Translating to algebraic expressions

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In summary, the conversation discusses how to calculate the amount of carpet needed for a room with dimensions of $p$ ft. by $x$ yards, using a carpet that is 2 ft. wide. The formula for finding the total area in square yards is $\frac{px}{3}$ and the formula for the area of the carpet in square yards is $\frac{2\ell}{3}$. By equating these two formulas, the minimum amount of carpet needed can be found, which is $\frac{px}{2}$.
  • #1
NotaMathPerson
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I am not able to express it symbolically I need your assistance. Thanks!

A room is $p$ ft. Long and $x$ yards in width; how many yards of carpet two ft. Wide will be required for the floor?
 
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  • #2
One yard is 3 feet, so each yard of carpet covers 6 square feet. Your total area is $\dfrac{px}{3}$ square feet.

So if $y$ is the number of yards of carpet, you need to solve for $y$ in:

$6y = \dfrac{px}{3}$, to get the minimum needed (it could be more if neither $p$ nor $\dfrac{x}{3}$ is evenly divisible by $2$ or $3$).
 
  • #3
Deveno said:
One yard is 3 feet, so each yard of carpet covers 6 square feet. Your total area is $\dfrac{px}{3}$ square feet.

So if $y$ is the number of yards of carpet, you need to solve for $y$ in:

$6y = \dfrac{px}{3}$, to get the minimum needed (it could be more if neither $p$ nor $\dfrac{x}{3}$ is evenly divisible by $2$ or $3$).

The answer in my book is
$\frac{px}{2}$

Why is that?
 
  • #4
The area $A_R$ of the room in square yards is:

\(\displaystyle A_R=\frac{px}{3}\)

For a length $\ell$ of carpet in yards, its area $A_C$ in square yards is:

\(\displaystyle A_C=\frac{2\ell}{3}\)

Equating the two areas:

\(\displaystyle \frac{px}{3}=\frac{2\ell}{3}\)

Solve for $\ell$:

\(\displaystyle \ell=\frac{px}{2}\)
 

Related to Translating to algebraic expressions

What is an algebraic expression?

An algebraic expression is a mathematical statement that contains variables, numbers, and operations such as addition, subtraction, multiplication, and division. It can also include exponents and parentheses.

How do you translate a word problem into an algebraic expression?

To translate a word problem into an algebraic expression, first identify the given information and what is being asked for. Then, use variables to represent the unknown quantities and write out the mathematical operations in the order they appear in the problem. Finally, simplify the expression if possible.

What are some common keywords used in word problems that indicate the need for an algebraic expression?

Some common keywords that indicate the need for an algebraic expression include "more than", "less than", "increased by", "decreased by", "multiplied by", "divided by", "total", "difference", and "product".

What is the difference between an algebraic expression and an equation?

An algebraic expression is a mathematical statement that contains variables, numbers, and operations, but does not have an equal sign. An equation, on the other hand, is a statement that shows the equality of two expressions, with an equal sign between them.

How can translating to algebraic expressions be useful?

Translating word problems into algebraic expressions can help us solve real-life problems and make sense of complex situations. It also allows us to perform calculations and find solutions more efficiently and accurately.

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