91Ó°ÊÓ

In the context of inverse variation, the variation constant (often denoted as k) is a crucial value. It links the two variables that vary inversely and helps us understand their relationship. When we know that one variable decreases as the other increases, the constant gives us a fixed value that quantifies this relationship.

For example, in the problem provided, we are given the situation where y varies inversely as x, and y equals 5 when x is 20. By substituting these values into the equation of inverse variation, we find the variation constant. In this case, the equation is:

• Step 1: Write the inverse variation equation: \( y = \frac{k}{x} \)


• Step 2: Substitute y = 5 and x = 20 into the equation: \( 5 = \frac{k}{20} \)


• Step 3: Solve for k by multiplying both sides by 20: \( 5 \times 20 = k \implies k = 100 \)

Here we find that k = 100 and this forms the core constant linking the inverse relationship between y and x.

An inverse relationship between two variables means that as one variable increases, the other decreases. This is the opposite of a direct relationship, where both variables move in the same direction.

In the example we're dealing with, the inverse relationship can be expressed through the equation: \( y = \frac{k}{x} \). Here, as x increases, y decreases, and vice versa. If x gets larger, y gets smaller because we are dividing a constant value (k) by a larger and larger number. Let's re-examine the initial conditions:

When x = 20, y = 5.

To find the new y value if x changes, we use the inverse relationship formula with our variation constant k = 100:

• Suppose x becomes 10: \( y = \frac{100}{10} \implies y = 10 \)


• Suppose x becomes 40: \( y = \frac{100}{40} \implies y = 2.5 \)

This shows how y varies directly in an inverse manner against x. Understanding these changes helps grasp the full width of how inverse relationships behave.

Algebraic equations in the context of variations often involve manipulating variables to understand relationships. In an inverse variation equation like \( y = \frac{k}{x} \), algebra is used to find unknown values and constant terms.

Let's go through each step of the problem again, focusing on the algebra involved:

• Identify the inverse variation: \( y = \frac{k}{x} \)


• Substitute the given values to form an equation: \( 5 = \frac{k}{20} \)


• Solve for k by isolating it: multiply both sides by 20 to get \( k = 100 \)

Once we have k, we create the full equation of variation: \( y = \frac{100}{x} \).

This equation can then be used to calculate either y or x when the other variable is known. The ability to transform and manipulate such equations is a fundamental skill in algebra.

Solving equations, particularly in variation problems, means isolating the unknown variable. In our inverse variation example, we encountered a problem where y varies inversely as x.

The steps to solve this type of equation are:

• Identify and write down the starting equation: \( y = \frac{k}{x} \)


• Substitute the known values for y and x: \( 5 = \frac{k}{20} \)


• Isolate k by performing an algebraic operation (multiplication in this case): \( 5 \times 20 = k \implies k = 100 \)


• Once k is found, write the final equation: \( y = \frac{100}{x} \)

Let's try solving an equation using this inverse variation formula: Suppose y = 2 and we want to know x. Using our formula:

• Start with: \( y = \frac{100}{x} \)


• Substitute y = 2: \( 2 = \frac{100}{x} \)


• To isolate x, multiply both sides by x and then divide both sides by 2:
  
  • \(2x = 100 \)
  
  
  • \(x = 50 \)

Breaking down each step and clearly isolating each variable helps demystify the process of solving algebraic equations.