Solve for x.

Question

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

 

If possible please help walk me through the steps.

Answer 1

1) Foil the denominator on the right.

 

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

 

2) Eliminate all pesky denominators by multiplying both sides by (x+2)(x-2).

 

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

 

3) Simplify the left.

 

3x^2 + 6x + x -2 = -4

3x^2 + 7x + 2 = 0

(3x + 1)(x+2) = 0

 

4) Solve both terms for x.

 

3x+1 = 0

3x = -1

x = -1/3

 

x+2 = 0

x = -2

Answer 2

 

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

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

 

3x² +6x +x-2 = -4

3x² +7x+2=0

3x² +6x +1x +2=0

3x(x+2) +1(x+2) =0

(x+2)(3x+1)=0

x+2=0  or 3x+1=0

x=-2 or x=-1/3

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