Problem 31 - Algebra and Trigonometry with Analytic Geometry

Question

Mixing a silver alloy A silversmith has two alloys, one containing \(35 \%\) silver and the other \(60 \%\) silver. How much of each should be melted and combined to obtain 100 grams of an alloy containing \(50 \%\) silver?

Step-by-Step Solution

Answer

Mix 40 grams of the 35% alloy with 60 grams of the 60% alloy.

Step 1: Define Variables

Let \( x \) represent the amount of the 35% silver alloy, and \( y \) represent the amount of the 60% silver alloy. We need to find the values of \( x \) and \( y \) such that their sum equals 100 grams.

Step 2: Set Up the Total Weight Equation

Since the total amount of alloy we want to obtain is 100 grams, we have the equation: \[ x + y = 100 \]

Step 3: Set Up the Silver Content Equation

We know that 35% of the alloy \( x \) is silver, and 60% of the alloy \( y \) is silver. We want the final mixture to contain 50% silver. This gives us the equation: \[ 0.35x + 0.60y = 0.50 \times 100 \] which simplifies to \[ 0.35x + 0.60y = 50 \]

Step 4: Solve the System of Equations

We have the system of equations: \[ \begin{align*} x + y &= 100 \ 0.35x + 0.60y &= 50 \end{align*} \] Substitute \( y = 100 - x \) from the first equation into the second equation: \[ 0.35x + 0.60(100 - x) = 50 \] Simplify and solve for \( x \): \[ 0.35x + 60 - 0.60x = 50 \] \[ -0.25x = -10 \] \[ x = 40 \]

Step 5: Find the Amount of the Second Alloy

Use \( y = 100 - x \) to find \( y \): \[ y = 100 - 40 = 60 \] So, the silversmith needs 60 grams of the second alloy.

Key Concepts

System of Equations

In a mixture problem like this one, we often deal with more than one unknown quantity. Here, the goal is to determine how much of two different silver alloys should be combined to achieve a new mixture with specific properties. To tackle this, we set up a system of equations. This method involves creating multiple equations that relate the known quantities to the unknowns.

• The first equation typically relates to the total amount of the mixtures. In our example, it's about the total weight of both alloys.
• The second equation relates to a specific property of interest, like the concentration of silver in this case.

By solving these equations together, we can pinpoint the exact amounts needed of each mixture. The key is that all equations must be satisfied simultaneously—hence why they form a 'system.' In this problem, they need to give us the exact mix that reaches both the total weight and the desired silver concentration.

Percentages

Percentages play a crucial role in mixture problems, as they help describe the concentrations of components within mixtures. In our problem, each alloy is described in terms of its percentage of silver.

• The first alloy has 35% silver, meaning that in every 100 grams of this alloy, 35 grams is silver.
• The second alloy has 60% silver.

Understanding how percentages break down into quantities allows you to convert these values into useful terms in your equations. Moreover, we need 50% silver in the final 100 grams of the mixture. By expressing percentages in decimal form (e.g., 0.35 for 35%), it becomes easier to set up equations and perform calculations.

Algebraic Substitution

Algebraic substitution is a key technique in solving systems of equations. It involves taking an equation and substituting terms in place of variables in another equation. This method simplifies solving for one of the unknowns.In our exercise:

• We started with the equation about the total weight: \( x + y = 100 \).
• This equation was rearranged to express \( y \) in terms of \( x \) as \( y = 100 - x \).
• The expression for \( y \) is then substituted into the silver content equation: \( 0.35x + 0.60y = 50 \).

By replacing \( y \) in the second equation, we create an equation with a single variable. Solving this gives a concrete value for \( x \), which can then be used to find \( y \) with the first equation. Substitution steps are essential for turning complex systems into manageable pieces.

Linear Equations

Linear equations are foundational in solving mixture problems. These equations are called 'linear' because they graph as straight lines. In this exercise:

• The total weight equation \( x + y = 100 \) represents a line on a graph where any point would have coordinates summing to 100.
• The silver content equation \( 0.35x + 0.60y = 50 \) is another line representing combinations of \( x \) and \( y \) with the desired 50 grams of silver.

The solution lies at the intersection of these two lines, i.e., where both conditions are satisfied. Linear equations can be either solved graphically or algebraically (as done with algebraic substitution). Each approach has its merits, but understanding that solutions are about balancing equations is key in tackling problems like this.