... have come across this problem ...
[MATH]\frac{6a}{a-b} - \frac{3b}{b-a} + \frac{5}{a^2 - b^2} [/MATH]
Hi Dale. Technically, the expression above is not a problem (it's just an expression) because you left out the instructions.
It's reasonable to assume that the book instructed you to combine the three algebraic ratios into a single ratio, but please include the complete exercise statement in the future so we know for sure what you're trying to do.
... This is where I get lost:
[MATH]\frac{6a}{a - b} = \frac{6a(-a - b)}{-a(a - b)(a + b)}[/MATH]
That equation is not true for all choices of a and b. You can confirm this, by assigning simple values for a and b, followed by evaluating each side.
Let a = 2 and b = 1
Your equation above evaluates as 12 = 6 (because of the extra factor of a in the denominator).
... If I am multiplying both numerator and denominator by the LCM, ... then [should] the fraction be
[MATH]\frac{6a(-a - b)(a + b)}{-a(a - b)(a + b)}[/MATH]?
No. Using a = 2 and b = 1, that ratio evaluates to 18 instead of 12.
In the numerator, it appears that -(a-b) may have been replaced with (-a-b) instead of (b-a). There is also an extra factor of a in the denominator.
I'd like to show you a different setup, where most of the work involves multiplying by the common denominator (LCM) on top -- in order to cancel all of the given denominators -- while the multiplication by the LCM on bottom waits "outside", for the final step.
Here's a symbolic example with different denominators D and d:
A/D + B/d
We combine these two algebraic ratios by first multiplying the entire expression by LCM/LCM. In this example, the LCM is D*d (which I type as Dd, below).
[math]\frac{1}{Dd} \bigg( \frac{Dd}{1} \cdot \frac{A}{D} + \frac{Dd}{1} \cdot \frac{B}{d} \bigg)[/math]
Note that the LCM on bottom has been written as the factor 1/LCM outside the grouping symbols, while the multiplication by LCM on top has been distributed inside the grouping symbols. Next, we simplify the expression inside the grouping symbols, and the result becomes the numerator of our answer. Multiplication by 1/LCM happens last, making LCM the denominator of our answer.
The very reason we multiply the original expression by the LCM
on top is to
cancel each denominator in the original expression. In the first multiplication, the denominator in A/D cancels with the factor D on top. In the second multiplication, the denominator in B/d cancels with the factor d on top. Here's the result of multiplying by the LCM on top:
[math]\frac{1}{Dd} \bigg( \frac{Ad + BD}{1} \bigg)[/math]
The last step is to carry out the multiplication by 1/(Dd). The answer is:
[math]\frac{Ad + BD}{Dd}[/math]
I hope you can follow this example, but let us know if you have questions.
Now, before you try these steps on your exercise, let's complete a sign simplification on the given expression. Doing this first will simplify the LCM.
The expression (b - a) is the same as -(a - b) because we can factor out a negative one.
Therefore, the subtraction of ratio (3b)/(b - a) can be thought of as subtracting -(3b)/(a-b) which is the same as adding (3b)/(a-b). We now have:
[MATH]\frac{6a}{a-b} + \frac{3b}{a-b} + \frac{5}{a^2 - b^2}[/MATH]
and the LCM is simply (a+b)(a-b).
[math]\frac{1}{(a+b)(a-b)} \bigg( \frac{(a+b)(a-b)}{1} \cdot \frac{6a}{a-b} + \frac{(a+b)(a-b)}{1} \cdot \frac{3b}{a-b} + \frac{(a+b)(a-b)}{1} \cdot \frac{5}{(a+b)(a-b)} \bigg)[/math]
Continue, by carefully canceling denominators inside the grouping symbols as you multiply. Simplify the result, by combining like terms. You can confirm whether your answer makes sense, by using a=2 and b=1 to compare your answer's value with the value of the given expression.
Please include your work so far, if you'd like to get more help. Cheers
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