Solve x/(x + 100) - (x + 25)/(x + 125) = 1/10. State the restrictions on x and say whether any real solution exists.

Step-by-step solution

1. 1

   Record the restrictions first. A denominator can never be zero, so before any algebra note

   $x\neq-100,\qquad x\neq-125.$

   Any candidate that hits one of these values would have to be thrown out at the end, so writing them down now is cheaper than discovering the problem later.

2. 2

   Combine the two fractions over the common denominator $(x+100)(x+125)$. Cross-multiplying the numerators:

   $\frac{x(x+125)-(x+25)(x+100)}{(x+100)(x+125)}=\frac{1}{10}.$

   Expanding each product separately,

   $x(x+125)=x^{2}+125x,\qquad (x+25)(x+100)=x^{2}+125x+2500.$

3. 3

   Watch the numerator collapse to a constant. The two expansions differ only by the constant term, so the $x^{2}$ and $125x$ pieces cancel completely:

   $\left(x^{2}+125x\right)-\left(x^{2}+125x+2500\right)=-2500.$

   The equation reduces to

   $\frac{-2500}{(x+100)(x+125)}=\frac{1}{10}.$

   This is the key structural fact: the left side is a constant over a quadratic, and because $-2500<0$ the left side can only be positive where the denominator is negative — i.e. for $-125<x<-100$.

4. 4

   Cross-multiply and form the quadratic. From $-25000=(x+100)(x+125)=x^{2}+225x+12500$,

   $x^{2}+225x+37500=0.$

   The discriminant is

   $\Delta=225^{2}-4(1)(37500)=50625-150000=-99375<0,$

   so the quadratic has no real roots and the original equation has no real solution.

5. 5

   Give the complex roots and confirm the geometric reason. Since $99375=625\cdot159$, $\sqrt{-99375}=25i\sqrt{159}$ and

   $x=\frac{-225\pm 25i\sqrt{159}}{2}.$

   Substituting either root back reproduces $\tfrac{1}{10}$ exactly. The real-variable reason for failure: on $(-125,-100)$ the product $(x+100)(x+125)$ has minimum value $-\left(\tfrac{25}{2}\right)^{2}=-156.25$, so $\frac{-2500}{(x+100)(x+125)}\ge 16$ there — never as small as $0.1$.