Ask your own question, for FREE!
Mathematics
OpenStudy (anonymous):

4x^2/(x^2-9) -2x/(x+3)=3/(x-3)?

7 years ago
OpenStudy (anonymous):

x=3/2 , -3

7 years ago
OpenStudy (anonymous):

(x+3)(x-3)[4x^2/(x+3)(x-3) - 2x/(x+3)] = [3/(x-3)] (x+3)(x-3) 4x^2 - [2x(x-3)] = [3(x+3)] 4x^2 - (2x^2 -6x) = 3x +9 4x^2-2x^2+6x = 3x+9 4x^2-2x^2+6x-3x-9 = 0 (4x^2-2x^2)+(6x-3x)-9 = 0 2x^2+3x-9=0 (2x-3)(x+3) =0

7 years ago
OpenStudy (anonymous):

The denominator of the LHS of the equation factorises into (x+3)(x-3). Multiply the whole equation by (x+3)(x-3) to get this: 4x^2 -2x (x+3) = 3(x-3) 4x^2 -2x^2 -6x =3x -9 2x^2 -6x = 3x -9 Collecting all the terms on the LHS of the equation, and thus reducing the RHS to zero. 2x^2 -6x -3x +9 =0 2x^2-9x +9 = 0 So, factorising 2x^2 -9x +9, ie, the LHS gives y ou (2x -3)(x - 3) for the LHS and zero for the RHS. So, (2x-3)(x-3) =0 Therefore, 2x-3 =0, and x-3 =0. So, x has two values, ie, 3/2 and 3. But lolking at the denominators, you'll find that if we put x=3 in x^2-9 gives us zero. Putting x=3 in the denominator (x-3) also equals zero. ie, putting x=3 in the equation gives us this: 4 (9)/(9-9) -2(3)/(3+3) = 3/(3-3) So here we have two denominators which become zero, when x=3. In other words, we are dividing by zero. Which is not acceptable. So x=3 is not acceptable as a solution. The only acceptable solution is therefore 2x=3, ie x=3/2. Put x=3/2 in your equation, and the LHS gives you: (9)(4)/(9-36) - (3)(2)/(3+6) ie, 36/-27 - 6/9 ie -4/3 -2/3, which equals -2. So for the LHS of the equation, x=3/2 gives us -2. The RHS of the equation (the orignal in the question) is: 3/(x-3) Wheln x=3/2, this becomes 6/(-3) which equals -2. But for x=3/2, the LHS is also -2. Hence although simplifying the equation gave us two answers (ie, 3 and 3/2), only x=3 is acceptable

7 years ago
OpenStudy (anonymous):

Sorry, the last two lines should read: "Hence although simplifying the equation gave us two answers (ie, 3 and 3/2), only x=3/2 is acceptable".

7 years ago
Can't find your answer? Make a FREE account and ask your own question, OR you can help others and earn volunteer hours!