What is the solution to the system of equations below?
2 answers:
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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 :
<span>96 = l x w </span>
we make the second equation of the following statements :
<em>The length of a rectangle is 2 foot less than 3 times its width.</em>
<span>so it becomes :
l = 3w - 2
</span>
<span>To solve, we can use the substitution method.
</span>
![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](https://tex.z-dn.net/?f=96%20%3D%20l%20%5Ctimes%20w%5C%5C96%3D%283w-2%29%20%5Ctimes%20w%5C%5C96%3D3w%5E%7B2%7D-2w%5C%5C3w%5E%7B2%7D-2w-96%3D0~~~~~~%5Bnow%5C%20factor%5C%20the%5C%20equation%5D%5C%5C%20%283w%2B16%29%28w-6%29%3D0)
<span><em>So if our width is 6, Now substitute the value of w = 6 into equation 2 </em></span>
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 :
<span>96 = l x w </span>
we make the second equation of the following statements :
<em>The length of a rectangle is 2 foot less than 3 times its width.</em>
<span>so it becomes :
l = 3w - 2
</span>
<span>To solve, we can use the substitution method.
</span>
<span><em>So if our width is 6, Now substitute the value of w = 6 into equation 2 </em></span>
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.