Part a: if (6^2)x = 1, what is the value of x? explain your answer. part b: if (6^0)x = 1, what are the possible values of x? explain your answer.

In part A you have 6^2 = 36, then you have to find out what times 36 will get you to 1, you realize that 36*1/36 = 1. X=1/36

In part B: if 6^0 = 1, and 1x=1, you will realize that 1*1*1 =1. X=1.

Thanks have a good day/night <3

(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