Question

How many extraneous solutions does the equation below have? StartFraction 2 m Over 2 m + 3 EndFraction minus StartFraction 2 m Over 2 m minus 3 EndFraction = 1 0 1 2 3

Answer 1

Answer:

A. on ed

Step-by-step explanation:

Answer 2

Answer:

0

Step-by-step explanation:

We have the fraction $\frac{2m}{2m+3} -\frac{2m}{2m-3}$

Step 1. Use LCM of the fraction, (2m+3)(2m-3), to simplify the fraction:

$\frac{(2m-3)(2m)-(2m+3)(2m)}{(2m+3)(2m-3)} =\frac{2m^2-6m-[2m^2+6m]}{(2m+3)(2m-3)} =\frac{2m^2-6m-2m^2-6m}{(2m+3)(2m-3)} =\frac{-12m}{(2m+3)(2m-3)}$

Step 2. Equate the resulting fraction to zero and solve for $m$:

$\frac{-12m}{(2m+3)(2m-3)} =0$

$-12m=0[(2m+3)(2m-3)]$

$-12m=0$

$m=\frac{0}{-12}$

$m=0$

Step 3. Replace the value in the original equation and check if it holds:

$\frac{-12m}{(2m+3)(2m-3)} =0$

$-12m=0$

Since $m=0$,

$-12(0)=0$

$0=0$

Since the only solution of the equation holds, the equation bellow doesn't have any extraneous solution