The length of a rectangle is 13 meters more than its width. The area of the rectangle is 48 square meters. Determine the length and width of the rectangle. (solve algebraically)

The length is 16 meters and the width is 3 meters.

Step-by-step explanation:

Since we know the length and w, we will represent the length by l and the width by w.

We know the length is 13 more than the width so we will represent it by the equation   l =  w + 13  .

The width will be represented by w.

We know the area of a rectangle is the length being multiplied by the width.

Since we know the area, multiply the length by the width and set it equal the area and solve for w.

w(w+ 13) = 48     Multiply on the left side

w^2 + 13w = 48       subtract 48 from both sides

         -48      -48

w^2 + 13w - 48 = 0  Now find two numbers that their product is -48 and their                                sum 13.

The numbers 16 and -3 works, because 16* -3 = -48  and  16 + (-3) = 13.

Rewrite the whole equation in this form.

w^2 + 16w -3w - 48 = 0  Now factor on the left side.

w(w + 16) -3(w+16) = 0      Factor out w+16

(w+16)( w-3) = 0   Now set each of the expressions to equal to 0 and solve for w.

w+ 16 = 0     or   w-3 = 0

  -16     -16              +3  +3

w = -16      or          w = 3

-16 can't be use to represent distance so w has to equal 3 and this means the width is 3.

The length is 13 meters more than the width so 3 + 13 will be the width.

l = 3 + 13

l = 16

add 5 subtract 1

step-by-step explanation:

answer: idk

step-by-step explanation: