What does it mean if my answer comes out negative?

A negative volume is physically impossible. It usually means the final concentration you asked for cannot be made by mixing those two solutions—for example, asking for 50% when mixing 10% and 30% solutions. The math is correct, but the problem itself has no real solution.

Do I have to use percentages for concentration?

No. Concentrations can be percentages, molarity, calories per ounce, grams per liter, or any other unit of strength. The key is that c_1, c_2, and c_f all use the same units, and v_1 and v_2 use the same volume units.

What if I want the final concentration to equal one of the original concentrations?

If c_f = c_1, then you use only solution 1 and none of solution 2. The formula still works: c_1 v_1 + c_2 \cdot 0 = c_1(v_1 + 0), which simplifies to c_1 v_1 = c_1 v_1. This tells you that any amount of solution 1 (and zero of solution 2) works.

Can I mix three solutions using this formula?

Yes. Extend it to three terms: c_1 v_1 + c_2 v_2 + c_3 v_3 = c_f(v_1 + v_2 + v_3). With three unknowns, you will need three equations total—the mixture formula and two additional constraints on the volumes—and then use Systems of Three Equations to solve.

Mixture Problems

Mixture Problems help you find how much of each different-strength solution to combine to achieve a target concentration.

c_1 v_1 + c_2 v_2 = c_f (v_1 + v_2)

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Type the problem. The solver will use Mixture Problems where Mixture Problems is the right tool, and tell you when it is not.

What each symbol means

What Mixture Problems takes

Symbol	Meaning
$c_1$	The strength of the first solution—a percentage, molarity, or similar measure—and not its volume; confusing these reverses which solution you use more of.
$c_2$	The strength of the second solution, using the same units as $c_1$ and $c_f$ (for example, all as percentages or all as molar concentrations).
$c_f$	The desired strength of the final mixture; this always falls between $c_1$ and $c_2$ when you mix two solutions of different strengths.
$v_1$	The volume (or mass or amount) of the first solution, measured in consistent units like liters or milliliters; using percentages here instead causes the equation to fail.
$v_2$	The volume of the second solution in the same units as $v_1$ , and the total volume of the mixture is always $v_1 + v_2$ , not just one of them.

When to use it

Use Mixture Problems when you need to combine two solutions of different concentrations to reach a specific final concentration.

Level

Usually taught in: Algebra I

Worked examples

1. Mix salt solutions with known volumes

Problem

How many liters of a 20% salt solution should be mixed with 30 liters of a 50% salt solution to make a 40% solution?

$20v_1 + 50(30) = 40(v_1 + 30)$
Substitute into the formula with $c_1 = 20$ , $c_2 = 50$ , $v_2 = 30$ , and $c_f = 40$ .
$20v_1 + 1500 = 40v_1 + 1200$
Compute $50 \times 30 = 1500$ on the left and expand $40(v_1 + 30) = 40v_1 + 1200$ on the right.
$300 = 20v_1$
Subtract $1200$ from both sides to get $20v_1 + 300 = 40v_1$ , then subtract $20v_1$ from both sides.
$v_1 = 15$
Divide both sides by 20.

Answer: $v_1 = 15 \text{ liters}$

We knew four of the five values in the formula, so we substituted them directly and solved for $v_1$ using two-step algebra. The answer makes sense: we need 15 liters of the weak solution and 30 liters of the strong solution to balance out to 40%.

2. Find volumes when the final amount is given

Problem

A scientist must create 500 mL of a 30% chemical solution by mixing a 10% solution with an 80% solution. How many milliliters of each should be used?

$10v_1 + 80v_2 = 30(500)$
Set up the mixture equation; the final amount of pure chemical is $30\% \times 500 = 15000$ units.
$10v_1 + 80v_2 = 15000$
Compute $30 \times 500 = 15000$ .
$v_1 + v_2 = 500$
Set up a second equation because the volumes must sum to 500 mL.
$v_1 = 500 - v_2$
Solve for $v_1$ in the volume equation.
$10(500 - v_2) + 80v_2 = 15000$
Substitute $v_1 = 500 - v_2$ into the mixture equation.
$5000 - 10v_2 + 80v_2 = 15000$
Distribute: $10 \times 500 = 5000$ .
$5000 + 70v_2 = 15000$
Combine like terms: $-10v_2 + 80v_2 = 70v_2$ .
$70v_2 = 10000$
Subtract 5000 from both sides.
$v_2 = \frac{1000}{7} \approx 142.86 \text{ mL}$
Divide by 70 and simplify the fraction.
$v_1 = 500 - \frac{1000}{7} = \frac{3500 - 1000}{7} = \frac{2500}{7} \approx 357.14 \text{ mL}$
Use the volume equation with $v_2$ ; convert 500 to $\frac{3500}{7}$ to subtract.

Answer: $v_1 = \frac{2500}{7} \text{ mL (approx. 357.14 mL)}, \quad v_2 = \frac{1000}{7} \text{ mL (approx. 142.86 mL)}$

When you know the final volume and all concentrations but not the individual volumes, you need two equations: one from the mixture formula and one for the total volume. Solving by Substitution eliminates one variable and leaves you with a single equation in one unknown.

