + cd = -a - d

And we know that da = 17 - b - c, so adding the respective sides gives:

ab + bc + cd + da = -a - d - b - c + 17

You'd written:

ab + bc + cd + da + 2(a + b + c + d) = 87

Therefore:

-a - b - c - d + 17 + 2(a + b + c + d) = 87

or

2(a + b + c + d) - (a + b + c + d) = 70

My brain is a bit fried; I'll try to revisit my papers later.

o_O
thanks a ton . I wanted to know how to solve it very badly . Thanks a lot for your time
 
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