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denis_gylaev

Member

Registered: 2015-03-19

Posts: 66

Really Hard Algebra Equations

I really don't know how to do these problems:

Suppose that ab = 7 and a^2b + ab^2 + a+b = 80. What is a^2+b^2?

Find the ordered quintuplet (a,b,c,d,e) that satisfies the system of equations

a  +  2b  +  3c  +  4d  +  5e  =  177,
2a  +  3b  +  4c  +  5d  +  e  =  154,
3a  +  4b  +  5c  +  d  +  2e  =  146,
4a  +  5b  +  c  +  2d  +  3e  =  138,
5a  +  b  +  2c  +  3d  +  4e  =  165.

Suppose  p+q+r = 7  and  p^2+q^2+r^2 = 9 . Then, what is the average (arithmetic mean) of the three products  pq ,  qr , and  rp ?

zetafunc

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Registered: 2014-05-21

Posts: 2,436

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Re: Really Hard Algebra Equations

For the first, you could factorise the LHS as (ab + 1)(a + b) = 8(a+b) = 80, i.e. a + b = 10. Then notice that (a + b)^2 = a^2 + 2ab + b^2.

For the second, there probably is some neat way of doing this, but you can do this easily with matrices (write in the form Ax = b, where A is a 5x5 matrix, x = (a,b,c,d,e), b = (177,154,146,138,165)).

For the third, take the equation p + q + r = 7 and multiply first by p, then q, then r to form 3 separate equations in p,q,r, then add them together. You can then re-arrange for pq + qr + rp.

Bob

Administrator

Registered: 2010-06-20

Posts: 10,626

Re: Really Hard Algebra Equations

hi denis_gylaev

Q2.  If you add all five equations together you'll get a value for a + b + c + d + e

If you subtract equation 2 from equation 1 you'll get a value for -a - b - c - d + 4e

So adding these gives e. 

Similarly you can the other letters by subtracting pairs of equations.

Q3.  p+q+r = 7  and  p^2+q^2+r^2 = 9

Bob

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