Wikipedia:Reference desk/Archives/Mathematics/2013 July 9

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July 9

Algebra

Given, $x+1/x+y+1/y=6$: Using this condition, there are two related but different questions:

i) $x/y+y/x=6$, then $x-1/x+y-1/y=?$

ii) $xy+1/xy=3$, then $x-y+1/y-1/x=?$

Another question from the same book, this one also I can't make it through: If $x^{2}+2=2x;x^{4}-x^{3}+x^{2}+1=?$

Now if you take out the values of x,y that's cheating, you have to find out the answers by manipulating the equations, not the long way. These are the questions of my book which I can't solve. Please help. 117.197.64.60 (talk) 13:20, 9 July 2013 (UTC)

I added <math> formatting for ease of interpretation – I hope you do not object. — Quondum 13:41, 9 July 2013 (UTC)

These questions seem to be aimed at building the ability to spot a way forward intuitively. In you last example, you should be able to spot that subexpressions in the latter part can be simplified using the former. We would generally try to get rid of variables, and a useful approach is often reduce the degree of a polynomial. A rather straightforward approach is to rewrite the first as $x^{2}-2x+2=0$, and then to add any multiple of the LHS to the unknown expression that will cancel its highest power. At worst you'll simplify it to a first-degree polynomial.

The first question presumably is aimed at a similar simple manipulation of the given equations to allow substitutions of subexpressions, with the result of simplifying the expression. This approach is often simpler than solving for one of the unknowns. — Quondum 14:06, 9 July 2013 (UTC)

The whole idea is to multiply all equations with the product xy. And then to factorize them so as to bring out s = x + y and p = xy. Like this:

x + ${\tfrac {1}{x}}$ + y + ${\tfrac {1}{y}}$ = 6 | * xy

x²y + y + xy² + x = 6 xy

(x + y)(xy + 1) = 6 xy

s (p + 1) = 6p

s = ${\tfrac {6p}{p+1}}\quad {\color {white}.}$ (1)

i)

${\tfrac {x}{y}}$ + ${\tfrac {y}{x}}$ = 6 | * xy

x² + y² = 6 xy | + 2 xy

(x + y)² = 8 xy

s² = 8p

p = ± ${\tfrac {s}{2{\sqrt {2}}}}\quad {\color {white}.}$ (2)

By replacing the value of p obtained from (2) into (1) we deduce the value of s, and then that of p.

x - ${\tfrac {1}{x}}$ + y - ${\tfrac {1}{y}}$ = z | * xy

x²y - y + xy² - x = z xy

(x + y)(xy - 1) = z xy

s (p - 1) = z * p

z = s ${\tfrac {p-1}{p}}$ = ...

Similarly for case ii). — 79.118.170.186 (talk) 16:37, 9 July 2013 (UTC)