Algebra Theory Misc equation solving

We will see multiple questions today and try to solve with "lesser" effort.
For e.g. without expanding or LCM or similar things.

Q1.

$$\frac{x - a - b}{c} + \frac{x - b - c}{a} + \frac{x - c - a}{b} = 3,$$

where $a, b, c$ are positive constants. Solve for $x$.
Solution:
Move 3 to LHS and Subtract 1 from each fraction.
(x-a-b)/c - 1 + (x-b-c)/a - 1 + (x-c-a)/b - 1  = 0
(x-a-b-c)(1/c + 1/a + 1/b) = 0
Since a,b,c are positive constants (1/c + 1/a + 1/b) can't be 0.
So x-a-b-c = 0 => 
x = a + b + c = Answer

Q2.
x+y+z = 2007
xy + yz + zx = 4011
x != 1, y != 1, z != 1
Find value of
1/(1-x) + 1/(1-y) + 1/(1-z).

Solution:

Numerator of the said expression is:
(1-y)(1-z) + (1-x)(1-z) + (1-y)(1-x)
Let's expand the terms:
(1-y)(1-x) = 1 + xy - x -y
(1-x)(1-z) = 1 + xz - x - z
(1-y)(1-z) = 1 + yz - z -y
Adding them we get:
3 + (xy + yz + xz) - 2(x + y + z)
= 3 + 4011 - 2.2007
= 0 = Answer.

Q3.

Find all positive integral solutions of

$$\frac{1}{x}+\frac{1}{y}=\frac{1}{7}$$Solution:
7x + 7y = xy
7x - xy + 7y = 0
x (7-y) + 7y - 49 = -49
x (7-y) + 7(y-7) = -49
x (7-y) - 7(7-y) = -49
(x-7)(7-y) = -49
(x-7)(y-7) = 49
Factors of 49:
1,49 | 7,7 | 49, 1
x-7 = 1, y-7=49 => x = 8, y = 56 
x-7 = 7, y-7=7 => x = y = 14 
y-7 = 1, x-7=49 => y = 8, x = 56 

Q4.
x + y = 5
y + z = 7
z + x = 4

Solve for x,y,z.

Solution:
Add all 3.
2(x+y+z) = 16
x+y+z = 8
Subtract first equation.
x+y+z  - (x+y)= 8 - 5
=> z =3, x = 1, y = 4

Q5.
ab/(a + b) = 1/3
bc/(b + c) = 1/4
ca/(c + a) = 1/5
Find value of :
24abc/(ab + bc  +ca)

Solution:
Invert each given equation, take LCM and add.
You will get:
(ac + bc + ab + ac + ab + bc)/abc = 12
2(ab + bc + ca) = 12abc
ab + bc + ca = 6abc
Answer = 4

Q6.
x + 1/y = 7/3
y + 1/z = 4
z + 1/x = 1

Find xyz.

Solution:
Add all to get:
x + y + z + 1/x + 1/y + 1/z = 22/3
Multiply all:
(x + 1/y)(y+1/z)(z+1/x) = 28/3
(xy + 1 + x/z + 1/yz)(z + 1/x) = 28/3
(xyz + z + x + 1/y + y + 1/x + 1/z + 1/xyz) = 28/3
xyz + 1/xyz + (x + y + z + 1/x + 1/y + 1/z) = 28/3
xyz + 1/xyz + (22/3) = 28/3
xyz + 1/xyz = 2
xyz = t => (t-1)^2 = 0 => t = 1 = xyz

Q7.

a,b,c are positive numbers such that abc = 1.
Solve for x in:
2ax/(ab + a + 1) + 2bx/(bc + c + 1) + 2cx/(ac + c + 1) = 1

Solution:
Quick solution, put a = b = c =1 => x = 1/2
Detailed solution:
c = 1/ab and replace in the given equation so that we have to deal with only 2 variables a,b.
Once you do that, all the fractions in the given equation will have the same denominator = (1 + a + ab).
And the equation becomes:
2x(1 + a  ab)/(1 + a + ab) = 1
=> x = 1/2  = answer.

Q8.

$$a_{n+1} = \frac{1}{1 + \frac{1}{a_n}}, \quad a_1 = 1$$

Find $a_1 a_2 + a_2 a_3 + \cdots + a_{2008} a_{2009}$

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Solution:
a_(n+1).(1 + a_n) = a_n
a_(n+1).a_n = a_n - a_(n+1)
So:
a1.a2 = a1 - a2
a2.a3 = a2 - a3
.....
a2008.a2009 = a2008 - a2009

Adding all:
Answer = a1 - a2009 = 1 - a2009.
But what's a2009?
Revisiting:
a_(n+1).a_n = a_n - a_(n+1)
Divide by a_(n+1).a_n
=> 1 = 1/a_(n+1) - 1/a_n
Let 1/a_n = b_n
=> b_(n+1) - b_n = 1
=> b_n is an A.P.
b_2009 = b1 + 2008 = 2009
=> a_2009 = 1/2009
Answer: 1 - 1/2009 = 2008/2009

Q9.

Find sum of all real $x$ satisfying

$$(3^x-27{)}^2+(5^x-625{)}^2=(3^x+5^x-652{)}^2$$Solution:
Let a = 3^x - 27 and b = 5^x - 625 then the equation becomes:
a^2 + b^2 = (a + b)^2
=> 2.a.b = 0
=> a = 0 or b = 0
=> 3^x = 27 or 5^x - 625
=> x = 3 or x = 4

Q10.

$$f(x) = \frac{x^{2010}}{x^{2010} + (1 - x)^{2010}}$$ $$f\left(\frac{1}{2011}\right)+f\left(\frac{2}{2011}\right)+\cdots +f\left(\frac{2010}{2011}\right)=?$$Solution:

Notice that f(x) + f(1-x) = 1
=> f(1/2011) + f(2010/2011) = 1
So the given expression becomes 1005.

Q11.

Let $m, n$ be integers such that $1 < m \leq n$.

$$f(m, n) = \left(1 - \frac{1}{m} \right) \left(1 - \frac{1}{m+1} \right) \left(1 - \frac{1}{m+2} \right) \cdots \left(1 - \frac{1}{n} \right)$$

If

$$d = f(2, 2049) + f(3, 2049) + \cdots + f(2049, 2049)$$

Find

$$\frac{d}{64}$$Solution:
Simplify f(m,n):
f(m,n) = (m-1)/m * (m)/(m+1) * .. (n-1)/n
= m-1/n
f(2,2049) = 1/2049
f(3,2049) = 2/2049
...
f(2049,2049) = 2048/2049
d = (1 + 2 ... 2048)/2049 = 1024
d/64 = 16 = answer.