Linear algebra deals with mathematical transformations that are linear. By definition they must preserve scalar multiplication and additivity.

T(u+v)= T(u) + T(v)

T(R*u)=r*T(u) Where "r" is a scalar

For example. T(x)=m*x where m is a scalar is a linear transform. Because

T(u+v)=m(u+v) = mu + mv = T(u) + T(v)

T(r*u)=m(r*u)=r*mu=r*T(u)

A consequence of this is that the transformation must pass through the origin.

T(x)=mx+b is not linear because it doesn't pass through the origin. Notice at x=0, the transformation is equal to "b", when it should be 0 in order to pass through the origin. This can also be seen by studying the additivity of the transformation.

T(u+v)=m(u+v)+b = mu + mv +b which cannot be rearranged as T(u) + T(v) since we are missing a "b". If it was mu + mv + b + b it would work because it could be written as (mu+b) + (mv+b) which is T(u)+T(v). But it's not, so we are out of luck.

Yes.

a/b / u/v / r/s = (a*v)/(b*u) / r/s = (a * v * s) / (b * u * r)

If:

v = u+at

Then:

-u = -v+at or u = v-at (by dividing all terms by -1)

a = (v-u)/t

t = (v-u)/a

Suppose you have the fractions p/q and r/s. Let the LCM of q and s be t.Then t is a multiple of q as well as of s

so let t= q*u and t = s*v

Then p/q = (p*u)/(q*u) = (p*u)/t

and r/s = (r*v)/(s*v) = (r*v)/t have the same denominators.

GRAB
GARB
GUV
GRUB
VUG
BURG
BRAG
GAB
BAG
BUG
BURA
BURR
VAU
VAR
ARB
GAUR
GUAR
URB
BAR
BUR
RUGA
RUB
BRA
BRR
AB
BA
RAG
RUG
GAR
AG
AR