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How do you solve #(3x )/ (x+1) = 12 / (x-1)#?

Answer 1

the answer is #x=(5+sqrt(41))/(2), (5-sqrt(41))/(2)#

first you have to cross multiply and you get: #12x+12=3x^2-3x# and now move all the terms to one side to make it equal to zero. #3x^2-15x-12=0# and you can take out #3# from all numbers so: #3(x^2-5x-4)=0# since you can't factor the parenthesis equation, you have to use the quadratic formula which is: #(-bsqrt(b^2-4ac))/(2a)# and put in the respectful numbers in:#-(-15)sqrt(4(3)(-12))/(2(3))# and now solve to get your answer

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Answer 2

Kennedy Adams

To solve the equation (3x)/(x+1) = 12/(x-1), we can cross-multiply to eliminate the fractions. This gives us 3x(x-1) = 12(x+1). Expanding both sides of the equation, we get 3x^2 - 3x = 12x + 12. Rearranging the terms, we have 3x^2 - 15x - 12 = 0. To solve this quadratic equation, we can factor it or use the quadratic formula. Factoring, we have (x-4)(3x+1) = 0. Setting each factor equal to zero, we get x-4 = 0 or 3x+1 = 0. Solving these equations, we find x = 4 or x = -1/3. Therefore, the solutions to the equation are x = 4 and x = -1/3.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.