Solve the equation, using any method. Be sure to check for extraneous solutions.

Mathematics

1 answer:

hichkok12 [17]3 years ago

7 0

Answer: No real solution.

Step-by-step explanation:

$\frac{2}{(2x+6)}-\frac{2}{x^2+5x+6}=\frac{3}{(x+3)}$

First solve $x^2+5x+6$

$\frac{2}{(2x+6)}-\frac{2}{(x+3)(x+2)}=\frac{3}{(x+3)}$

$\frac{[2(x+3)(x+2)]-[2(2x+6)]}{(2x+6)(x+3)(x+2)} = \frac{3}{(x+3)}$

$\frac{[(2x+6)(x+2)]-(4x-12)}{(2x+6)(x+3)(x+2)} = \frac{3}{(x+3)}$

$\frac{(2x^2+4x+6x+12)-(4x-12)}{(2x+6)(x+3)(x+2)} = \frac{3}{(x+3)}$

$\frac{2x^2+10x+12-4x+12}{(2x+6)(x+3)(x+2)} = \frac{3}{(x+3)}$

$\frac{2x^2+6x+24}{(2x+6)(x+3)(x+2)} = \frac{3}{(x+3)}$

Past this point, we can't solve $2x^2+6x+24$ and obtain real values of x. Therefore, there are no real solutions.

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