A rectangular pool is 20 feet wide and 50 feet long. A deck used for sunning surrounds the pool. The deck is the same width all the way around the pool. The total area of the deck is 456 square feet. How wide is the deck around the pool?

  3 feet

Step-by-step explanation:

Let x represent the width of the deck. The deck adds to the pool dimensions on both sides, so the area of the site including pool and deck will be ...

  A = (20 +2x)(50 +2x) = 1000 +140x +4x^2

The area of the pool alone is

  A = (20)(50) = 1000

So, the area of the deck is the difference:

  456 = (4x^2 +140x +1000) -(1000)

Dividing by 4 and subtracting 114, we have ...

  x^2 +35x -114 = 0

  (x -3)(x +38) = 0

Solutions to this are x = 3 and x = -38. The positive solution is the one of interest.

The deck is 3 feet wide.

(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

imma say 16%. sorry if its wrong. i'm doing the same stuff right now

step-by-step explanation: