5.3 Systems of Linear Equations

Foundations of $2 \times 2$ Linear Systems

A system of linear equations in two variables consists of two or more linear equations containing the same set of variables. On the ACCUPLACER Quantitative Reasoning, Algebra, and Statistics (QAS) test, you will primarily solve $2 \times 2$ systems written in standard form:

$\begin{cases} a_1 x + b_1 y = c_1 \\ a_2 x + b_2 y = c_2 \end{cases}$

A solution to a linear system is an ordered pair $(x, y)$ that satisfies both equations simultaneously. Geometrically, this solution corresponds to the point of intersection of the two straight lines on a Cartesian coordinate plane.

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Algebraic Solving Methods

While graphing provides geometric intuition, exact solutions on the ACCUPLACER are computed using two algebraic techniques: the Substitution Method and the Elimination (Addition) Method.

1. The Substitution Method

The substitution method is ideal when at least one variable in either equation has a coefficient of $+1$ or $-1$.

Substitution Algorithm:
Step 1: Isolate one variable with a coefficient of ±1 in one of the equations.
Step 2: Substitute that resulting algebraic expression into the OTHER equation.
Step 3: Solve the resulting single-variable linear equation.
Step 4: Back-substitute the numerical value into the isolated expression from Step 1 to find the second variable.
Step 5: Check the ordered pair (x, y) in both original equations.

Worked Example (Substitution): $\begin{cases} x - 2y = 5 \quad \text{(Eq. 1)} \\ 3x + 4y = 5 \quad \text{(Eq. 2)} \end{cases}$

1. Isolate $x$ in Eq. 1: $x = 2y + 5$.
2. Substitute $(2y + 5)$ for $x$ in Eq. 2: $3(2y + 5) + 4y = 5$
3. Solve for $y$: $6y + 15 + 4y = 5 \implies 10y + 15 = 5 \implies 10y = -10 \implies y = -1$
4. Back-substitute $y = -1$ into the expression for $x$: $x = 2(-1) + 5 = -2 + 5 = 3$
5. Ordered pair solution: $(3, -1)$. (Check: Eq. 1: $3 - 2(-1) = 5$ (true); Eq. 2: $3(3) + 4(-1) = 9 - 4 = 5$ (true)).

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2. The Elimination (Addition) Method

The elimination method is optimal when equations are given in standard form $Ax + By = C$ and coefficients are integers without an isolated variable.

Elimination Algorithm:
Step 1: Write both equations in standard form Ax + By = C with aligned terms.
Step 2: Multiply one or both equations by non-zero constants so that the coefficients of ONE variable are exact opposites (e.g., +6 and -6).
Step 3: Add the two equations vertically to eliminate that target variable.
Step 4: Solve the resulting single-variable linear equation.
Step 5: Back-substitute the known value into either original equation to find the second variable.
Step 6: Check the ordered pair in both original equations.

Worked Example (Elimination): $\begin{cases} 3x + 4y = 7 \quad \text{(Eq. 1)} \\ 5x - 2y = 29 \quad \text{(Eq. 2)} \end{cases}$

1. Target variable $y$: Eq. 1 has $+4y$ and Eq. 2 has $-2y$. Multiply Eq. 2 by $2$ to create $-4y$: $2 \cdot (5x - 2y) = 2 \cdot (29) \implies 10x - 4y = 58$
2. Add the modified Eq. 2 to Eq. 1: $\begin{array}{rl} 3x + 4y &= 7 \\ +\quad 10x - 4y &= 58 \\ \hline 13x + 0y &= 65 \end{array}$
3. Solve for $x$: $13x = 65 \implies x = 5$
4. Back-substitute $x = 5$ into Eq. 1 to solve for $y$: $3(5) + 4y = 7 \implies 15 + 4y = 7 \implies 4y = -8 \implies y = -2$
5. Ordered pair solution: $(5, -2)$.

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Method Selection Guide

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Geometric & Algebraic Classification of Linear Systems

When attempting to solve a linear system, the variable terms may completely cancel out. The algebraic outcome directly reveals the geometric relationship between the lines:

Analytical Examples of Inconsistent & Dependent Systems

• Inconsistent System (Parallel Lines / No Solution): $\begin{cases} 2x - 4y = 6 \\ x - 2y = 8 \end{cases}$ Multiply second equation by $-2$: $-2x + 4y = -16$. Adding yields $0 = -10$ (False) $\implies \text{No Solution } (\emptyset)$.
• Dependent System (Identical Lines / Infinite Solutions): $\begin{cases} 3x - 6y = 12 \\ -x + 2y = -4 \end{cases}$ Multiply second equation by $3$: $-3x + 6y = -12$. Adding yields $0 = 0$ (True) $\implies \text{Infinitely Many Solutions}$.

