Problem 251 - Elementary Algebra

Question

In the following exercises, translate to a system of equations and solve. Marissa wants to blend candy selling for \(\$ 1.80\) per pound with candy costing \(\$ 1.20\) per pound to get a mixture that costs her \(\$ 1.40\) per pound to make. She wants to make 90 pounds of the candy blend. How many pounds of each type of candy should she use?

Step-by-Step Solution

Answer

30 pounds of \(1.80 candy, 60 pounds of \)1.20 candy.

Step 1 - Define Variables

Let x be the number of pounds of candy costing \(1.80 per pound. Let y be the number of pounds of candy costing \)1.20 per pound.

Step 2 - Set Up the Equations

We know two things: the total weight of the mixture and the total cost per pound of the mixture. This gives us the following equations:1. Total weight: \[ x + y = 90 \]2. Total cost per pound:\[ 1.80x + 1.20y = 1.40 \times 90 \]

Step 3 - Simplify the Cost Equation

Simplify the second equation to make it easier to solve:\[ 1.80x + 1.20y = 126 \]

Step 4 - Solve for One Variable

Solve the first equation for y:\[ y = 90 - x \]

Step 5 - Substitute in the Cost Equation

Substitute \( y = 90 - x \) into the simplified cost equation:\[ 1.80x + 1.20(90 - x) = 126 \]

Step 6 - Solve for x

Distribute and combine like terms:\[ 1.80x + 108 - 1.20x = 126 \]\[ 0.60x + 108 = 126 \]Subtract 108 from both sides:\[ 0.60x = 18 \]Divide both sides by 0.60:\[ x = 30 \]

Step 7 - Find y

Substitute \( x = 30 \) back into the equation \( y = 90 - x \): \[ y = 90 - 30 \]\[ y = 60 \]

Step 8 - Verify the Solution

Substitute \( x = 30 \) and \( y = 60 \) back into the original equations to ensure they hold true:1. Total weight: \[ 30 + 60 = 90 \]2. Total cost per pound:\[ 1.80(30) + 1.20(60) = 54 + 72 = 126 \]

Key Concepts

Linear Equations

Linear equations are equations where the highest power of the variable is always one. They take the form of \[ ax + b = c \], where 'a', 'b', and 'c' are constants, and 'x' is the variable. In our candy blending problem, the equations \[ x + y = 90 \] and \[ 1.80x + 1.20y = 126 \] are linear equations. Here, 'x' and 'y' represent pounds of different candies, and the constants are the costs and total weight. Linear equations are straight lines when graphed. They are easy to handle because you can add, subtract, multiply and divide them without worrying about changing the nature of the equation.

In most real-world problems, linear equations help link different conditions or constraints to form a system of equations. These systems can then be solved to find the values of the variables, which in our case are the pounds of two types of candy. Understanding linear equations is, therefore, foundational for solving many types of algebraic problems.

Mixture Problems

Mixture problems involve combining two or more substances to create a blend with particular properties. In our problem, Marissa wants to mix two types of candy to achieve a specific cost for the blend. There are two main aspects to consider: the total weight of the mixture and the cost per pound of the mixture.

To set up the mixture problem, we define variables for the quantities of each component being combined. Next, we write equations that express the relationship between these quantities. In our case, we know:

• Total weight: \[ x + y = 90 \]
• Total cost per pound: \[ 1.80x + 1.20y = 1.40 \times 90 \]

These equations take into account both the weight and the cost factors, providing a comprehensive equation system. Mixture problems like this often appear in various fields, such as chemistry, cooking, and manufacturing, where precise combinations are crucial.

Solving Equations

Solving equations means finding the value of the unknown variables that make the equation true. For a system of linear equations, we typically use substitution or elimination methods. In our candy mixture problem, we used substitution to solve the system.

Here’s a brief recap of how we solved it:

• First, we defined our variables as 'x' and 'y'.
• Then, we set up two equations based on total weight and cost per pound.
• After simplifying the cost equation, we solved the first equation for 'y'. For example, \[ y = 90 - x \]
• We then substituted 'y = 90 - x' into the cost equation, which gave us a new equation with just 'x' as the variable.
• After simplifying and solving for 'x', we found \[ x = 30 \]
• Substituting 'x'= 30 back into 'y = 90 - x' gave us \[ y = 60 \]

The final step is always verifying the solution by substituting x = 30' and 'y = 60' back into the original equations to ensure they hold true. Solving these equations accurately is key to ensuring the solution is correct and valid for the problem.