The length of a rectangle is 5 m less than twice the width, and the area of the rectangle is # 33 m^2#. What are the dimensions of the rectangle?

Let #x=# the width of the rectangle.

Then the length is:

#(2x-5)m#

Area is length x width:

#(2x-5)*x=33=> 2x^2-5x-33=0#

Factor:

#(2x-11)(x+3)=0=>x=-3 and x=11/2# ( negative root not applicable )

Dimensions of rectangle:

Length#= 2(11/2)-5=color(blue)(6)#

Width #= color(blue)(11/2)#

We can start by listing what we know and what we don't know.

• #l = 2w - 5#
• #a = 33#
• #w = ?#
• #a = lw#

We know that if A = B and A = C, then B must equal C. We can use this basic logical principle to write ourselves an equation that we can use to find the solution.

if #a = 33 and a = lw# then #lw = 33#
#lw = 33#

We can then use the same principle to simplify this equation so that it uses only one variable. If #l = 2w - 5#, then #lw = (2w-5)w#, so we can say that
#(2w-5)w = 33#.

After that, you need to simplify, factor, and evaluate:

#2w^2 - 5w = 33#
#2w^2 - 5w - 33 = 0#
#2w^2 - 5w - 33 = (2w-11)(w+3)#
#(2w-11)(w+3) = 0#

#((2w-11)(w+3) = 0)/(w+3)# or #((2w-11)(w+3) = 0)/(2w-11)#

#2w - 11 = 0# or #w+3 = 0#
#2w = 11# or #w = -3#
#w = 5.5# or #w = -3#

It is impossible for a rectangle to have a width of -3 meters, so we can eliminate that option. Now we know that our width is 5.5 meters.

Now we can plug the width into our length equation to find the length.

#l = 2(5.5) - 5#
#l = 6#

To check our answers, we can plug our length and width back into our #lw = 33# equation to make sure that #lw# actually equals 33.

#6*5.5 = 33#
#33 = 33#