combine the following rational expressions
(2x+2)/(3x^2+6x+3)+x-7/3x^2-3
(2x + 3)/3(x + 1)^2 + (x - 7)/3(x^2 - 1)
(2x + 3)/3(x + 1)^2 +
(x - 7)/3(x - 1)(x + 1)
LCD = 3(x + 1)^2(x - 1)
Take it from here.
After many reminders for past posts, you must know by now that brackets are critical for
these type of problems, so
(2x+2)/(3x^2+6x+3)+x-7/(3x^2-3)
= 2(x+1)(3)(x^2 + 2x + 1)/(3(x^2 - 1))
= 6(x+1)(x+1)(x+1)/(3(x+1)(x-1))
= 2(x+1)^2 /(x-1) , x ≠ ±1
reiny
this 2(x+1)(x+1)/x-1 the answer.
oops , let's try that again, should write these out first.
assuming you meant:
(2x+2)/(3x^2+6x+3)+ (x-7)/(3x^2-3)
= 2(x+1)/((3)(x+1)(x+1)) + (x-7)/(3(x+1)(x-1))
LCD is 3(x+1)(x+1)(x-1)
= ( 2(x+1) + (x-7)(x+1) )/( (3(x+1)(x+1)(x-1) )
= (2x+2 + x^2 - 6x - 7)/( 3(x+1)(x+1)(x-1) )
= x^2 - 4x - 5)/ (3(x+1)(x+1)(x-1) )
= (x+1)(x-5) / (3(x+1)(x+1)(x-1) )
= (x-5) / (3(x+1)(x-1) ) , x ± 1
To combine the given rational expressions, we need to find a common denominator and then add or subtract the resulting fractions.
First, let's simplify each expression individually. Starting with the first expression:
(2x + 2) / (3x^2 + 6x + 3)
We can see that both the numerator and denominator can be divided by 2, giving us:
(x + 1) / (3x^2 + 6x + 3)
Now let's simplify the second expression:
(x - 7) / (3x^2 - 3)
Both the numerator and denominator have a common factor of (x - 1), which we can cancel out:
(x - 7) / (3(x^2 - 1))
Next, let's factor the denominator of the second expression using the difference of squares formula:
(x - 7) / (3(x - 1)(x + 1))
Now we have both expressions simplified. To combine them, we need a common denominator. In this case, the common denominator is (3(x - 1)(x + 1)).
Let's rewrite each fraction with the common denominator:
[(x + 1) / (3x^2 + 6x + 3)] + [(x - 7) / (3(x - 1)(x + 1))]
To add these fractions, the numerators need to be added together and placed over the common denominator:
[(x + 1) + (x - 7)] / (3(x - 1)(x + 1))
Simplifying the numerator by combining like terms:
[2x - 6] / (3(x - 1)(x + 1))
So, the combined expression is (2x - 6) / (3(x - 1)(x + 1)).