Problem 14 - Algebra Form and Function

Question

In Exercises 14-17, assume the two quantities are directly proportional to each other. If \(r=36\) when \(s=4\), find \(r\) when \(s\) is 5.

Step-by-Step Solution

Answer

Answer: When s = 5, the value of r is 45.

Step 1: Write the proportionality equation

Since r and s are directly proportional, we can write the equation as: \(r = k \times s,\) where k is the constant of proportionality.

Step 2: Find the constant of proportionality (k)

To find the value of k, we are given that r = 36 when s = 4. Plug these values into the equation: \(36 = k \times 4,\) Now, solve for k by dividing both sides of the equation by 4: \(k = \frac{36}{4} = 9,\)

Step 3: Find the value of r when s = 5

We now know the value of the constant of proportionality k. With k = 9, we can rewrite our proportionality equation as: \(r = 9 \times s,\) Now, we can find r when s = 5 by plugging s = 5 into the equation: \(r = 9 \times 5 = 45,\) So when s is 5, the value of r is 45.

Key Concepts

constant of proportionality

In many mathematical and real-world problems, two quantities can change in relation to each other in a consistent manner. Direct proportionality is one such relationship where if one quantity increases, the other one also increases at a constant rate. This rate is what we call the "constant of proportionality".

To express this mathematically, we use an equation like \( r = k \times s \), where \( k \) is the constant of proportionality. It acts as a multiplier that adjusts the size of one variable in direct response to changes in another. For example, if you double \( s \), \( r \) will also double, provided \( k \) remains constant.

• This means that for every unit increase in \( s \), the increase in \( r \) is \( k \) times bigger.
• Understanding and finding \( k \) is crucial to solving direct proportionality problems.
• In our exercise, once we find \( k = 9 \), it allows us to predict values of \( r \) for any given \( s \).

algebraic equations

Algebra is a branch of mathematics that deals with symbols and the rules for manipulating these symbols to solve equations. An equation is a mathematical statement that asserts the equality of two expressions. When dealing with direct proportions, algebraic equations help us create a relationship between the two variables of interest.

The equation for direct proportionality, \( r = k \times s \), is a simple yet powerful tool. Let's break it down:

• \( r \) and \( s \) are variables representing quantities that change.
• \( k \) is the constant of proportionality, representing the fixed rate of change between \( r \) and \( s \).

Such equations help not only in mathematics but also in various practical applications, like determining the speed of a vehicle (where distance is proportional to time) or converting currencies (where one currency might be a fixed multiple of another). By mastering algebraic equations, students are equipped to model and solve these kinds of real-world problems.

problem solving in algebra

Solving problems in algebra often involves following a series of logical steps to isolate variables and find solutions. The process applied in our exercise is a great illustration of effective problem-solving in algebra.

Firstly, identify what you are given and what you need to find. Then, express the problem in terms of an equation. This step involved writing \( r = k \times s \).

Next, use the known values to find any unknown constants. In our example, using \( r = 36 \) and \( s = 4 \), we calculated \( k = 9 \).

Once we know all constants, substitute the new value you want to investigate into the equation to find the unknown variable. By plugging \( s = 5 \) into \( r = 9 \times s \), we determined that \( r = 45 \).

• Break down the problem into manageable parts.
• Use logical reasoning to connect each step.
• Always double-check your final equations and solutions. Consistent practice will enhance your problem-solving skills in algebra.