5.5 Solve Mixture Applications with Systems of Equations

Mixture Applications with Systems of Equations

Mixture problems ask you to combine two or more things (coins, investments, solutions) and figure out how much of each you need. They show up constantly in algebra because they translate messy real-world situations into clean systems of two equations with two unknowns.

Every mixture problem follows the same core pattern: one equation tracks the total quantity, and the other tracks the total value (or interest, or concentration). Once you see that pattern, these problems become very manageable.

Systems for Coin and Ticket Mixtures

Coin and ticket problems give you a total count of items and a total dollar value, then ask you to find how many of each type you have.

Setting up the equations:

• Let $x$ = the number of one item (e.g., nickels)
• Let $y$ = the number of the other item (e.g., dimes)
• Write one equation for the total count: $x + y = \text{total number of items}$
• Write one equation for the total value: $(\text{value of } x) \cdot x + (\text{value of } y) \cdot y = \text{total value}$

Example: You have 100 coins, all nickels and dimes, worth $7.50 total.

• Count equation: $x + y = 100$
• Value equation: $0.05x + 0.10y = 7.50$

Solving step-by-step (substitution):

1. Solve the first equation for $y$: $y = 100 - x$

2. Substitute into the second equation: $0.05x + 0.10(100 - x) = 7.50$

3. Distribute: $0.05x + 10 - 0.10x = 7.50$

4. Combine like terms: $-0.05x + 10 = 7.50$

5. Subtract 10 from both sides: $-0.05x = -2.50$

6. Divide by $-0.05$: $x = 50$

7. Substitute back: $y = 100 - 50 = 50$

So there are 50 nickels and 50 dimes. Always check: $0.05(50) + 0.10(50) = 2.50 + 5.00 = 7.50$. It works.

Ticket problems follow the exact same structure. If adult tickets cost $12 and child tickets cost $8, just replace the coin values with ticket prices.

Interest Rate Problems with Investments

These problems involve splitting money between two accounts (or loans) with different interest rates. The structure is identical to coin problems, but now "value" means "interest earned."

Setting up the equations:

• Let $x$ = amount invested at one rate (e.g., 5%)
• Let $y$ = amount invested at another rate (e.g., 3%)
• Total investment equation: $x + y = \text{total amount}$
• Total interest equation: $(\text{rate}_1)(x) + (\text{rate}_2)(y) = \text{total interest}$

Example: You invest $10,000 total in two accounts. One earns 5% annually, the other earns 3%. After one year, you earn $425 in interest. How much is in each account?

• $x + y = 10{,}000$
• $0.05x + 0.03y = 425$

Solving step-by-step (elimination):

1. Multiply the first equation by $-0.03$: $-0.03x - 0.03y = -300$

2. Add it to the second equation: $0.05x + 0.03y + (-0.03x - 0.03y) = 425 + (-300)$

3. Simplify: $0.02x = 125$

4. Divide by $0.02$: $x = 6{,}250$

5. Substitute back: $y = 10{,}000 - 6{,}250 = 3{,}750$

Check: $0.05(6{,}250) + 0.03(3{,}750) = 312.50 + 112.50 = 425$. Confirmed.

A common mistake here is forgetting to convert percentages to decimals. 5% is $0.05$, not $5$.

Concentration Problems in Solutions

Concentration problems involve mixing two solutions with different percentages of some substance (salt, acid, alcohol) to get a final mixture with a target concentration.

Setting up the equations:

• Let $x$ = amount of the first solution (e.g., liters of 20% salt solution)
• Let $y$ = amount of the second solution (e.g., liters of 30% salt solution)
• Total volume equation: $x + y = \text{total volume of final mixture}$
• Concentration equation: $(\text{concentration}_1)(x) + (\text{concentration}_2)(y) = \text{amount of substance in final mixture}$

The right side of the concentration equation equals the desired concentration times the total volume. For a 50-liter mixture that's 25% salt: $0.25 \times 50 = 12.5$ liters of salt.

Example: Mix a 20% salt solution with a 30% salt solution to get 50 liters of 25% salt solution.

• $x + y = 50$
• $0.20x + 0.30y = 12.5$

Solving step-by-step (substitution):

1. From the first equation: $y = 50 - x$

2. Substitute: $0.20x + 0.30(50 - x) = 12.5$

3. Distribute: $0.20x + 15 - 0.30x = 12.5$

4. Combine like terms: $-0.10x + 15 = 12.5$

5. Subtract 15: $-0.10x = -2.5$

6. Divide by $-0.10$: $x = 25$

7. Substitute back: $y = 50 - 25 = 25$

Check: $0.20(25) + 0.30(25) = 5 + 7.5 = 12.5$ liters of salt. That's $12.5 / 50 = 0.25 = 25\%$. Correct.

Notice that mixing equal parts of 20% and 30% gives you exactly 25%, which is the midpoint. If the target concentration were closer to 30%, you'd need more of the 30% solution. This kind of reasoning can help you check whether your answer makes sense.

Problem-Solving Tips

The universal pattern across all three problem types:

• Equation 1: total quantity (count, dollars invested, volume)
• Equation 2: total "weighted" quantity (value, interest, amount of substance)

Once you identify which numbers go where, the algebra is the same every time.

Avoiding common errors:

• Always convert percentages to decimals before writing equations
• Make sure both equations use the same units (don't mix cents and dollars)
• After solving, plug your answers into both original equations to verify
• If your answer is negative or doesn't make sense in context (like a negative number of coins), recheck your setup