How do you solve #\frac { 7x - 49} { 3x ^ { 2} + 6x - 72} + \frac { 1} { x + 6} = \frac { 5} { 3x - 12}#?

Question

Ethan Adkins · Accepted Answer

#x=91/5#

First factor the terms in the denominators:

#\frac { 7x - 49} { 3x ^ { 2} + 6x - 72} + \frac { 1} { x + 6} = \frac { 5} { 3x - 12}#

#(7x-49)/(3(x + 6)(x - 4))+1/(x+6)=5/(3(x-4))#

So we have common factors of #3(x + 6)(x - 4)# in the denominator, now we multiply the whole thing times the common denominator:

#3(x + 6)(x - 4)[(7x-49)/(3(x + 6)(x - 4))+1/(x+6)=5/(3(x-4))]#

now cancel out and multiply through using the distributive property:

#7x-49 + 3(x-4) = 5(x+6)#

#7x-49 + 3x-12 = 5x+30#

#10x -61 =5x+30#

#5x=91#

#x=91/5#

Eva Adamson · Answer

undefined To solve the equation $ \frac{7x - 49}{3x^2 + 6x - 72} + \frac{1}{x + 6} = \frac{5}{3x - 12} $:

1. Factor the denominators of the fractions, if possible.
2. Find a common denominator for the fractions.
3. Combine the fractions into one.
4. Solve for $ x $ by simplifying and rearranging the equation.
5. Check for extraneous solutions.

Let's go through these steps:

1. Factor the denominators of the fractions:
$$ 3x^2 + 6x - 72 = 3(x^2 + 2x - 24) = 3(x + 6)(x - 4) $$

2. Find a common denominator: It is $ (3x + 6)(x - 4) $.

3. Rewrite the equation with the common denominator:
$$ \frac{7x - 49}{(3x + 6)(x - 4)} + \frac{1}{x + 6} = \frac{5}{3(x - 4)} $$

4. Combine the fractions:
$$ \frac{(7x - 49) + (1)(3x + 6)}{(3x + 6)(x - 4)} = \frac{5}{3(x - 4)} $$
$$ \frac{7x - 49 + 3x + 6}{(3x + 6)(x - 4)} = \frac{5}{3(x - 4)} $$
$$ \frac{10x - 43}{(3x + 6)(x - 4)} = \frac{5}{3(x - 4)} $$

5. Cross multiply and solve the resulting equation:
$$ (10x - 43)(3(x - 4)) = 5(3x + 6)(x - 4) $$
$$ 30x^2 - 160x + 171 = 15x^2 - 30x - 60x + 120 $$
$$ 15x^2 - 130x + 51 = 0 $$

Now, you can solve this quadratic equation using factoring, completing the square, or the quadratic formula to find the values of $ x $.