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Single rational expressions help!! Very confusing 

Guest Nov 4, 2017

Best Answer 

 #1
avatar+7181 
+1

The goal is to get a common denominator and combine the fractions into one single fraction.

 

Notice that  (a - c)  is in the denominator of the first fraction, but not in the second.  So we need to multiply and divide the second fraction by  (a - c) . That is, multiply it by  (a - c)/(a - c)

 

\(\frac{1}{(2a-b)(a-c)}+\frac{1}{(b-c)(b-2a)} \\~\\ =\frac{1}{(2a-b)(a-c)}+\frac{1(a-c)}{(b-c)(b-2a)(a-c)}\)

 

Now we need to multiply the first fraction by  (b - c)/(b - c) .

 

\(=\frac{1(b-c)}{(b-c)(2a-b)(a-c)}+\frac{1(a-c)}{(b-c)(b-2a)(a-c)}\)

 

Now, notice that the term (2a - b)  is  -1 multiplied by  (b - 2a) .

So let's multiply the second fraction by  -1/-1 .

 

\(=\frac{b-c}{(b-c)(2a-b)(a-c)}+\frac{(-1)(a-c)}{(b-c)(-1)(b-2a)(a-c)} \\~\\ =\frac{b-c}{(b-c)(2a-b)(a-c)}+\frac{-a+c}{(b-c)(-b+2a)(a-c)} \\~\\ =\frac{b-c}{(b-c)(2a-b)(a-c)}+\frac{-a+c}{(b-c)(2a-b)(a-c)}\)        And distribute it.

 

Now the denominators are the same and we can add the fractions together.

 

\(=\frac{b-c+-a+c}{(b-c)(2a-b)(a-c)} \\~\\ =\frac{b-a}{(b-c)(2a-b)(a-c)}\)           -c  and  +c  cancel

hectictar  Nov 4, 2017
 #1
avatar+7181 
+1
Best Answer

The goal is to get a common denominator and combine the fractions into one single fraction.

 

Notice that  (a - c)  is in the denominator of the first fraction, but not in the second.  So we need to multiply and divide the second fraction by  (a - c) . That is, multiply it by  (a - c)/(a - c)

 

\(\frac{1}{(2a-b)(a-c)}+\frac{1}{(b-c)(b-2a)} \\~\\ =\frac{1}{(2a-b)(a-c)}+\frac{1(a-c)}{(b-c)(b-2a)(a-c)}\)

 

Now we need to multiply the first fraction by  (b - c)/(b - c) .

 

\(=\frac{1(b-c)}{(b-c)(2a-b)(a-c)}+\frac{1(a-c)}{(b-c)(b-2a)(a-c)}\)

 

Now, notice that the term (2a - b)  is  -1 multiplied by  (b - 2a) .

So let's multiply the second fraction by  -1/-1 .

 

\(=\frac{b-c}{(b-c)(2a-b)(a-c)}+\frac{(-1)(a-c)}{(b-c)(-1)(b-2a)(a-c)} \\~\\ =\frac{b-c}{(b-c)(2a-b)(a-c)}+\frac{-a+c}{(b-c)(-b+2a)(a-c)} \\~\\ =\frac{b-c}{(b-c)(2a-b)(a-c)}+\frac{-a+c}{(b-c)(2a-b)(a-c)}\)        And distribute it.

 

Now the denominators are the same and we can add the fractions together.

 

\(=\frac{b-c+-a+c}{(b-c)(2a-b)(a-c)} \\~\\ =\frac{b-a}{(b-c)(2a-b)(a-c)}\)           -c  and  +c  cancel

hectictar  Nov 4, 2017

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