Question

solve the equation.\\(\frac{c + 1}{3c + 9} + \frac{c}{2c + 6} = \frac{3}{4c + 12}\\)\\(c = \boxed{}\\)

Explanation:

Step1: Factor denominators

$3c+9=3(c+3)$, $2c+6=2(c+3)$, $4c+12=4(c+3)$
Equation becomes: $\frac{c+1}{3(c+3)} + \frac{c}{2(c+3)} = \frac{3}{4(c+3)}$

Step2: Eliminate denominators

Multiply all terms by $12(c+3)$ (LCD of 3,2,4):
$4(c+1) + 6c = 9$

Step3: Expand and simplify left side

$4c + 4 + 6c = 9$
$10c + 4 = 9$

Step4: Isolate variable term

$10c = 9 - 4$
$10c = 5$

Step5: Solve for c

$c = \frac{5}{10} = \frac{1}{2}$
(Verify $c
eq -3$, which holds here)

Answer:

$\frac{1}{2}$