We can solve this problem using a system of linear equations. Let's define:

• $x$ = ounces of Food A
• $y$ = ounces of Food B
• $z$ = ounces of Food C

Given nutritional content per ounce:

Food	Calories	Protein (g)	Vitamin C (mg)
A	100	8	100
B	300	14	350
C	200	13	150

The meal should contain:

• 1200 calories
• 80 grams of protein
• 1150 milligrams of Vitamin C

Step 1: Set up the system of equations

1. Calories Equation:
   $100x + 300y + 200z = 1200$

2. Protein Equation:
   $8x + 14y + 13z = 80$

3. Vitamin C Equation:
   $100x + 350y + 150z = 1150$

Step 2: Solve the system

We will solve this system using algebraic methods (substitution or elimination).

1. Divide the first equation by 100:
   $x + 3y + 2z = 12$

2. Rewrite the protein equation:
   $8x + 14y + 13z = 80$

3. Divide the third equation by 50:
   $2x + 7y + 3z = 23$

Now, we solve for $x, y, z$:

Step 3: Solve for one variable

From equation (1):
$x = 12 - 3y - 2z$

Substituting $x$ into equations (2) and (3):

$8(12 - 3y - 2z) + 14y + 13z = 80$

$2(12 - 3y - 2z) + 7y + 3z = 23$

Simplify both:

$96 - 24y - 16z + 14y + 13z = 80$

$24 - 6y - 4z + 7y + 3z = 23$

$-10y - 3z = -16$

$y - z = -1$

Solving for $y$:

$y = z - 1$

Substituting into $-10y - 3z = -16$:

$-10(z - 1) - 3z = -16$

$-10z + 10 - 3z = -16$

$-13z = -26$

$z = 2$

$y = 1$

Now, solve for $x$:

$x = 12 - 3(1) - 2(2)$

$x = 12 - 3 - 4$

$x = 5$

Final Answer:

• Food A: 5 ounces
• Food B: 1 ounce
• Food C: 2 ounces

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Would you like a more detailed step-by-step breakdown or have any questions?

Related Questions:

1. How do you solve systems of equations using substitution and elimination?
2. What are other methods to solve systems of equations (e.g., matrix methods)?
3. How does nutrition analysis help in diet planning?
4. How can we use linear algebra in real-life applications?
5. Can we solve this problem using matrix inversion?

Tip: When solving systems of equations, always check your solution by substituting values back into the original equations.