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Applied Systems Word Problems

1. Mixture Problems

Mixture problems combine two items with different concentrations (or prices) to produce a mixture with a target concentration.

The Mixture System Template:

$\begin{cases} \text{Quantity Eq: } & x + y = \text{Total Quantity} \\ \text{Value/Amount Eq: } & c_1 x + c_2 y = c_{\text{final}}(\text{Total Quantity}) \end{cases}$

Worked Example (Mixture): A laboratory technician needs to prepare $50\text{ liters}$ of a $36\%$ saline solution. In stock, the technician has a $20\%$ saline solution and a $60\%$ saline solution. How many liters of each should be mixed?

1. Define variables: Let $x = \text{liters of } 20\%$ solution, $y = \text{liters of } 60\%$ solution.
2. Set up the system: $\begin{cases} x + y = 50 \quad \text{(Total Volume)} \\ 0.20x + 0.60y = 0.36(50) = 18 \quad \text{(Pure Saline)} \end{cases}$
3. From Eq. 1, $x = 50 - y$. Substitute into Eq. 2: $0.20(50 - y) + 0.60y = 18$ $10 - 0.20y + 0.60y = 18$ $10 + 0.40y = 18 \implies 0.40y = 8 \implies y = \frac{8}{0.40} = 20\text{ liters}$
4. Compute $x$: $x = 50 - 20 = 30\text{ liters}$. The technician needs $30\text{ L}$ of $20\%$ solution and $20\text{ L}$ of $60\%$ solution.

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2. Total Value / Ticket / Coin Problems

Worked Example (Ticket Sales): A community theater sold $400\text{ total tickets}$ for a weekend musical, collecting a total of $\$3,800$. Adult tickets cost $\$12$ each and student tickets cost $\$7$ each. How many adult tickets and student tickets were sold?

1. Define variables: Let $A = \text{adult tickets}$, $S = \text{student tickets}$.
2. Set up the system: $\begin{cases} A + S = 400 \quad \text{(Total Tickets)} \\ 12A + 7S = 3800 \quad \text{(Total Revenue)} \end{cases}$
3. Multiply Eq. 1 by $-7$ to eliminate $S$: $-7A - 7S = -2800$
4. Add to Eq. 2: $(12A - 7A) + (7S - 7S) = 3800 - 2800$ $5A = 1000 \implies A = 200$
5. Find $S$: $200 + S = 400 \implies S = 200$. The theater sold $200\text{ adult tickets}$ and $200\text{ student tickets}$.

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3. Economic Break-Even Analysis

Break-even occurs when Total Revenue equals Total Cost: $R(x) = C(x)$, or when Profit $P(x) = R(x) - C(x) = 0$.

Worked Example: A small artisanal candle maker incurs fixed monthly overhead costs of $\$1,200$ and a variable production cost of $\$4$ per candle. The candles sell for $\$16$ each.

1. Formulate Cost and Revenue functions: $C(x) = 1200 + 4x$ $R(x) = 16x$
2. Set $R(x) = C(x)$: $16x = 1200 + 4x$ $12x = 1200 \implies x = 100\text{ candles}$
3. Break-even revenue: $R(100) = 16(100) = \$1,600$. The business breaks even when producing and selling $100\text{ candles}$.

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Common Pitfalls & ACCUPLACER Exam Traps

1. Solving for the Wrong Variable: ACCUPLACER questions often ask for the value of $y$, the sum $x + y$, or the product $xy$, rather than $x$. Always re-read the prompt after finding $x$.
2. Sign Distribution Errors in Elimination: When subtracting one equation from another, distribute the negative sign to every term on both sides: $-(5x - 2y = 11) \implies -5x + 2y = -11$.
3. Failure to Check in BOTH Equations: A calculated pair may satisfy one equation while violating the second due to an arithmetic slip. Always check both.
4. Confusing Inconsistent and Dependent Terminology: Remember: "Inconsistent" means incompatible lines that never touch (0 solutions); "Dependent" means overlapping identical lines (infinitely many solutions).
5. Unit Alignment in Mixture Problems: When writing concentration equations, ensure percentages are converted to decimals (e.g., $25\% = 0.25$) and applied to the total mixture weight or volume.