Solve the equation 1 over 3x + 9 - 2 over x + 3 = 2

Answer 1

To solve the equation 1/(3x + 9) - 2/(x + 3) = 2, we need to first find a common denominator for the fractions on the left side of the equation.

The common denominator for 3x + 9 and x + 3 is (3x + 9)(x + 3) = 3(x + 3)(x + 3) = 3(x^2 + 6x + 9) = 3x^2 + 18x + 27.

Now we can rewrite the equation with the common denominator:

1/(3x + 9) - 2/(x + 3) = 2 becomes:

(x + 3) / [3(x^2 + 6x + 9)] - 2(3) / [3(x^2 + 6x + 9)] = 2

Now we have:

(x + 3 - 6) / [3(x^2 + 6x + 9)] = 2

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

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

Now we multiply both sides by 3(x + 3)^2:

(x - 3) = 6(x + 3)^2

Expand the right side:

x - 3 = 6(x^2 + 6x + 9)

x - 3 = 6x^2 + 36x + 54

Rearrange the equation to set it equal to zero:

6x^2 + 35x + 57 = 0

Now we have a quadratic equation that we can solve using the quadratic formula:

x = [-b ± sqrt(b^2 - 4ac)] / 2a

In this case, a = 6, b = 35, and c = 57.

Plugging in the values:

x = [-35 ± sqrt(35^2 - 4(6)(57))] / 2(6)

x = [-35 ± sqrt(1225 - 1368)] / 12

x = [-35 ± sqrt(-143)] / 12

Since the square root of a negative number is not a real number, this equation does not have a real solution.