4.3 Solve Mixture Applications with Systems of Equations

Mixture and Financial Applications of Systems of Equations

Systems of equations let you find unknown quantities in real-world scenarios like combining chemical solutions, splitting investments, or analyzing business costs. In each case, the setup is the same: translate the word problem into two equations with two unknowns, then solve. This section covers the three most common application types you'll see.

Systems of Equations for Mixtures

Mixture problems ask you to combine two solutions with different concentrations to get a final mixture at some target concentration. The unknowns are almost always how much of each solution to use.

Setting up the system:

1. Define your variables. Let $x$ = amount of solution 1 and $y$ = amount of solution 2 (in liters, gallons, etc.).
2. Write a total amount equation using any information about the final quantity. For example, if you need 100 liters total: $x + y = 100$
3. Write a concentration equation by tracking the actual substance (salt, acid, etc.) in each solution. Multiply each amount by its concentration (as a decimal): $0.20x + 0.30y = 0.25(100)$

The left side adds up the substance contributed by each solution. The right side is the desired concentration times the total amount.

1. Solve the system using substitution or elimination.

Example walkthrough: You need 100 liters of a 25% acid solution. You have a 20% solution and a 30% solution.

• Total equation: $x + y = 100$
• Acid equation: $0.20x + 0.30y = 25$
• From the first equation, $x = 100 - y$. Substitute into the second: $0.20(100 - y) + 0.30y = 25$, which gives $20 - 0.20y + 0.30y = 25$, so $0.10y = 5$ and $y = 50$.
• Then $x = 50$.

Verify by checking the concentration: $0.20(50) + 0.30(50) = 10 + 15 = 25$. That's 25 out of 100 liters, or 25%. It checks out.

Interest Calculations with Equations

These problems typically involve splitting money between two accounts or investments at different interest rates. You know the total amount invested and the total interest earned, and you need to find how much went into each account.

For Intermediate Algebra, you'll use the simple interest formula:

$I = Prt$

where $I$ is interest earned, $P$ is the principal (amount invested), $r$ is the annual interest rate (as a decimal), and $t$ is time in years. Most textbook problems set $t = 1$, which simplifies things to $I = Pr$.

Setting up the system:

1. Let $P_1$ = amount invested at rate 1 and $P_2$ = amount invested at rate 2.
2. Write a total principal equation: $P_1 + P_2 = \text{total invested}$
3. Write a total interest equation: $r_1 P_1 + r_2 P_2 = \text{total interest earned}$
4. Solve the system.

Example: You invest $8,000 total. Part goes into an account earning 4% and the rest into one earning 7%. After one year, you earn $470 in total interest.

• $P_1 + P_2 = 8000$
• $0.04P_1 + 0.07P_2 = 470$
• From the first equation, $P_1 = 8000 - P_2$. Substitute: $0.04(8000 - P_2) + 0.07P_2 = 470$, giving $320 - 0.04P_2 + 0.07P_2 = 470$, so $0.03P_2 = 150$ and $P_2 = 5000$.
• Then $P_1 = 3000$.

So $3,000 was invested at 4% and $5,000 at 7%.

Cost-Revenue Analysis Using Equations

These problems model a business scenario where you need to find the break-even point, the number of units where revenue exactly equals cost (no profit, no loss).

Two key formulas:

• Total cost: $C = F + vx$, where $F$ is the fixed cost (rent, equipment), $v$ is the variable cost per unit (materials, labor), and $x$ is the number of units produced.
• Total revenue: $R = px$, where $p$ is the price per unit and $x$ is the number of units sold.

Finding the break-even point:

Set $C = R$ and solve for $x$:

$F + vx = px$

$F = px - vx$

$F = x(p - v)$

$x = \frac{F}{p - v}$

Example: A company has fixed costs of $12,000, a variable cost of $8 per unit, and sells each unit for $20.

$x = \frac{12000}{20 - 8} = \frac{12000}{12} = 1000$

They need to sell 1,000 units to break even. Selling more than 1,000 units generates profit; fewer means a loss.

Profit at any production level is $\text{Profit} = R - C = px - (F + vx)$.

Key Problem-Solving Strategies

• Algebraic modeling is the core skill here: read the problem carefully, identify what's unknown, define variables, and translate each piece of information into an equation. If you have two unknowns, you need two equations.
• Unit consistency matters. If one quantity is in milliliters and another in liters, convert before setting up equations.
• Weighted averages show up in both mixture and interest problems. The final concentration (or overall interest rate) is a weighted average of the individual rates, weighted by how much of each you're using. That's exactly what the concentration and interest equations represent.