## What is Direct Variation?

Direct variation expresses a relationship between two variables. In direct variation situations, the increase in the independent variable leads to an increase in the dependent variable.

**Examples:**

- A driver travels at a rate of 60 miles per hour. As the number of hours increases, the distance covered will also increase.
- A student earns a wage of $7.50 per hour making sandwiches at a local sandwich store. The longer his working hours, the larger his wage will be.

Direct variation could represent a simple linear situation or more complex situations dealing with polynomial examples.

**Examples:**

- \(y\) varies directly as \(x\)
- \(y\) varies directly as \(h^{2}\)

If a direct variation, there is no intercept on the independent or dependent axes.

Direct variation has a constant of variation, \(k\), also called a constant of proportionality that helps establish the consistent relationship between the variables in question. The constant of proportionality is helpful in calculating the value of the dependent variable when the value of the independent variable is changed

## Solving Direct Variation Problems

To solve direct variation problems, follow these steps:

- Express the relationship between the variables in the form \(y = kx\)
- Find the constant of proportionality, \(k\)
- Rewrite the equation representing the relationship with the actual value of \(k\)
- Use the new equation to solve for any missing values.

## Direct Variation Sample Problems

**Example 1:** **If \(x\) varies directly as \(y\) and \(x = 12\) when \(y = 18\), what is the value of \(y\) when \(x = 36\)?**

**Solution:**

**First, express the relationship in the form \(y = kx\):**

\(18 = k \times 12\)

**Solve for \(k\):**

Divide both sides by \(12,\) the coefficient of \(p\):

\(\frac{18}{12} = \frac{k \times 12}{12}\)

\(\frac{18}{12} = k\)

**Simplify the fraction:**

\(\frac{18 \div 6}{12 \div 6} = \frac{3}{2} = k\)

So, \(k = \frac{3}{2}\)

**The equation that expresses the relationship between \(x\) and \(y\) is:**

\(y = \frac{3}{2}x\)

**Using the equation \(y = \frac{3}{2}x\), if \(x = 36\),**

\(y= \frac{3}{2} \times 36 = \frac{3 \times 36}{2} = \frac{108}{2} = 54\)

**So, if \(x = 36, y = 54\)**

**Example 2:** **If \(y\) varies directly as \(x\) squared and \(y = 54\) when \(x = 3\), what is the value of \(y\) when \(x = 7\)?**

**Solution:**

**First, express the relationship in the form \(y = kx^{2}\) (from the question:**

\(54 = k \times 3^{2}\)

\(54 = k \times 9\)

**Solve for \(k\):**

Divide both sides by \(9,\) the coefficient of \(p\):

\(\frac{54}{9} = \frac{k \times 9}{9}\)

\(\frac{54}{9} = k\)

**Simplify the fraction:**

\(\frac{54 \div 9}{9 \div 9} = 6 = k\)

So, \(k = 6\)

**The equation that expresses the relationship between \(x\) and \(y\) is:**

\(y = 6x^{2}\)

**Using the equation \(y = 6x^{2}\), if \(x = 7\),**

\(y= 6 \times 7^{2} = 6 \times 49 = 294\)

**So, if \(x = 7, y = 294\)**

**Example 3: The cost of a bag of grains is directly proportional to the weight of the bag. If a customer spends $22.50 on grains purchased at a rate of $1.50 per pound, how many pounds of the grain did the customer purchase?**

Solution:

**Identify the variables**

Cost is directly proportional to weight

**Express the relationship in the form \(y = kx\)**

\(c = kw\)

**Substitute with values from the question**

\(22.50 = k \times 1.50\)

**Find the value of k**

\(\frac{22.50}{1.50} = \frac{k \times 1.50}{1.50}\)

\(15 = k\)

The customer purchased \(15\) pounds of grain.