For what value of c is the relation a function?

StartSet (2, 8), (12, 3), (c, 4), (negative 1, 8), (0, 3) EndSet

{ c  ∣  c ≠  2 , 12, -1, 0, c ∈ R }

Step-by-step explanation:

Considering the set

As we know that

So, the value of c must not be equal to 2, 12, -1, 0 i.e. c ≠ 2, 12, -1, 0

Therefore,

                   { c  ∣  c ≠  2 , 12, -1, 0, c ∈ R }

(0,-3) see attached graph

step-by-step explanation:

a. to solve the system, graph the two equations.

x = y+3 becomes y=x-3 has m=1/1 and b=-3. start at (0,-3) on the y-axis and plot a point. then move up 1 unit and 1 unit to the right. plot a point and connect.

y=-4x+3 has m =-4/1 and b=-3. start at (0,-3) on the y-axis and plot a point. then move down four units and to the right 1 unit. plot a point and connect.

the solution is the point (x,y) where the two lines intersect. by the graph below it is (0,-3).

b. to solve by substitution, substitute y = -4x-3 into x= y+3

x= (-4x-3) +3

x=-4x

5x=0

x=0

now substitute x = 0 back in to find y.

y= -4(0)-3

y=0-3

y=-3

solution is (0,-3)

c. use elimination by adding the equations together.

x-y=3

4x+y=-3

5x=0

x=0

then substitute into one equation to find y.

y=-4(0)-3

y=-3