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

The equation has 0 extraneous solutions.

Step-by-step explanation:

As per the equation given it is solved by:

To determine the number of extraneous solutions, we need to solve the equation.

Multiplying both sides by the denominators eliminates the fractions:

2m - 2m = 10(2m + 3)(2m - 3)

Simplifying the equation:

0 = 10(4m² - 9)

Setting each factor equal to zero:

4m² - 9 = 0

Applying the quadratic formula:

m = ±√(9/4)

Since there are no extraneous solutions, the equation has 0 extraneous solutions.

Answer 2

0

Step-by-step explanation:

We have the fraction
`(2m)/(2m+3) -(2m)/(2m-3)`

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

`((2m-3)(2m)-(2m+3)(2m))/((2m+3)(2m-3)) =(2m^2-6m-[2m^2+6m])/((2m+3)(2m-3)) =(2m^2-6m-2m^2-6m)/((2m+3)(2m-3)) =(-12m)/((2m+3)(2m-3))`

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

`(-12m)/((2m+3)(2m-3)) =0`

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

`-12m=0`

`m=(0)/(-12)`

`m=0`

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

`(-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