Question

It’s a precalculus holiday equation. Need to simplify to find the “secret message” can anyone help???

Answer 1

Take the expression in chunks:

>>

$((X+A)(X-A)M+A^2M)E=((X^2-A^2)M+A^2M)E=X^2ME$

>>

$\dfrac{\frac{R+X}X+\frac X{R-X}}{\frac X{R-X}}=\dfrac{\frac{R+X}X}{\frac X{R-X}}+1$

$\dfrac{\frac{(R+X)(R-X)}{X(R-X)}}{\frac{X^2}{X(R-X)}}=\dfrac{(R+X)(R-X)}{X^2}=\dfrac{R^2-X^2}{X^2}=\dfrac{R^2}{X^2}-1$

$\implies\dfrac{\frac{R+X}X+\frac X{R-X}}{\frac X{R-X}}=\dfrac{R^2}{X^2}$

>>

Note that $(a+b)^2-(a-b)^2=(a^2+2ab+b^2)-(a^2-2ab+b^2)=4ab$. So

$\displaystyle\frac{4AS}3\sum_{\rm cyc}\frac1{X((E+Y)^2-(E-Y)^2)}=\frac{4AS}3\sum_{\rm cyc}\frac1{4XEY}=\frac{AS}3\sum_{\rm cyc}\frac1{XEY}$

where the cyclic sum notation means

$\displaystyle\sum_{\rm cyc}f(x,y,z)=f(x,y,z)+f(z,x,y)+f (y,z,x)$

In other words, we take the sum over all possible cycles of the sequence of arguments to the summand. The second square-bracketed chunk reduces to

$\displaystyle\frac{AS}3\sum_{\rm cyc}\frac1{XEY}=\frac{AS}3\left(\dfrac1{XEY}+\dfrac1{EYX}+\dfrac1{YXE}\right)=\dfrac{AS}3\dfrac3{XEY}=\dfrac{AS}{XEY}$

So to recap, we've reduced the starting expression to

$X^2ME\left(\dfrac{R^2}{X^2}-\dfrac{AS}{XEY}\right)Y$

>>

$\dfrac{R^2}{X^2}-\dfrac{AS}{XEY}=\dfrac{R^2EY-ASX}{X^2EY}$

Finally, we have

$X^2ME\dfrac{R^2EY-ASX}{X^2EY}Y=M(R^2EY-ASX)$

Then distributing $M$ and rewriting to decode the message, we have

$MERRY-XMAS$