Word problem - linear equation.

Click For Summary
SUMMARY

The discussion centers on solving a word problem involving a rectangular swimming pool and its surrounding cement walk. The pool's length is defined as twice its width, leading to the equations where the width (x) is 28.5 ft and the length is 57 ft. However, there is a discrepancy with the book's answer of 30 ft by 60 ft, which suggests an area of 784 ft² for the walk, contradicting the problem's stated area of 748 ft². The participants confirm that the calculations are correct based on the given conditions.

PREREQUISITES
  • Understanding of linear equations and algebraic manipulation
  • Knowledge of area calculations for rectangles
  • Familiarity with word problem interpretation in mathematics
  • Ability to work with single-variable equations
NEXT STEPS
  • Review algebraic techniques for solving linear equations
  • Practice solving word problems involving area and perimeter
  • Explore the implications of variable definitions in geometric contexts
  • Learn about common pitfalls in interpreting mathematical problems
USEFUL FOR

Students learning algebra, educators teaching word problems, and anyone interested in improving their problem-solving skills in mathematics.

paulmdrdo1
Messages
382
Reaction score
0
please help me with this problem

please use single variable only.

The length of a rectangular swimming pool is twice its width. The pool is surrounded by a cement walk 4ft wide. If the area of the walk is 748 ft^2, determine the dimensions of the pool.

let x = width of the pool
2x = length of the pool

the dimensions of the pool and cement walk combined

x+8 = width
2x+8 = length

the area of the pool plus the area of the cement walk is equal to the whole area.

$(2x+8)(x+8)=748+2x^2$

$x= 28.5$

the dimensions of the pool is 28.5 ft by 57 ft.

but the answer in my book is 30 ft. by 60 ft. why is that? where did I miss?

regards!

please use one variable only.
 
Mathematics news on Phys.org
The area of the walk (using your variables) is:

$$2(4x+4(2x+8))=748$$

$$4(6x+16)=748$$

$$6x+16=187$$

$$6x=171$$

$$x=28.5$$

I agree with you.
 
If the answer is 30 ft x 60 ft then the area of the walk would be 784, not 748. Did you read the question correctly?
 

Similar threads

  • · Replies 2 ·
Replies
2
Views
2K
Replies
1
Views
2K
  • · Replies 5 ·
Replies
5
Views
3K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 7 ·
Replies
7
Views
3K
  • · Replies 6 ·
Replies
6
Views
3K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 15 ·
Replies
15
Views
3K
Replies
1
Views
2K