How would you solve x for 2bx + 3ax = b^2 - a^2?

Question

anonymous · Answer

OOOH OUCH...I would do this:
$$\eqalign{
&2bx+3ax=b^2-a^2 \
&(2b+3a)x=b^2-a^2 \
&x=\frac{b^2-a^2}{2b+3a}

}$$

anonymous · Answer

Yep got it thanks @KeithAfasCalcLover

anonymous · Answer

Anytime @AnImEfReaK !

anonymous · Answer

I don't believe you can actually solve for it for real. Its actually a literal equation which happens to take the form of what is called an infinite elliptic cone. However, that's not important here. The only thing I really know to do here is move everything to one side first.
So,
a^2-b^2+2bx+3ax=0
Then, we can sort of factor by grouping. It will look like this.
(a^2-b^2)+(2bx+3ax)=0
(a+b)(a-b)+x(2b+3a)=0
Now, move (a+b)(a-b) to the right hand side
x(2b+3a)=-(a+b)(a-b)
Divide by (2b+3a).
x=-(a+b)(a-b)/(2b+3a)

anonymous · Answer

Basically the same explanation as KeithAfasCalcLover only looking at it from a longer route

anonymous · Answer

Yeah expansion.

anonymous · Answer

But for the simplest form, you would always multiply out the numerator before calling it finished.