What is the solution to the equation (3 / m+3) - (m/3-m) = (m^2+g / m^2-g)

m = 3

m = 6

all real numbers

no solution

No solution.

Step-by-step explanation:

I must correct your given equation because it must read like this:

(3 / (m+3)) - (m / (3-m)) = (m∧2+9) / (m∧2-9)

Now we can solve it:

First we will transfer the fraction from the right to the left side of the equation.

(3 / (m+3)) - (m / (3-m)) - (m∧2+9) / (m∧2-9) = 0

We will change binomial sign (3-m) = - (m-3) and that change sign of the second fraction from - to +, also we will change (m∧2-9) which is difference squared to (m-3) (m+3) and we get:

(3 / (m+3)) + (m / (m-3)) - (m∧2+9) / (m-3) (m+3) = 0

The smallest common denominator for this equation is (m+3) (m-3)

when we multiply entire equation with (m+3) (m-3) we get:

3· (m-3) + m· (m+3) - (m∧2+9) = 0 = > 3m-9+m∧2+3m-m∧2-9=0 = >

6m-18=0 = > 6m=18 = > m=18/6 = 3 = > m=3

But we have conditions (m+3) (m-3) ≠ 0 = > m+3≠0 and m-3≠0 = > m≠-3 and m≠3

We conclude that this equation does not have solution.

Good luck!