Answer :
for example,area = a, length = l, width = w
First we write two equations. The easier equation is the area equation, which we know to be
a = l x w
So that the first equation is :
96 = l x w
we make the second equation of the following statements :
The length of a rectangle is 2 foot less than 3 times its width.
so it becomes :
l = 3w - 2
To solve, we can use the substitution method.
[tex]96 = l \times w\\96=(3w-2) \times w\\96=3w^{2}-2w\\3w^{2}-2w-96=0~~~~~~[now\ factor\ the\ equation]\\ (3w+16)(w-6)=0[/tex]
[tex]3w+16=0\\3w=-16\\w= -\frac{16}{3}~~~~\Rightarrow~~~answer\ doesnt\ make\ sense\ , negative\ width. [/tex]
[tex]w-6=0\\\boxed{w=6~~~~~~~\Rightarrow~~~~positive\ answer\ for\ width\ makes\ sense } [/tex]
So if our width is 6, Now substitute the value of w = 6 into equation 2
[tex]l=3w-2\\l=3(6)-2\\l=18-2\\l=16[/tex]
So "w" = 6 and "l" = 16, and if we multiply them together, we get the correct area, 96. So our dimensions are 6 by 16.
First we write two equations. The easier equation is the area equation, which we know to be
a = l x w
So that the first equation is :
96 = l x w
we make the second equation of the following statements :
The length of a rectangle is 2 foot less than 3 times its width.
so it becomes :
l = 3w - 2
To solve, we can use the substitution method.
[tex]96 = l \times w\\96=(3w-2) \times w\\96=3w^{2}-2w\\3w^{2}-2w-96=0~~~~~~[now\ factor\ the\ equation]\\ (3w+16)(w-6)=0[/tex]
[tex]3w+16=0\\3w=-16\\w= -\frac{16}{3}~~~~\Rightarrow~~~answer\ doesnt\ make\ sense\ , negative\ width. [/tex]
[tex]w-6=0\\\boxed{w=6~~~~~~~\Rightarrow~~~~positive\ answer\ for\ width\ makes\ sense } [/tex]
So if our width is 6, Now substitute the value of w = 6 into equation 2
[tex]l=3w-2\\l=3(6)-2\\l=18-2\\l=16[/tex]
So "w" = 6 and "l" = 16, and if we multiply them together, we get the correct area, 96. So our dimensions are 6 by 16.
