Algebraic Fraction: How to Solve and Simplify?

[mathlearn]

Hi,


\(\displaystyle \frac{3 }{ 2(x+1)} + \frac{1}{x+1} = \frac{5}{6}\)
Can you help me on this problem. I'm not sure on how to solve this and please be kind enough to explain your steps little bitMany Thanks:)

 

[topsquark]

Hi,


\(\displaystyle \frac{3 }{ 2(x+1)} + \frac{1}{x+1} = \frac{5}{6}\)
Can you help me on this problem. I'm not sure on how to solve this and please be kind enough to explain your steps little bitMany Thanks:)

Hint: What is the common denominator for these fractions?

-Dan

 

[mathlearn]

not sure might be (x+1) or 2,1?? :)

 

[topsquark]

not sure might be (x+1) or 2,1?? :)

The common denominator will contain 3 terms: (x + 1), 2(x + 1), and 6. So it must have one factor of (x + 1) and a factor of 6. (The 6 = 2 x 3, so it contains the 2 automatically.) So you are looking for 6 (x + 1), right? Can you take it from there?

-Dan

 

[phymat]

Just multiply the numerator and denominator of second term of LHS by 2 to equate it's denominator to the first term of LHS.

\(\displaystyle \frac{3}{2(x+1)} + \frac{1}{x+1} = \frac{5}{6}\)

\(\displaystyle \frac{3}{2(x+1)} + \frac{{\color{red}{2}} \cdot 1}{{\color{red}{2}} \cdot (x+1)} = \frac{5}{6}\)

\(\displaystyle \frac{5}{2(x+1)} = \frac{5}{6}\)

\(\displaystyle x=2\)

Verify it by substituting the value of $x$ to be 2 in the LHS.

 

[mathlearn]

Many Thanks (Smile)