Max@Math Revolution wrote:[Math Revolution GMAT math practice question]

If 1/x(x-1)+1/x(x+1)=1/(x+1)(x-1), what is the value of x?

A. √2
B. √3
C. 0
D. 4
E. No solution

$$\frac{1}{{x\left( {x - 1} \right)}} + \frac{1}{{x\left( {x + 1} \right)}} = \frac{1}{{\left( {x + 1} \right)\left( {x - 1} \right)}}\,\,\,\,\,\,\,\,\,\,\,\left[ {\,x \notin \left\{ { - 1,0,1} \right\}\,\,\,{\text{implicitly}}\,} \right]$$
$$? = x$$
$$\frac{{1 \cdot \boxed{\left( {x + 1} \right)}}}{{x\left( {x - 1} \right) \cdot \boxed{\left( {x + 1} \right)}}} + \frac{{1 \cdot \boxed{\left( {x - 1} \right)}}}{{x\left( {x + 1} \right) \cdot \boxed{\left( {x - 1} \right)}}} = \frac{{1 \cdot \boxed{\,x\,}}}{{\left( {x + 1} \right)\left( {x - 1} \right) \cdot \boxed{\,x\,}}}\,\,\,\,\, \Rightarrow \,\,\,\,\,\frac{{2x}}{{x\left( {x - 1} \right)\left( {x + 1} \right)}} = \frac{x}{{x\left( {x - 1} \right)\left( {x + 1} \right)}}$$
$$\mathop \Rightarrow \limits^{x\, \ne \,\,0} \,\,\,\frac{2}{{\left( {x - 1} \right)\left( {x + 1} \right)}} = \frac{1}{{\left( {x - 1} \right)\left( {x + 1} \right)}}\,\,\,\,\, \Rightarrow \,\,\,\,{\text{impossible}}$$

The correct answer is therefore (E).


This solution follows the notations and rationale taught in the GMATH method.

Regards,
Fabio.

=>
1/x(x-1)+1/x(x+1)=1/(x+1)(x-1)
=> (x+1) + (x-1) = x, by multiplying both sides with x(x+1)(x-1)
=> 2x = x
=> x = 0.
However, x = 0 is an erroneous solution since it gives both fractions on the right-hand side denominators of 0.

Therefore, the answer is E.
Answer: E