Home > Math Group > Question and Answer

Question:

Answers:

1. 
   krishna-agrawala Teacher
   Graduate School

   eNotes Editor

   I believe the given equation is meant to be:

   x/(x - 4) = 5x/(x^2 - x - 12) - 3/(x + 3)

   To solve this equation we multiple every term by (x + 4)*(x + 3), which is equal to (x^2 - x - 12).

   Thus the equation becomes:

   x*(x + 3) = 5x - 3*(x - 4)

   Therefore:

   x^2 + 3x = 5x -3x +12

   Therefore: x^2 + x - 12 = 0

   Therefore: x^2 - 3x + 4x - 12 = 0

   Therefore: x*(x - 3) + 4*(x - 3) = 0

   Therefore: (x - 3)*(x + 4) = 0

   Therefore: (x - 3) = 0 or (x + 4) = 0

   Therefore:  x  = 3 or -4

   Rate answer:

   • Flag as inappropriate

   Posted by krishna-agrawala on Tuesday November 10, 2009 at 5:33 AM
   
   

2. 
   neela Teacher
   Graduate School

   eNotes Editor

   x/(x-4)=5x/(x2-x-12)-3/(x+3).

   To find x.

   Solution:

   The expression, x^2-x-12 is a denominator on the right  is equal to (x-4)(x+3) on factorisation.

   Therefore, the LCM of the denominators (x-4), (x^2-x-12) and (x+3) appearing in the given equation  is (x-4)(x-3).

   Multiplying by the LCM both sides of the equation, we get:

   [x/(x-4)]*(x-4)(x+3)=[ 5x/(x^2^2-x-12)]*(x-4)(x+3)-[3/(x+3)]*(x-4)(x+3), which simplifies to:

   x(x+3)=5x-3(x-4)

   x^2+3x=5x-3x+12

   x^2+3x-5x+3x-12=0

   x^2+x-12=0..........................................(1)

   We know that a quadratic equation like ax^2+bx+c=0 has the solution: x= [-b+sqrt(b^2-4ac)]/(2a) or

   x = [-b-sqrt(b^2-4ac)]/(2a)

   In equation (1), a = 1, b = 1 and c= = -12. Therefore,

   Therefore, x = { -1+sqrt[1^2-4*1(-12)] }/(2*1) or

   x = { -1-sqrt[1^2-4*1(-12)] }/(2*1)

   So, x= [-1+sqrt(1+48)]/2 = (-1+7)/2 =  3 or

   x = (-1-7)/2 = -4

   Therefore,  x = 3 or x=-4 are the possible solutions.

    

   Rate answer:

   • Flag as inappropriate

   Posted by neela on Tuesday November 10, 2009 at 6:39 AM