Answer:
The length is 16.1 ft and the width is 4.7 ft
Step-by-step explanation:
Let
x -----> the total length of the tables
y -----> the width of the tables
we know that
The area is equal to
[tex]A=xy[/tex]
[tex]A=75\ ft^{2}[/tex]
so
[tex]75=xy[/tex] -----> equation A
[tex]x=3y+2[/tex] -----> equation B
substitute equation B in equation A
[tex]75=(3y+2)y[/tex]
[tex]3y^{2} +2y-75=0[/tex]
Solve the quadratic equation by graphing
The solution is [tex]y=4.7\ ft[/tex]
Find the value of x
[tex]x=3(4.7)+2=16.1\ ft[/tex]
therefore
The length is 16.1 ft and the width is 4.7 ft
Answer:
The length and width of the serving table is 16.1 ft and 4.7 ft respectively.
Step-by-step explanation:
Consider the provided information.
Let the width of the table is x and length of the table is y.
The total length of the tables will be two more than three times the width.
This can be written as:
y = 2+3x
The area of the table is 75 ²ft
The area of rectangle is:
length × width = Area
Substitute width = x and length = 2+3x in above formula.
(x)(2+3x) = 75
2x+3x²-75 = 0
3x²+2x-75 = 0
The above equation is in the form of ax²+bx+c=0. Now use the quadratic formula to find the root of the equation.
[tex]x_{1,2}=\frac{-b\pm\sqrt{b^2-4ac}}{2a}[/tex]
Substitute a=3, b=2 and c=-75 in above formula.
[tex]x_{1,2}=\frac{-2\pm\sqrt{2^2-4(3)(-75)}}{2(3)}[/tex]
[tex]x_{1,2}=\frac{-2\pm\sqrt{904}}{6}[/tex]
[tex]x_{1,2}=\frac{-2\pm30.07}{6}[/tex]
[tex]x_{1}=\frac{-2+30.07}{6}[/tex]
Ignore the negative value of x as width should be a positive number.
[tex]x=4.7\ ft[/tex]
Now substitute the value of x in y = 2+3x.
y = 2+3(4.7)
y = 16.1 ft
Hence, the length and width of the serving table is 16.1 ft and 4.7 ft respectively.
