Fractional Equations #4

Algebra Level 4

Given that the value of $x$ that satisfies the equation

$\dfrac{x+1}{x+2}+\dfrac{x+8}{x+9}=\dfrac{x+2}{x+3}+\dfrac{x+7}{x+8}$

can be expressed in the form $-\dfrac{m}{n}$ , where $m$ and $n$ are coprime, positive integers, find the value of $m+n$ .

2 solutions

Kartik Sharma
Oct 26, 2014

$\frac{(x+2) -1}{x+2} + \frac{(x+9)-1}{x+9} = \frac{(x+3)-1}{x+3} + \frac{(x+8)-1}{x+8}$

$2 - \frac{1}{x+2} - \frac{1}{x+9} = 2 - \frac{1}{x+3} - \frac{1}{x+8}$

$\frac{1}{x+8} - \frac{1}{x+9} = \frac{1}{x+2} -\frac{1}{x+3}$

$\frac{1}{(x+8)(x+9)} = \frac{1}{(x+2)(x+3)}$

${x}^{2} + 5x + 6 = {x}^{2} + 17x + 72$

$12x + 66 = 0$

$x = \frac{-11}{2}$

Exactly the same!!!

Anik Mandal - 6 years, 7 months ago

U Z
Oct 26, 2014

$\frac{ x + 2 -1}{x +2} + \frac{ x + 9 -1}{x + 9} = \frac{ x + 3 -1}{x + 3} + \frac{ x + 8 -1}{x + 8}$

$2 - ( \frac{1}{x +2} + \frac{1}{x + 9}) = 2 - ( \frac{1}{x +3} + \frac{1}{x +8})$

$\frac{ 2x + 11}{(x + 2)(x + 9)} = \frac{ 2x + 11}{(x + 3)(x + 8)}$

therefore

$(2x + 11)( \frac{1}{(x + 2)(x + 9)} - \frac{1}{(x + 3)(x + 8)} = 0$

hence $x = \frac{ -11}{2} or x^{2} + 11x + 24 - x^{2} + 11x + 18 =0$

therefore $x = \frac{ -11}{2}$

Just 2 minutes faster!

Kartik Sharma - 6 years, 7 months ago