Answer :
Answer:
0
Step-by-step explanation:
We have the fraction [tex]\frac{2m}{2m+3} -\frac{2m}{2m-3}[/tex]
Step 1. Use LCM of the fraction, (2m+3)(2m-3), to simplify the fraction:
[tex]\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)}[/tex]
Step 2. Equate the resulting fraction to zero and solve for [tex]m[/tex]:
[tex]\frac{-12m}{(2m+3)(2m-3)} =0[/tex]
[tex]-12m=0[(2m+3)(2m-3)][/tex]
[tex]-12m=0[/tex]
[tex]m=\frac{0}{-12}[/tex]
[tex]m=0[/tex]
Step 3. Replace the value in the original equation and check if it holds:
[tex]\frac{-12m}{(2m+3)(2m-3)} =0[/tex]
[tex]-12m=0[/tex]
Since [tex]m=0[/tex],
[tex]-12(0)=0[/tex]
[tex]0=0[/tex]
Since the only solution of the equation holds, the equation bellow doesn't have any extraneous solution
The number of extraneous solution the equation has is 1
Given:
StartFraction 2 m Over 2 m + 3 EndFraction minus StartFraction 2 m Over 2 m minus 3 EndFraction
[tex] = \frac{2m}{2m + 3} - \frac{2m}{2m - 3} [/tex]
solve by finding the lcm first
[tex] = \frac{2m(2m - 3) - 2m(2m + 3)}{(2m + 3) (2m - 3)} [/tex]
= 4m² - 6m - 4m² - 6m / 4m² - 6m + 6m - 9
= - 12m / 4m² - 9
- equate to 0
0 = 12m / (4m² - 9)
- cross product
0 × (4m² - 9) = 12m
0 = 12m
m = 0/12
m = 0
Therefore, the number of extraneous solution the equation has is 1
Learn more about extraneous solution:
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