Start with

We observe that both fractions are not defined if
. So, we will assume
.
We multiply both numerator and denominator of the first fraction by 3 and we sum the two fractions:

We multiply both sides by
:

We move everything to one side and solve the quadratic equation:

We check the solution:

which is true
I'm thinking this is what the problem looks like:

. The first thing to do is to move the

over to the other side because it has a common denominator with the other side. Doing that and at the same time combining them over their common denominator looks like this:

. The best way to solve for x now is to cross-multiply to get 3(4-x)=-4(x-4). Distributing through the parenthesis is 12 - 3x = -4x + 16. Solving for x gives us x = 4. Of course when we sub a 4 back in for x we get real problems, don't we? Dividing by zero breaks every rule in math that there ever was! So, yes, the solution is extraneous.
I think the answer would be C but I am not certain.
Answer:
OGH and TCA
Step-by-step explanation:
that's quite easy...
the congruency criterion is ASA or Angle side Angle.
i.e. a side lying between two Angles.
so we have
GH = FN= CA
now, G=I= C
and H=N=A
it's pretty clear that G-GH-H = C-CA-A
(just to explain)
hence, the answer