Problem 51 - Algebra and Trigonometry

Question

A researcher performs an experiment to test a hypothesis that involves the nutrients niacin and retinol. She feeds one group of laboratory rats a daily diet of precisely 32 units of niacin and 22,000 units of retinol. She uses two types of commercial pellet foods. Food A contains 0.12 unit of niacin and 100 units of retinol per gram. Food B contains 0.20 unit of niacin and 50 units of retinol per gram. How many grams of each food does she feed this group of rats each day?

Step-by-Step Solution

Answer

200 grams of Food A and 40 grams of Food B.

Step 1: Define Variables

Let \( x \) be the number of grams of Food A and \( y \) be the number of grams of Food B that the researcher feeds the rats.

Step 2: Set Up Equations for Niacin

The niacin from Food A is \( 0.12x \) and from Food B is \( 0.20y \). We need a total of 32 units of niacin: \[ 0.12x + 0.20y = 32. \]

Step 3: Set Up Equations for Retinol

The retinol from Food A is \( 100x \) and from Food B is \( 50y \). We need a total of 22,000 units of retinol: \[ 100x + 50y = 22000. \]

Step 4: Solve the System of Equations

We now have two equations:1. \( 0.12x + 0.20y = 32 \)2. \( 100x + 50y = 22000 \)First, simplify the second equation by dividing by 50: \[ 2x + y = 440. \]Equation 1 remains: \[ 0.12x + 0.20y = 32. \]

Step 5: Solve Equation for One Variable

From the simplified equation \( 2x + y = 440 \), express \( y \) in terms of \( x \): \[ y = 440 - 2x. \]

Step 6: Substitute Back into First Equation

Substitute \( y = 440 - 2x \) into the first equation:\[ 0.12x + 0.20(440 - 2x) = 32. \]

Step 7: Simplify and Solve for x

Distribute \( 0.20 \):\[ 0.12x + 88 - 0.40x = 32. \]Simplify the equation:\[ -0.28x = -56. \]Solving for \( x \):\[ x = 200. \]

Step 8: Solve for y

Substitute \( x = 200 \) back into the equation \( y = 440 - 2x \):\[ y = 440 - 2(200). \] This simplifies to:\[ y = 40. \]

Step 9: Verify the Solution

Verify that \( x = 200 \) and \( y = 40 \) satisfy both original equations: 1. \( 0.12(200) + 0.20(40) = 32 \) correctly holds.2. \( 100(200) + 50(40) = 22000 \) correctly holds.The solution is verified.

Key Concepts

Linear Equations

Linear equations are equations where variables are raised to the power of one. They form a straight line when graphed on a coordinate plane. These equations are often used to represent real-world situations.
In our problem, we have two linear equations that correspond to the nutrients: niacin and retinol.

• The first equation: \( 0.12x + 0.20y = 32 \), represents the total amount of niacin needed each day.
• The second: \( 100x + 50y = 22000 \), captures the total retinol requirement.

Linear equations can help model situations where two variables are related in a consistent way, allowing researchers to predict or plan accordingly. In this context, linear equations help determine the combination of Food A and Food B to meet nutritional goals.

Substitution Method

The substitution method involves solving one of the equations for a variable and then substituting that expression into the other equation. This method effectively reduces the number of unknowns, making it easier to solve a system of equations.
In this problem, we demonstrate the substitution method by the following process:

• First, simplify the equation \( 2x + y = 440 \) to express \( y \) in terms of \( x \): \( y = 440 - 2x \).
• Substitute this expression into the first equation: \( 0.12x + 0.20(440 - 2x) = 32 \).

This method conveniently eliminates one variable, allowing us to solve for the other variable easily. It is particularly useful when one of the equations is already solved or can be simply manipulated to express one variable in terms of the other.

Algebraic Manipulation

Algebraic manipulation is the process of rearranging equations, and it's a key step in solving systems of equations. This involves distributing, combining like terms, and isolating variables.
Let's see how it's applied:

• In \( 0.12x + 0.20(440 - 2x) = 32 \), distribute to get \( 0.12x + 88 - 0.40x = 32 \).
• Combine like terms to simplify: \( -0.28x + 88 = 32 \).
• Finally, isolate \( x \) by moving terms: \( -0.28x = -56 \).

Through these steps, we derive \( x = 200 \). This algebraic manipulation helps simplify complex equations, making it simpler to find solutions systematically. Regardless of the complexity, employing these steps effectively reduces intricate equations into simpler ones that are easier to solve.