Precalc problem- finding all real solutions

Question

Can someone help me find the solutions of this equation? I tried a few times but I can't seem to get the right answer.

 

1 +          2x         =  2     +  4   
      (x + 7)(x + 6)   x+7     x+6                 

Edit: Sorry, the formatting was a bit odd. Those spaces should have division symbols. So the problem is [1]+[2x ÷ (x+7)(x+6)] = [2 ÷ (x + 7)] + [4 ÷ (x + 6)]

Answer 1

It appears to me that, after CLEARING FRACTIONS [MULTIPLYING BY (X+6)(X+7)] you will have

(x+6)(x+7) + 2x = 2(x+6) + 4(x+7)

and then, you eventually get x² + 9x + 2 = 0...YOU SHOULD check this, of course.

his quadratic has two irrational zeros, which you can find using the quadratic formula.

Answer 2

Multiply both sides by (x + 7)(x + 6), simplify, and solve for x.

 

(x + 7)(x + 6) + 2x = 2(x + 6) + 4(x + 7)

 

(x² + 13x + 42) + 2x = 2x + 12 + 4x + 28

 

x² + 9x + 2 = 0

 

Can you solve this quadratic equation in x and finish from here?

Answer 3

If you multiply both sides of the equation by (x+7)(x+6) you can reduce the expression to x² + 9x + 2 = 0

 

That won't factor, but you can solve using the quadratic equation.

 

Hope that helps!