The tables represent two linear functions in a system. what is the solution to this system? a. (-13/3, -25) b. (-14/3, -54) c. (–13, –50) d. (–14, –54)

first function                                        2nd function

slope =(y2-y1)/(x2-x1)                          slope =(y2-y1)/(x2-x1)

  =(14-2)/(3-0)                                        m = (-3--12/(3-0)

12/3                                                               m=(-3+12)/3

4                                                                   m =9/3 =3

y = 4x+2                                                    y = 3x+-12

set these two equations equal

4x+2 = 3x-12

subtract 3x

x+2 = -12

subtract 2 from each side

x = -14

y = 3x-12

y =3*(-14)-12

y = -42-12

y = -54

ChoiceD

answer:

the position of equilibrium will shift to the left.

step-by-step explanation:

the general equation for the dissociation of a weak base is

b + h₂o   ⇌   bh⁺ + oh⁻

according to le châtelier's principle, when a stress is applied to a system at equilibrium, the system will respond in a way that tends to relieve the stress.

if you apply a stress by adding more bh⁺ or oh⁻,   the system will respond by moving in a direction that gets rid of the added ions.

the position of equilibrium will shift to the left, forming more of the undissociated base.