Problem 44 - Algebra 1

Question

You are in charge of bringing breakfast for your scout troop. You buy 6 bagels and 8 donuts for a total of \(4.10. Then you decide to buy 3 extra of each for a total of \)1.80. How much did each bagel and donut cost?

Step-by-Step Solution

Answer

Each bagel costs $0.35 and each donut costs $0.25.

Step 1: Represent the problem with equations

Let's denote the cost of a bagel as \(b\) and the cost of a donut as \(d\). According to the problem statement, we can form these two equations: 6b + 8d = 4.10 (equation 1 presenting the first purchase) and 3b + 3d = 1.80 (equation 2 presenting the second purchase).

Step 2: Modify the equations to solve for the variables

Multiply equation 2 by 2: 6b + 6d = 3.60. Now, we have two equations with the same coefficient for \(b\): 6b + 8d = 4.10 and 6b + 6d = 3.60.

Step 3: Subtract one equation from the other

Subtract the second equation from the first one to isolate \(d\): 2d = 0.50. Divide each side by 2 to get the value for \(d\): \(d\) = 0.25.

Step 4: Substitute the value of d into one of the equations

Now, plug \(d = 0.25\) into equation 2: 3b + 3*0.25 = 1.80. Therefore, 3b + 0.75 = 1.80. Subtract 0.75 from both sides: 3b = 1.05. Divide by 3 to solve for \(b\): \(b\) = 0.35.

Key Concepts

Linear Equations

Linear equations are fundamental in understanding how the relationships between variables work in algebra. In this exercise, we have two linear equations that model the situation of buying bagels and donuts. These are equations of the form \( ax + by = c \), where \(a\), \(b\), and \(c\) are constants, and \(x\) and \(y\) are variables representing the quantities we want to find. Here, the equation \( 6b + 8d = 4.10 \) describes the cost of 6 bagels and 8 donuts, while \( 3b + 3d = 1.80 \) describes the cost of 3 more of each. Linear equations often form straight lines when graphed. However, in this context, they help us determine the exact cost of one bagel and one donut through algebraic manipulation. Understanding that these equations are linear because each term is either a constant or the product of a constant and a single variable makes it easier to solve them through methods like substitution or elimination.

Variable Substitution

Variable substitution is a powerful method in solving systems of equations, especially when the equations are given in linear form. The idea is to express one variable in terms of another and then substitute it into the other equation. For instance, after subtracting the modified second equation from the first, we found that \(2d = 0.50\). Solving this gives \(d = 0.25\). At this point, we have one of our variables isolated. By substituting \(d\) back into one of our original equations, such as \(3b + 3d = 1.80\), we can solve for the other variable, \(b\). After calculation, substituting \(d = 0.25\) gives us \(3b = 1.05\), leading us to find \(b\). This method reduces the complexity by transforming a system of equations into a simpler one, where each equation has only one variable.

Equation Solving

Solving equations involves finding the values of variables that make the equation true. In our exercise, we've demonstrated this by solving for \(d\) and \(b\) through a series of steps. Initially, it involved eliminating one of the variables by matching coefficients, a method known as elimination. By carefully subtracting one equation from another, we isolate \(d\). Solving for \(d\) was straightforward once it was isolated, by simply dividing both sides of the equation by its coefficient. The next step was to substitute \(d\) into the other equation to solve for \(b\).

• Identify similar terms or coefficients to simplify calculations.
• Eliminate unnecessary terms by subtraction or addition of equations.
• Solve for each variable by isolating and simplifying.

This structured approach simplifies the process, making it easier to follow and ensuring accuracy in getting the correct values for \(b\) and \(d\), hence understanding the respective costs of the bagels and donuts.