Question

A biologist wants to create a diet from fish and meal containing 183 grams of protein and 93 grams of carbohydrate per day. If fish contains $70\mathrm{\%}$ protein and 10 \%h carbohydrate, and meal contains $30\mathrm{\%}$ protein and $60\mathrm{\%}$ carbohydrate, how much of each food is required each day?

Short answer

210 grams of fish and 120 grams of meal are required each day.

Step-by-step solution

Identify Variables

Let $x$ be the amount of fish, and $y$ be the amount of meal, both in grams. We need to determine the values of $x$ and $y$ such that the dietary requirements are satisfied.

Set Up Protein Equation

From the problem, fish provides $70\mathrm{\%}$ of protein while meal provides $30\mathrm{\%}$. So, the equation for protein is $0.7x+0.3y=183$.

Set Up Carbohydrate Equation

Considering carbohydrates, fish provides $10\mathrm{\%}$ and meal provides $60\mathrm{\%}$. Hence, the equation for carbohydrates is $0.1x+0.6y=93$.

Solve the System of Equations

We have the following system of equations:1. $0.7x+0.3y=183$2. $0.1x+0.6y=93$Solve the system using substitution or elimination method.

Elimination Method: Adjust Equations

Multiply the second equation by 7 to help eliminate variable $x$:$$0.7x+0.3y=183$$$$0.7x+4.2y=651$$

Subtract to Eliminate $x$

Subtract the first equation from the modified second equation:$(0.7x+4.2y)-(0.7x+0.3y)=651-183$ Which simplifies to:$3.9y=468$.

Solve for $y$

Divide both sides of the equation by 3.9 to find $y$:$$y=\frac{468}{3.9}=120$$.

Substitute $y$ into Equation for $x$

Substitute $y=120$ into the first equation:$0.7x+0.3(120)=183$$0.7x+36=183$Subtract 36:$0.7x=147$

Solve for $x$

Divide by 0.7:$$x=\frac{147}{0.7}=210$$.

Conclusion

The biologist needs 210 grams of fish and 120 grams of meal each day to meet the dietary requirements.

Key concepts

Protein and Carbohydrate Requirements

When creating a diet, it's essential to balance protein and carbohydrates to meet daily requirements. Each nutrient plays a vital role in our bodies. Proteins are the building blocks of muscles and tissues, while carbohydrates provide the energy needed for daily activities.
In the context of the biologist's problem, ensuring 183 grams of protein and 93 grams of carbohydrate per day results in a precise dietary balance.
Fish and meal are chosen because of their distinct nutrient profiles:

• Fish is an excellent source of protein, providing 70% protein content.
• Meal is higher in carbohydrates, offering 60% carbohydrate content, but it still contributes to protein with 30%.

The goal is to determine the exact grams of fish and meal required to achieve these nutrient targets efficiently. Knowing how to calculate the right amounts can help in planning a balanced diet.

Solving Systems of Equations

Understanding systems of equations is crucial when faced with multiple constraints that need to be satisfied simultaneously. A system of equations includes more than one equation that relates different unknowns.
In our exercise, we have two equations due to the dual requirements for protein and carbohydrates. These can be represented as:

• Protein equation: $0.7x+0.3y=183$
• Carbohydrate equation: $0.1x+0.6y=93$

The goal of such systems is to find values for the variables, in this case, $x$ and $y$ (representing the amounts of fish and meal), that satisfy both equations simultaneously. By solving these equations, you effectively balance both aspects of the diet, creating a harmonious result that meets all specified needs.

Substitution and Elimination Methods

Two effective methods for solving systems of equations are substitution and elimination.
Substitution involves solving one equation for a single variable and then substituting this expression into the other equation. This method is beneficial when one equation can be easily expressed in terms of one variable.
Elimination, on the other hand, is more suitable when you have aligned coefficients as seen in our exercise.
Here's how elimination works in this scenario:

• The protein and carbohydrate equations are first manipulated to align coefficients for one of the variables. This usually involves multiplying one or both equations by a constant to make the coefficients of one variable identical.
• Next, subtract or add the equations to eliminate one variable, simplifying the system to a single-variable equation.

In this exercise, once the $x$ variable is eliminated, you solve for $y$, then substitute $y$ back into one of the original equations to find $x$. This method can simplify the process, especially when dealing with fraction coefficients or harder substitutions.