MHB Algebraic Fraction: How to Solve and Simplify?

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To solve the algebraic fraction equation $$\frac{3 }{ 2(x+1)} + \frac{1}{x+1} = \frac{5}{6}$$, the common denominator is determined to be 6(x + 1). By adjusting the second term's numerator to match the first term's denominator, the equation simplifies to $$\frac{5}{2(x+1)} = \frac{5}{6}$$. Solving this leads to the solution x = 2. Verification of the solution can be done by substituting x back into the original equation. The final answer is confirmed as correct.
mathlearn
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Hi,


$$\frac{3 }{ 2(x+1)} + \frac{1}{x+1} = \frac{5}{6}$$
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:)
 
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mathlearn said:
Hi,


$$\frac{3 }{ 2(x+1)} + \frac{1}{x+1} = \frac{5}{6}$$
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
 
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not sure might be (x+1) or 2,1?? :)
 
mathlearn said:
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
 
Just multiply the numerator and denominator of second term of LHS by 2 to equate it's denominator to the first term of LHS.

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

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

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

$$x=2$$

Verify it by substituting the value of $x$ to be 2 in the LHS.
 
Many Thanks (Smile)
 
Seemingly by some mathematical coincidence, a hexagon of sides 2,2,7,7, 11, and 11 can be inscribed in a circle of radius 7. The other day I saw a math problem on line, which they said came from a Polish Olympiad, where you compute the length x of the 3rd side which is the same as the radius, so that the sides of length 2,x, and 11 are inscribed on the arc of a semi-circle. The law of cosines applied twice gives the answer for x of exactly 7, but the arithmetic is so complex that the...

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