Question

Table \(7.3\) shows the values of \(x\) and \(y\). Given that \(y\) is proportional to \(x\) (a) find an equation connecting \(y\) and \(x\) (b) calculate the value of \(y\) when \(x=36\) (c) calculate the value of \(x\) when \(y=200\) $$ \begin{array}{lclllc} \hline x & 5 & 10 & 15 & 20 & 25 \\ y & 22.5 & 45 & 67.5 & 90 & 112.5 \\ \hline \end{array} $$

Short answer

Answer: The values of y when x = 36 is 162, and the value of x when y = 200 is approximately 44.44.

Step-by-step solution

1. Understanding Proportionality

We are given that, \(y\) is proportional to \(x\). This means \(y=kx\) where \(k\) is the constant of proportionality.

2. Find the constant of proportionality (\(k\)) using the given values of \(x\) and \(y\)

Utilize the given values of \(x\) and \(y\) from the table to find the value of \(k\). For example, use the data from the first row, where \(x = 5\) and \(y = 22.5\). Plug these values into the equation \(y = kx\): \(22.5 = k(5)\) Now divide both sides of the equation by 5 to find out what \(k\) is. \(k = \dfrac{22.5}{5} = 4.5\)

3. Write the equation connecting \(y\) and \(x\)

Now that we've found the value of \(k\), we can write the equation connecting \(y\) and \(x\): \(y = 4.5x\)

4. Calculate the value of \(y\) when \(x = 36\)

To calculate the value of \(y\) when \(x=36\), substitute \(x=36\) in the equation we found in step 3: \(y = 4.5(36) = 162\) So, when \(x = 36\), the value of \(y\) is \(162\).

5. Calculate the value of \(x\) when \(y=200\)

Next, to calculate the value of \(x\) when \(y=200\), substitute \(y=200\) in the equation we found in step 3: \(200 = 4.5x\) Now, divide both sides of the equation by \(4.5\) to find the value of \(x\): \(x = \dfrac{200}{4.5} \approx 44.44\) So, when \(y = 200\), the value of \(x\) is approximately \(44.44\).

Key concepts

Constants of Proportionality

In mathematics, when we say one quantity is directly proportional to another, it means that as one quantity increases, the other changes at a constant rate. This constant rate is known as the constant of proportionality. For example, if we take the equation \( y = kx \), \( k \) is the constant of proportionality. This term establishes a direct relationship between the two quantities, \( x \) and \( y \).
In practical scenarios, finding \( k \) helps us understand how much of one quantity is equivalent to another. In our exercise, we used the pair \( (5, 22.5) \), representing \( x = 5 \) and \( y = 22.5 \). By setting up our equation from \( y = kx \) or \( 22.5 = k(5) \), we calculated \( k \) as \( 4.5 \) by dividing \( 22.5 \) by \( 5 \).
This consistent relationship simplifies problem-solving tremendously, allowing predictions and calculations even as variables change.

Linear Equations

Linear equations represent a straight-line relationship between two variables. In our equation \( y = 4.5x \), both variables \( y \) and \( x \) are linked linearly, where changes in \( x \) lead to proportional changes in \( y \). The equation follows the form \( y = mx + b \), where \( b = 0 \) because the line passes through the origin, and \( m \) is the slope, equivalent to the constant of proportionality \( k \).
Linear equations like this one allow us to calculate easily how one value affects another. For instance, given any \( x \), we can find \( y \) by multiplying \( x \) by \( 4.5 \). Conversely, if \( y \) is provided, we can compute \( x \) by rearranging to \( x = \frac{y}{4.5} \). These calculations are straightforward due to the predictable changes that result from direct proportions in linear equations.

Mathematical Problem Solving

Solving mathematical problems often involves recognizing and applying patterns like direct proportionality. In the exercise given, the main problem-solving process starts with understanding what direct proportionality means for \( x \) and \( y \). Once this concept is clear, we locate the constant of proportionality \( k \), then form and utilize the related linear equation.
Problem-solving with linear equations involves substituting known values to find unknowns. For instance, when \( x = 36 \), we determined \( y \) by inserting 36 into our equation. Similarly, to find \( x \) when \( y = 200 \), we rearranged the equation and solved for \( x \).
This structured approach helps break down complex-sounding problems into manageable steps, using calculations that stem from understanding the basic relationships between numbers. Such a skill is valuable across all mathematics and real-world applications.