Respuesta :
Let
X-----------> is the length of rectangle
y----------> is the width of rectangle
we know that
x=y+3-----------------> equation 1
A=x*y--------> x*y=154----------> equation 2
substituting 1 in 2
(y+3)*y=154-----------> y²+3y-154=0
y²+3y-154=0
using a graph tool----------> to calculate the quadratic equation
see the attached figure
the solution is
y=11 m
x=y+3--------> x=11+3---------> x=14 m
the answer is
the width is 11 m
X-----------> is the length of rectangle
y----------> is the width of rectangle
we know that
x=y+3-----------------> equation 1
A=x*y--------> x*y=154----------> equation 2
substituting 1 in 2
(y+3)*y=154-----------> y²+3y-154=0
y²+3y-154=0
using a graph tool----------> to calculate the quadratic equation
see the attached figure
the solution is
y=11 m
x=y+3--------> x=11+3---------> x=14 m
the answer is
the width is 11 m

To get this answer you would use the formula A = lw to find the missing dimension.
You will substitite in what you know the area is and an expression for the length.
154 = w (3 + w)
154 = 3w + w^2
Create an equivalent equation equal to 0.
w ^2 + 3w -154 =0
(w - 11 ) (w + 14) = 0
w would be either 11 or -14 for this to be true.
Only 11 is a possible dimension, so the width is 11 meters.
You will substitite in what you know the area is and an expression for the length.
154 = w (3 + w)
154 = 3w + w^2
Create an equivalent equation equal to 0.
w ^2 + 3w -154 =0
(w - 11 ) (w + 14) = 0
w would be either 11 or -14 for this to be true.
Only 11 is a possible dimension, so the width is 11 meters.