Problem 46 - Algebra 1: Concepts and Skills

Question

The formula \(d=\frac{n}{2}+26\) relates nozzle pressure \(n\) (in pounds per square inch) and the maximum distance the water reaches \(d\) (in feet \()\) for a fire hose with a certain size nozzle. Solve for \(n\) to find how much pressure is needed to reach a fire 50 feet away.

Step-by-Step Solution

Answer

The pressure needed to reach a fire 50 feet away is 48 pounds per square inch.

Step 1: Understand the Given Formula

The given formula is \(d = \frac{n}{2} + 26\), where \(n\) is the nozzle pressure and \(d\) is the maximum distance the water reaches. Our goal is to rearrange this formula to solve for \(n\) to find how much pressure is needed to reach a fire 50 feet away.

Step 2: Rearrange the Formula to Solve for \(n\)

Start by subtracting 26 from both sides of the equation to isolate the term with \(n\) on one side: \(d - 26 = \frac{n}{2}\). Now, to solve for \(n\), multiply both sides of the equation by 2: \(n = 2(d - 26)\).

Step 3: Substitute \(d\) with 50

We are asked to find the pressure needed to reach a fire 50 feet away. Substituting \(d\) with 50 in our rearranged formula will give us: \(n = 2(50 - 26)\).

Step 4: Compute for \(n\)

Perform the operations as indicated to obtain the value of \(n\): \(n = 2 \times 24 = 48\) pounds per square inch.

Key Concepts

Algebraic Expressions

In algebra, we often use algebraic expressions to represent relationships between different quantities. An algebraic expression is a combination of numbers, variables, and arithmetic operations. In the exercise, the expression \(d=\frac{n}{2}+26\) is a perfect example. This expression includes the division of the variable \(n\) by 2, and the addition of a constant, 26.

Understanding how to read and manipulate these expressions is crucial. For instance, \(\frac{n}{2}\) tells us that the nozzle pressure \(n\) is halved, representing the proportional relationship between pressure and distance in this scenario. The constant 26 can be interpreted as the base distance that the water reaches without any pressure. Algebraic expressions allow us to create a formula that we can use to calculate one variable in terms of another. It's a fundamental aspect of solving real-world problems using algebra.

Isolating Variables

When solving algebraic equations, isolating the variable means manipulating the equation in such a way that the variable we want to solve for stands alone on one side of the equals sign. It's like we're giving the variable its own personal space. This is what we aim for when we're rearranging equations.

In our example, we rearranged the formula to isolate \(n\) by subtracting 26 from both sides to push it away from the variable term. This left us with \(d - 26 = \frac{n}{2}\). After this, multiplying both sides by 2 removed the fraction, finally giving \(n\) its own space as \(n = 2(d - 26)\). Isolating the variable is an essential step in finding solutions to equations and understanding how each part of the equation affects the overall result.

Substitution Method

The substitution method is a powerful tool in algebra that allows us to find the value of unknown variables. Once we've isolated the variable, we can substitute specific numbers in place of other variables to calculate its value. This is particularly useful with formulas that involve two or more variables.

In our exercise, the distance \(d\) is 50 feet. By substituting 50 for \(d\) in the isolated variable equation \(n = 2(d - 26)\), we make the equation much more straightforward to solve. Substitution turns the abstract into the concrete, giving us \(n = 2(50 - 26)\). This demonstrates the substitution method's power to simplify equations and reveal the values we are after, which is essential whenever we face complex relationships in algebraic problems.