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Chapter 3: Problem 73

Bakery. Gigi's Cupcakes offers a gift box with six cupcakes for $$ 15.99 .\( Gigi's also sells cupcakes individually for \)3 each. Gigi's sold a total of 256 cakes one Saturday for a total of $$ 701.67$ in sales (excluding tax). How many six-cupcake gift boxes were included in that day's sales total?

Short Answer

Expert verified
33 six-cupcake gift boxes were sold.

Step by step solution

01

- Define Variables

Let the number of six-cupcake gift boxes sold be \( x \). Let the number of individual cupcakes sold be \( y \). We need to set up equations based on the given information.
02

- Write Equations

From the problem, we have two pieces of information:1. The total number of cupcakes sold: \[ 6x + y = 256 \ \] 2. The total revenue from the sales of cupcakes: \[ 15.99x + 3y = 701.67 \ \]
03

- Solve for One Variable

First, solve Equation 1 for \( y \): \[ y = 256 - 6x \ \]
04

- Substitute into Second Equation

Substitute \( y = 256 - 6x \) into Equation 2: \[ 15.99x + 3(256 - 6x) = 701.67 \ \]
05

- Simplify the Equation

Distribute and simplify: \[ 15.99x + 768 - 18x = 701.67 \ -2.01x + 768 = 701.67 \ -2.01x = 701.67 - 768 \ -2.01x = -66.33 \ \]
06

- Solve for \( x \)

Divide both sides by -2.01 to find \( x \): \[ x = \frac{-66.33}{-2.01} \ = 33 \ \]
07

- Verify the Solution

To verify, we substitute \( x = 33 \) back into the first equation: \[ 6(33) + y = 256 \ 198 + y = 256 \ y = 58 \ \] Also check the revenue equation: \[ 15.99(33) + 3(58) = 701.67 \ 527.67 + 174 = 701.67 \ \] Both equations hold true, confirming the solution.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

systems of linear equations
When dealing with algebra, it's common to encounter systems of linear equations. This means you have more than one equation that work together. Each equation has the same variables, which means they represent relationships between the same quantities. In the given problem, we had two equations from the total number of cupcakes sold and the total revenue from sales. To solve these equations, both equations must be true at the same time. That means finding values for the variables that satisfy all given equations.
substitution method
One powerful way to solve systems of linear equations is the substitution method. It involves solving one of the equations for one variable, and then substituting that expression into the other equation. This way, you can solve for one variable easily. In our bakery example, we solved the first equation for y:
  • \[y = 256 - 6x\]
By substituting this expression in the second equation, we only needed to solve for x, making the problem simpler. This method can often reduce complex systems to simpler single-variable equations. Once we find the value for one variable, we use it to find the value of the other.
simplifying equations
Simplifying equations is crucial when solving algebra problems. After substituting one equation into another, you often get more complex expressions that need to be simplified. This means combining like terms and isolating the variable. For example, after substitution in our problem, we had:
  • \[15.99x + 3(256 - 6x) = 701.67\]
  • Simplifying it gave us: \[-2.01x + 768 = 701.67\]
Breaking it down further allowed us to isolate x and solve for it:
  • \[x = \frac{-66.33}{-2.01} = 33\]
Always remember to break down the equation into simpler parts and solve step-by-step.
defining variables
In word problems, defining your variables clearly is the first step to solving the problem effectively. This means identifying what each variable represents. In our cupcake problem, we defined:
  • \(x\) as the number of six-cupcake gift boxes sold
  • \(y\) as the number of individual cupcakes sold
With clear definitions, it鈥檚 easier to set up equations based on the problem statement. This clarity helps to avoid confusion, especially when dealing with multi-step solutions or more complex problems. Always take a moment to define and understand your variables before diving into solving equations.

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