SOLUTION: A rectangular garden is to be constructed, and it was decided that the length should be 10 more than the width. if the total area enclosed will be 180 ft^2, set up and solve an equ

Algebra ->  Polynomials-and-rational-expressions -> SOLUTION: A rectangular garden is to be constructed, and it was decided that the length should be 10 more than the width. if the total area enclosed will be 180 ft^2, set up and solve an equ      Log On


   

Question 319085: A rectangular garden is to be constructed, and it was decided that the length should be 10 more than the width. if the total area enclosed will be 180 ft^2, set up and solve an equation to find the dimensions of the garden.
Please help, this is like a foreign language to me!!!!!
Answer by Fombitz(32388) About Me  (Show Source):
You can put this solution on YOUR website!
First name the variables.
L-length of the rectangular garden
W-width of the rectangular garden
.
.
.
Now translate the words to equations,
"length should be 10 more than the width"
1.L=10%2BW
"total area enclosed will be 180 ft^2"
What's the area of a rectangle???
In terms of its length and width, the area is,
A=L%2AW
2.L%2AW=180
Now you have two independent equations with two unknowns, you can solve for each.
Substitute eq. 1 into eq. 2,
%2810%2BW%29W=180
10W%2BW%5E2=180
W%5E2%2B10W-180=0
Use the quadratic formula,
W=+%28-b+%2B-+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Aa%29+
W=+%28-10+%2B-+sqrt%28+10%5E2-4%2A1%2A%28-180%29+%29%29%2F%282%2A1%29+
W=+%28-10+%2B-+sqrt%28+100%2B720+%29%29%2F%282%29+
Only the positive result makes sense in this case.
W=+%28-10+%2B+sqrt%28820+%29%29%2F%282%29+
W=+%28-10+%2B+2%2Asqrt%28+205+%29%29%2F%282%29+
highlight%28W=+-5+%2B+sqrt%28+205+%29%29+ ft
Then from above,
L=10%2BW
highlight%28L=5%2Bsqrt%28205%29%29 ft