I have to find with what value ''m'' those two equations are equivalent?
+ x + (6-5m)x = 0
and +2(m-3)x + -7m +12 = 0
I found that with m=3 ... I did this: + x + (6-5m)x = +2(m-3)x + -7m +12
Not sure if this is the right way?
m=2 doesn't work.
as the multipliers of "x" must be equal.
which is the exact same as the two constants being equal
(the constant in one equation is zero).
Then to find the 2 solutions for "m",
find the factors of 12 that sum to -7
in order to factor the quadratic in "m".