MHB What Is the Fastest Method to Solve This Complex Fraction Equation?

AI Thread Summary
The discussion focuses on finding the quickest method to solve the complex fraction equation. One suggested approach involves multiplying both sides by the common denominator, while another emphasizes simplifying the fractions by identifying common factors. A detailed breakdown of the simplification process is provided, leading to a final equation that can be solved for x. The conclusion reached is that the solution for x is 2(n-m-k). This method streamlines the problem-solving process significantly.
NotaMathPerson
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Hello! I just want to solve this exercise using shortest way possible.

$\frac{k-n}{2m+x}+\frac{m-n}{2k+x}=\frac{k+m-2n}{k+m+x}$

Because when I tried it is so lengthy. Maybe you can teach me how to go about it faster. Thanks
 
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NotaMathPerson said:
Hello! I just want to solve this exercise using shortest way possible.

$\frac{k-n}{2m+x}+\frac{m-n}{2k+x}=\frac{k+m-2n}{k+m+x}$

Because when I tried it is so lengthy. Maybe you can teach me how to go about it faster. Thanks
The only way I can think of is to multiply both sides by (2m + x)(2k + x)(k + m + x). I don't see any other way to do the problem.

-Dan
 
NotaMathPerson said:
Hello! I just want to solve this exercise using shortest way possible.

$\frac{k-n}{2m+x}+\frac{m-n}{2k+x}=\frac{k+m-2n}{k+m+x}$

Because when I tried it is so lengthy. Maybe you can teach me how to go about it faster. Thanks

The approach suggested by topsquark is a way, and you can also try to find the common factors so to simplify things in a more manageable and easier way, like this:

$$\frac{k-n}{2m+x}+\frac{m-n}{2k+x}=\frac{(k-n)+(m-n)}{k+m+x}$$

$$\frac{k-n}{2m+x}+\frac{m-n}{2k+x}=\frac{k-n}{k+m+x}+\frac{m-n}{k+m+x}$$

$$\frac{k-n}{2m+x}-\frac{k-n}{k+m+x}=\frac{m-n}{k+m+x}-\frac{m-n}{2k+x}$$

$$\left(k-n\right)\left(\frac{1(k+m+x)-(2m+x)}{(2m+x)(k+m+x)}\right)=\left(m-n\right)\left(\frac{1(2k+x)-1(k+m+x)}{(k+m+x)(2k+x)}\right)$$

Since $$k+m+x\ne 0$$, we have:

$$\left(k-n\right)\left(\frac{k+m+x-2m-x}{2m+x}\right)=\left(m-n\right)\left(\frac{2k+x-k-m-x}{2k+x}\right)$$

$$\left(k-n\right)\left(\frac{k-m}{2m+x}\right)=\left(m-n\right)\left(\frac{k-m}{2k+x}\right)$$

And since $k\ne m$, we get:

$(k-n)(2k+x)=(m-n)(2m+x)$

Solve the above equation for $x$ we get $x=2(n-m-k)$.
 
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