2

Can $(x+1)(2x-1)$ be the LCM of this biquadratic equation

$$\frac{5x-1}{x+7}=\frac{3x+1}{x+5}$$

k170
  • 9,045
common grackle
  • 45
  • 1
    The function $\text{lcm}$ takes numbers as inputs, not equations. Please clarify what you mean using the right terms. – user26486 Dec 21 '14 at 18:08
  • $\text{lcm}(x+5,x+7)\neq (x+1)(2x-1), \forall x\in\mathbb Z$. This can easily be proved using the fact that $\text{lcm}(a,b)=\frac{ab}{\gcd(a,b)}$. – user26486 Dec 21 '14 at 18:19

1 Answers1

1

The lowest common denominator (LCD) of rational fractions $$\frac{5x-1}{x+7}\quad {\rm and} \quad \frac{3x+1}{x+5}$$ is the lowest common multiple (LCM) of denominator polynomials $$ x+7 \quad {\rm and} \quad x+5, $$ which is $$ (x+7)(x+5).$$ So, formally, the original equality of rational fractions can be re-written as: $$\frac{(5x-1)(x+5)}{(x+7)(x+5)}= \frac{(3x+1)(x+7)}{(x+5)(x+7)}.$$

ir7
  • 6,249