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