3. Coffee shop blend: mixing daily roasts

Problem

A coffee shop blends light roast (15 calories per ounce) with dark roast (20 calories per ounce) to create a house blend with exactly 17.5 calories per ounce. If they make 48 ounces, how many ounces of each roast are needed?

$15v_1 + 20v_2 = 17.5(48)$
Set up the mixture equation; each ounce of light roast contributes 15 calories, each ounce of dark roast contributes 20 calories.
$15v_1 + 20v_2 = 840$
Compute $17.5 \times 48 = 840$ .
$v_1 + v_2 = 48$
The two roasts must sum to 48 ounces.
$v_1 = 48 - v_2$
Solve for $v_1$ in terms of $v_2$ .
$15(48 - v_2) + 20v_2 = 840$
Substitute into the mixture equation.
$720 - 15v_2 + 20v_2 = 840$
Distribute: $15 \times 48 = 720$ .
$720 + 5v_2 = 840$
Combine like terms: $-15v_2 + 20v_2 = 5v_2$ .
$5v_2 = 120$
Subtract 720 from both sides.
$v_2 = 24$
Divide by 5.
$v_1 = 48 - 24 = 24$
Use the volume equation to find $v_1$ .

Answer: $v_1 = 24 \text{ oz of light roast}, \quad v_2 = 24 \text{ oz of dark roast}$

Real-world mixing problems often provide the final volume and all three concentrations, making this a standard Solving by Substitution problem. Here the answer is symmetric—equal amounts of each—because 17.5 is exactly halfway between 15 and 20.

Common mistakes

Where Mixture Problems usually goes wrong

The mistake	Why it is wrong	The fix
Writing the formula as $c_1 + c_2 = c_f$ (or $c_f = \frac{c_1 + c_2}{2}$ ), treating it as a simple average of concentrations.	The final concentration is a weighted average, not an unweighted average; the volumes matter because you mix different amounts of each solution.	Always use the full formula $c_1 v_1 + c_2 v_2 = c_f(v_1 + v_2)$ and include both volumes and concentrations.
Swapping concentrations and volumes, such as using volume numbers in place of concentration values or vice versa.	Concentrations and volumes are different quantities with different meanings; putting a volume where a concentration belongs (or the reverse) breaks the equation and gives nonsensical answers.	Before substituting, label which values are concentrations (strength, percentage, or molarity) and which are volumes (liters, ounces, etc.), then match them to the correct symbols in the formula.
Using only one volume on the right side, writing $c_1 v_1 + c_2 v_2 = c_f v_1$ or $c_f v_2$ instead of $c_f(v_1 + v_2)$ .	The final concentration applies to the entire mixture, which has total volume $v_1 + v_2$ ; using only one volume ignores part of the solution.	The right side must always be $c_f$ times the total volume: $c_f(v_1 + v_2)$ , not just $c_f$ times one of the original volumes.

Tips and when to use something else

Check that your final concentration lies between the two original concentrations; if not, you either set up the equation wrong or the problem asks for something impossible.
Keep units consistent: if $c_1$ is a percentage, then $c_2$ and $c_f$ must also be percentages; if $v_1$ is liters, then $v_2$ must be liters too.
When you have the final volume and want to find the mixing ratio, use Solving by Substitution or Solving by Elimination instead of guessing values.
For problems with three or more solutions, extend the formula to $c_1 v_1 + c_2 v_2 + c_3 v_3 = c_f(v_1 + v_2 + v_3)$ and use Systems of Three Equations.

Related formulas

Linear EquationSolve for the value of x in any linear equation by rearranging to isolate x, finding the unique solution where the expression equals zero.Solving by SubstitutionSolving by Substitution solves a system of equations by replacing one variable with its equivalent expression, turning two equations into one.Solving by EliminationSolve a system of two linear equations by adding or subtracting to eliminate one variable, then solve for the remaining variable.Systems of Three EquationsSystems of three equations let you find three unknowns from three relationships, the standard approach when a real-world situation has multiple constraints.

Frequently asked questions

What does it mean if my answer comes out negative?: A negative volume is physically impossible. It usually means the final concentration you asked for cannot be made by mixing those two solutions—for example, asking for 50% when mixing 10% and 30% solutions. The math is correct, but the problem itself has no real solution.
Do I have to use percentages for concentration?: No. Concentrations can be percentages, molarity, calories per ounce, grams per liter, or any other unit of strength. The key is that $c_1$ , $c_2$ , and $c_f$ all use the same units, and $v_1$ and $v_2$ use the same volume units.
What if I want the final concentration to equal one of the original concentrations?: If $c_f = c_1$ , then you use only solution 1 and none of solution 2. The formula still works: $c_1 v_1 + c_2 \cdot 0 = c_1(v_1 + 0)$ , which simplifies to $c_1 v_1 = c_1 v_1$ . This tells you that any amount of solution 1 (and zero of solution 2) works.
Can I mix three solutions using this formula?: Yes. Extend it to three terms: $c_1 v_1 + c_2 v_2 + c_3 v_3 = c_f(v_1 + v_2 + v_3)$ . With three unknowns, you will need three equations total—the mixture formula and two additional constraints on the volumes—and then use Systems of Three Equations to solve.

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Reviewed 2026-09-18