Problem 92 - Introductory and Intermediate Algebra for College Students 4th

Question

The formula $$p=15+\frac{5 d}{11}$$ describes the pressure of sea water, \(p,\) in pounds per square foot, at a depth of d feet below the surface. Use the formula to solve. At what depth is the pressure 20 pounds per square foot?

Step-by-Step Solution

Answer

The depth at which the pressure is 20 pounds per square foot is 11 feet.

Step 1: Substitution

Given that the pressure (\(p\)) equals 20 pounds per square foot, we substitute the value of \(p\) into the equation to get: \(20 = 15 + \frac{5d}{11}\)

Step 2: Solving for d

We can solve for \(d\) by first moving 15 from the right-hand side of the equation to the left-hand side, making it negative, to give: \(20 - 15 = \frac{5d}{11}\). This simplifies to: \(5 = \frac{5d}{11}\). Now, if we cross multiply the equation, we have: \(5*11 = 5d\). Dividing both sides by 5 allows us to find the value of \(d\).

Step 3: Find the depth

When you divide both sides by 5, you get: \(\frac{5*11}{5}=d\), which simplifies to \(d=11\).

Key Concepts

Algebraic Manipulation

Understanding algebraic manipulation is essential when working with equations in mathematics, especially in physics and engineering where formulas like the pressure of sea water are frequently used. Algebraic manipulation involves rearranging and simplifying expressions without changing their value. It’s like rearranging the pieces of a puzzle to get a clearer picture.

In the context of the given exercise, algebraic manipulation was necessary to isolate the variable 'd' (depth) and make the equation easier to solve. Starting with the initial equation of the pressure of sea water, \( p=15+\frac{5d}{11} \), and knowing that the pressure \( p \) is 20 pounds per square foot, you substitute and rearrange terms to solve for 'd'. The crux of algebraic manipulation lies in performing operations that simplify the equation, such as subtracting 15 from both sides to isolate the fraction containing 'd'. This kind of manipulation is foundational for solving many problems in algebra.

Solving Equations

Solving equations is a fundamental aspect of mathematics, involving finding the value(s) of the variable(s) that make the equation true. It's a process of determination and often, a bit of detective work. Equations can range from simple linear equations to more complex polynomials and differential equations.

In our sea water pressure formula, solving the equation means finding at what depth ('d') the pressure is exactly 20 pounds per square foot. This process involves a systematic approach to isolate the variable in question. After algebraic manipulation, you end with a simpler equation, \( 5 = \frac{5d}{11} \). From here, you apply arithmetic operations, such as multiplying both sides by 11 and then dividing by 5, to solve for 'd'. Through these steps, the seemingly complex relationship between pressure and depth becomes tangible, arriving at the solution that \( d=11 \) feet.

Substitution Method

The substitution method is a technique used to solve systems of equations, as well as individual equations with a single variable, by replacing variables with their equivalent values. This method is particularly useful when an initial expression is too complex to solve directly and needs simplification.

In our example, the substitution method comes into play when we are given that the pressure is a known value, specifically 20 pounds per square foot, and we ‘substitute’ this value in place of the variable 'p' in our original formula. With this substitution, \( 20 = 15 + \frac{5d}{11} \), the equation becomes one in which the sole unknown is 'd'. Substituting known values simplifies the algebraic process, leading to more straightforward solutions. This method not only makes the problem more manageable but also highlights the importance of understanding the relationships within an equation.