solving and non permissible values answer as soon as you can please!

Question

Solve and state the non-permissable values

 

   ( x+4)/( x² +5x+ 6) = ( x+2 )/(x²+2x-3) - (x-1)/(x² +1x-2)

   

 

Thanks so much!!

Answer 1

 (x+4)/( x²+5x+6) = (x+2)/(x²+2x-3) - (x-1)/(x²+x-2)

 

Factoring our denominators we get

 

(x+4)/[(x+3)(x+2)] = (x+2)/[(x+3)(x-1)] - (x-1)/[(x+2)((x-1)]

 

Since we cannot divide by zero we have non-permissible values at x = -3, -2, 1

 

We can eliminate (x-1) from the last fraction since it is in both numerator and denominator

 

(x+4)/[(x+3)(x+2)] = (x+2)/[(x+3)(x-1)] - 1/(x+2)

 

A common denominator for everything would be (x+3)(X+2)(x-1) so multiply everything by that

 

[(x+3)(X+2)(x-1)][(x+4)/[(x+3)(x+2)]] = [(x+3)(X+2)(x-1)][(x+2)/[(x+3)(x-1)]

 + [(x+3)(X+2)(x-1)][- 1/(x+2)]

 

= (x-1)(x+4) = (x+2)(x+2) - [(x+3)(x-1)]

 

x² + 3x - 4= x² + 4x + 4 - (x² + 2x -3)

x² + 3x - 4 = x² + 4x + 4 - x² - 2x + 3

x² + 3x - 4 = 2x + 7

 

Let's make a quadratic by moving everything to the left side

 

x² + x -11 = 0

There are no factors of -11 that will add to 1 so we need to use the quadratic equation

 

a = 1, b = 1, c = -11

x = [-b +- √(b²-4ac)]/2a

 

x = [-1 +- √(1-4(1)(-11))]/2(1)

x = [-1 +-  √(45)]/2

x = [-1 +- 3√5]/2

 

x = [-1 + 3√5]/2   and    x = [-1 - 3√5]/2