Problem 147 - Algebra and Trigonometry

Question

Three of the richest comedians in the United States are Larry David (creator of Seinfeld), Matt Groening (creator of The Simpsons), and Trey Parker (co- creator of South Park). Larry David is worth \(\$ 450\) million more than Trey Parker. Matt Groening is worth \(\$ 150\) million more than Trey Parker. Combined, the net worth of these three comedians is \(\$ 1650\) million (or \(\$ 16.5\) billion). Determine how much, in millions of dollars, each of these comedians is worth. (Source: petamovies.com) (Section 1.3, Example 1)

Step-by-Step Solution

Answer

Trey Parker is worth \$350 million, Larry David is worth \$800 million, and Matt Groening is worth \$500 million.

Step 1 - Define the Variables

Let's label the variables \(x\), \(y\), and \(z\) to represent the net worths of Trey Parker, Larry David, and Matt Groening respectively. Then we can write down the three equations represented by the problem: \(y = x + 450\), \(z = x + 150\), and \(x + y + z = 1650\).

Step 2 - Substitute the Values of \(y\), \(z\) Into the Last Equation

We rewrite the equation \(x + y + z = 1650\) by replacing \(y\) and \(z\): \(x + (x + 450) + (x + 150) = 1650\). Simplifying this, we get \(3x + 600 = 1650\).

Step 3 - Solve for \(x\)

To find the value of \(x\), we subtract 600 from both sides and then divide by 3. This gives us \(x = 350\). So, Trey Parker is worth \$350 million.

Step 4 - Find the Values of \(y\), \(z\)

Now that we know \(x\), we can find \(y\) and \(z\) by substiting \(x\) into the first and second equations. This gives us \(y = 350 + 450 = 800\) and \(z = 350 + 150 = 500\). So, Larry David is worth \$800 million, and Matt Groening is worth \$500 million.

Key Concepts

Algebraic Expressions

At the heart of many word problems in algebra, especially those involving systems of equations, we find algebraic expressions. An algebraic expression is a mathematical phrase that can contain ordinary numbers, variables (like x or y), and operators (such as add, subtract, multiply, and divide). In our case, the expressions represent the net worth of comedians in millions of dollars.

For example, when we say Larry David is worth \$450\ million more than Trey Parker, we express this relationship algebraically as \(y = x + 450\). Here, \(y\) represents Larry David's net worth, \(x\) represents Trey Parker's net worth, and 450 is the dollar amount in millions separating their net worth. Such expressions are crucial as they form the building blocks of our equations that define the relationships among different quantities.

Variable Substitution

Variable substitution is a powerful technique used to simplify equations and solve them more easily. It involves replacing one variable with another expression that represents the same value. In the context of our problem, we already have the equations \(y = x + 450\) and \(z = x + 150\), which define the worth of Larry and Matt relative to Trey's worth, represented by \(x\).

When we substitute these expressions into the combined net worth equation, \(x + y + z = 1650\), we're able to transform it into a single variable equation. After substitution, we get \(x + (x + 450) + (x + 150) = 1650\), a much simpler equation to solve. This is a prime example of how substituting variables can help us manage and solve systems of equations with greater ease.

Solving Linear Equations

Solving linear equations is a fundamental skill in algebra that involves finding the values of variables that make the equation true. These equations often represent straight-line relationships between two quantities and can be solved using various methods, such as isolation, substitution, or graphing.

In our exercise, after we have substituted the values for \(y\) and \(z\), the equation simplifies to \(3x + 600 = 1650\). Solving for \(x\) requires isolating it on one side of the equation. This is done by first subtracting 600 from both sides, yielding \(3x = 1050\), and then dividing by 3 to get \(x = 350\). Once we know the value of \(x\), we can substitute it back into the expressions for \(y\) and \(z\) to find the net worth of each comedian. Understanding how to manipulate and solve these equations is crucial for solving a wide array of problems in algebra.