A rectangular shelf has a perimeter of 30 inches and an area of 50 square inches. What are the dimensions of the shelf

Answer 1

Perimeter=2(x&y) =30 where x=length, y=width

Area = xy = 50

two equations 2 unknowns, the combination for a unique solution.

2x + 2y = 30

x+y=15

x=15-y

substitute

xy=50

(15-y)y=50

15y-y²=50

y²-15y+50=0

use factoring, completing the square or quadratic formula

try factoring

(y-5)(y-10)=0

y=5 or 10, like the old five & dime stores

so x = 5 or 10.

Answer (x,y) = (5.10) or (10,5)

Answer 2

P=2l+2w

30=2l+2w

15=l+w

l=15-w

A=lw

50=l*w

this is basically as far as you need to go but your teacher would want you to put down all of the work

50=lw

50=(15-w)(w)

50=15w-w^2

w^2-15w+50=0

the question now is what two numbers multiply to make 50 and add to make 15 ?

this is exactly the question when you got the equations 15=l+w and 50=lw

w^2-15w+50=0

50=2*5*5=(2*5)*(5)=10*5

(w-10)(w-5)=0

w-10=0

w=10

w-5=0

w=5

the length is 10 inches and the width is 5 inches (you could have it the other way around also but the shelf would be sticking out which makes no sense)

Answer 3

sides are x and y

perimeter = 2x + 2y

area = x*y

solve the pair of equations to get x and y

Answer 4

We can translate this into an algebraic system of equations to solve

If we call 2 of the sides x and 2 of the sides y, we can create 2 equations based on the information given for perimeter and area:

30= 2x+2y

and

50=xy

As usual, to solve this system, rewrite one of the equations to equal x (or to equal y),

then substitute what x equals into the other equation everywhere x appears

Once you have solved for y, you can go back to the rewritten equation to solve for x

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