Problem 16 - Algebra 2

Question

Shopping Suppose you want to fill nine 1-lb tins with a holiday snack mix. You plan to buy almonds for \(\$ 2.45 / 1 \mathrm{b}\) , peanuts for \(\$ 1.85 / 1 \mathrm{b}\) , and raisins for \(\$ .80 / 1 \mathrm{b}\) .You have \(\$ 15\) and want the mix to contain twice as much of the nuts as of the raisins by weight. How much of each ingredient should you buy? a. Writing Explain how each equation in the system at the right relates to the problem. What does each variable represent? b. Solve the system. $$ \left\\{\begin{array}{l}{x+y+z=9} \\ {2.45 x+1.85 y+0.8 z=15} \\ {x+y=2 z}\end{array}\right. $$

Step-by-Step Solution

Answer

After solving the system, you should buy x pounds of almonds, y pounds of peanuts, and z pounds of raisins for the given amounts, where the exact quantities are determined by the calculated values for x, y, and z.

Step 1: Define Variables

Let's define the variables for the quantities of the ingredients we need to purchase: let x be the number of pounds of almonds, y be the number of pounds of peanuts, and z be the number of pounds of raisins to be bought.

Step 2: Understand the Equations

The first equation, x + y + z = 9, represents the total weight of the mix which is 9 lbs. The second equation, 2.45x + 1.85y + 0.8z = 15, corresponds to the total cost of the ingredients amounting to $15. The last equation x + y = 2z indicates the mix should contain twice as much of the nuts (almonds and peanuts combined) as of the raisins by weight.

Step 3: Solve for z

Using the third equation, x + y = 2z, we can express z in terms of x and y: z = 1/2(x + y).

Step 4: Substitute z into the First Equation

Replace z in the first equation with the expression from Step 3: x + y + 1/2(x + y) = 9. Then solve for x and y.

Step 5: Simplify and Solve for y in Terms of x

After simplifying the equation from Step 4, we find 3/2(x + y) = 9. Dividing both sides by 3/2 gives x + y = 6. We can now solve for y in terms of x: y = 6 - x.

Step 6: Substitute y and z into the Second Equation

Using our expressions for y and z from Steps 3 and 5, we substitute into the second equation 2.45x + 1.85y + 0.8z = 15 to find x.

Step 7: Solve for x

After substitution, we solve for x, which will give us the amount of almonds we are to purchase.

Step 8: Calculate y Using the Value of x

Using the value of x obtained in Step 7, calculate the value of y by substituting x into y = 6 - x.

Step 9: Calculate z Using the Value of x and y

Finally, calculate the value of z using the initial expression z = 1/2(x + y) with the found values of x and y.

Key Concepts

Linear Equations

When learning about linear equations in an algebra class, it's important to understand that they are the building blocks for modeling relationships between quantities. A linear equation is an algebraic equation where each term is either a constant or the product of a constant and a single variable. Linear equations are used to describe lines in a coordinate system, and they follow the format of \( ax + by = c \), where \( a \) and \( b \) are coefficients, \( x \) and \( y \) are variables, and \( c \) is the constant.

For example, in our shopping problem scenario, we have the linear system:
\[\begin{align*}x + y + z &= 9 \2.45x + 1.85y + 0.8z &= 15 \x + y &= 2z\end{align*}\]
Each equation represents a different constraint on the purchase: the total weight, the total cost, and the weight relationship between nuts and raisins. Simplifying and manipulating these equations, just like we would manipulate numbers with basic arithmetic, allows us to find the values of \( x \), \( y \), and \( z \) that satisfy all constraints simultaneously.

Algebraic Problem Solving

In the realm of algebraic problem solving, a structured approach is crucial to dissect complex problems into manageable parts. The first step is usually to define the variables, as we did with \( x \), \( y \) and \( z \) representing the amount of almonds, peanuts, and raisins respectively. This allows us to translate words into mathematical expressions, making it possible to apply computational techniques.

Once we have our equations set up, as shown in the problem, we can start solving for the unknowns through substitution or elimination methods. The goal is to reduce the system to simpler equations that can be solved directly. It may involve several stages of reworking and substituting until the values are isolated. In our example, by substituting and eliminating variables systematically, we find the quantities of each ingredient that can be bought within the budget, meeting the weight constraint for the holiday snack mix.

Educational Algebra

Using educational algebra in contexts such as creating a holiday snack mix makes learning algebra relevant and engaging. Applying algebraic concepts to real-life situations develops critical thinking and problem-solving skills. When students can contextualize algebra by solving for quantities of ingredients, or budgeting consumables, it enhances their understanding and retention of mathematical concepts.

Moreover, providing clear explanations and simplifying each step is important for educational algebra to ensure that key concepts are understood, not just memorized. Breaking down a problem, like determining the quantities of almonds, peanuts, and raisins to buy, into smaller, solvable parts helps students grasp each stage of the algebraic process. This also demonstrates the utility of algebra beyond the classroom, preparing students for analytical challenges they may encounter in everyday life.