The Timpson Lyceum · Intermediate Algebra (2023)

Lecture 8 (Intermediate Algebra) Solving Systems of Equations

Overview

This lecture extends the substitution techniques for two-equation systems to larger and more applied settings. First it tackles three equations in three unknowns, showing that the same "solve-and-substitute" idea reduces a 3×3 system to a familiar 2×2 system. It then applies two-variable systems to three classic word-problem types: value (coin/ticket) problems, interest (investment) problems, and mixture problems. The unifying theme is that the algebra is mechanical once the equations are set up — the real skill is translating a situation into equations and always checking the answer by substituting back.

Key Concepts

Systems in three variables. A solution must satisfy all equations at once. To solve uniquely you need as many equations as unknowns — three equations for three unknowns. Geometrically each equation is a plane in space, and three planes generally meet at a single point $(x, y, z)$, just as two lines generally meet at a single point.

Solution strategy (3×3 → 2×2 → 1 variable).

• Pick any one equation and solve it for one variable (say $z$).
• Substitute that expression into the other two equations (never back into the one you used — that gives a circular identity).
• You now have a 2×2 system in the remaining two variables; solve it by substitution as before.
• Back-substitute to recover all three values.

Special cases (number of solutions). When the algebra collapses, the leftover statement tells you which case you have:

• A consistent value like $x = a,\ y = b,\ z = c$ → exactly one solution.
• An always-true identity like $4 = 4$ → infinitely many solutions (planes share a line or coincide).
• A contradiction like $4 = -6$ → no solution (parallel/non-intersecting planes).

Value (coin and ticket) problems. Build two equations:

• a count equation (total number of items), and
• a value equation (each item's worth times how many of it).

Working in cents avoids decimals: a quarter contributes $25$, a dime $10$. For example, $q + d = 11$ (count) and $25q + 10d = 185$ (value).

Interest (investment) problems. Interest earned on a principal $P$ at annual rate $r$ is $P\cdot r$, where a percentage is written as a decimal ($10\% = 0.10$). With two accounts, write a total-interest equation, $P_1 r_1 + P_2 r_2 = \text{total}$, plus a relationship between the rates (e.g. "$4\%$ more" means $r_2 = r_1 + 0.04$).

Mixture problems. Track the amount of the active ingredient (methane, butterfat, dollar-value of coffee) before and after mixing. For each component, (volume) × (concentration) = amount of ingredient, and the amounts must balance: $V_1 c_1 + V_2 c_2 = (V_1 + V_2)\, c_{\text{final}}$ The final volume is simply the sum of what you started with and what you added.

Always check. These problems involve a lot of arithmetic, so substitute the final values back into an original equation to confirm a true statement.

Worked Example

Solve the 3×3 system

Solve the third equation for $z$, then substitute it into the first two to get a 2×2 system:

(The $y$ terms cancel in the first reduced equation, giving $x$ at once.) Now back-substitute to find $y$ and $z$:

Check the solution $(x, y, z) = (2,\ -3,\ 1)$ in the first equation: $3(2) + 2(-3) - 1 = 6 - 6 - 1 = -1,$ which matches, confirming the system is consistent. The solution is $(2,\ -3,\ 1)$.

Summary

• A unique solution needs as many independent equations as unknowns; geometrically, three planes meet at one point.
• Solve a 3×3 system by solving one equation for a variable and substituting into the other two, reducing it to a 2×2 system.
• Collapsing to $a = a$ means infinitely many solutions; collapsing to a false statement means no solution.
• Value problems use a count equation and a value equation (work in cents to avoid decimals).
• Interest problems use $\text{interest} = P\cdot r$ (rates as decimals) plus any stated relationship between the rates.
• Mixture problems balance the amount of active ingredient: $V_1c_1 + V_2c_2 = (V_1+V_2)c_{\text{final}}$.
• Always substitute your answers back into an original equation to catch arithmetic mistakes.