express the perimeter as a single fraction in simplest form

Question

the length of a rectangle is (4x+8)/(x²-16) and its width is (x+3)/(x+4)
find the perimeter as a single fraction in simplest form!!

Answer 1

length=(4x+8)/(x²-16)

factor the terms of the length

4(x+2)/(x-4)(x+4)

width=(x+3)/(x+4)

P=2l+2w

P=2(4)(x+2)/(x-4)(x+4) + 2*(x+3)/(x+4)

we need a common denominator

P=2(4)(x+2)/(x-4)(x+4) + 2*(x+3)(x-4)/(x-4)(x+4)

P=(8x+16)/(x-4)(x+4) + (2x²-2x-24)/(x-4)(x+4)

P=[(8x+16)+(2x²-2x-24)]/(x-4)(x+4)

P=(8x+16+2x²-2x-24)/(x-4)(x+4)

P=2x²+6x-8)/(x-4)(x+4)

P=2(x²+3x-4)/(x-4)(x+4)

P=2(x+4)(x-1)/(x-4)(x+4)

cancel the (x+4)'s

P=2(x-1)/(x-4)

Answer 2

Hi Farheen,

 

Perimeter of a rectangle = 2( length+ Width)

    

                                    = 2 [4x+8 /(x+4)(x-4)  + (x+3) / (x+4)]

 

                                    = 2[ 4x + 8 +(x+3)(x-4)  / (x+4)(x-4)

 

                                     = 2[ 4x + 8 +x² -4x+3x-12 /(x+4)(x-4)

 

                                     = 2[x² +3x-4 /(x+4)(x-4)]

 

                                     =2 [(x+4)(x-1) /(x+4)(x-4)]

 

                                     =2 (x-1) /(x-4)

 

                                      = 2x -2 /x-4