“the length of the rectangle is 5 centimeters less than twice width, it’s area is 63 square centimeter. find the dimension
of the rectangle!
The width is
the length is”
---------------------------------------In[5]:=
$Line = 0;
From the problem statement:
In[1]:=
length == 2 * width - 5;
The area of the rectangle is:
In[2]:=
length * width Š 63;
Solve the simultaneous equations for the length and width respectively.
In[3]:=
Solve@ 8length == 2 width - 5, length * width Š 63<, 8length , width< D
9
Out[3]=
::length ® - 14, width ® -
>, 8length ® 9, width ® 7<>
2
We want the second set of values for the length and width. Plug those values into the simultaneous equations to check
the validity of the length and width answers. True means that the LHS is equal to the RHS, else, False is returned.
In[4]:=
Out[4]=
8length == 2 width - 5, length * width Š 63< . %7@@2DD
8True, True<