The area of a rectangle is 42 yd^2 , and the length of the rectangle is 11 yd less than three times the width, how do you find the dimensions length and width?

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Answer 1 · 2016-02-23T18:07:35

The dimensions are as follows:
Width#(x) = 6# yards
Lenght #(3x -11) =7# yards

Area of rectangle #=42# square yards.

Let the width #=x# yards.

The length is 11 yards less than thrice the width:

Length #= 3x -11# yards.

Area of rectangle #=# length #xx# width

#42 = (3x-11) xx (x)#

#42 = 3x^2 - 11x#

# 3x^2 - 11x- 42= 0#

We can Split the Middle Term of this expression to factorise it and thereby find the solutions.

# 3x^2 - 11x- 42=3x^2 - 18x +7x- 42 #

# = 3x(x-6) + 7(x-6)#

# (3x-7)(x-6) # are the factors, which we equate to zero in order to obtain #x#

Solution 1:
#3x- 7 =0, x = 7/3# yards (width).
Length #= 3x -11 = 3 xx (7/3 )-11 = -4 # yards, this scenario is not applicable.

Solution 2:

#x-6 =0, x=6# yards (width) .

Length #= 3x -11 = 3 xx 6-11 = 7 # yards (length).