6.4 Systems of Linear Equations & Word-to-Algebra Translation

6.4 Systems of Linear Equations & Word-to-Algebra Translation

Translating word problems into formal mathematical equations and solving systems of linear equations simultaneous equations are essential high-level skills tested on the AFQT Mathematics Knowledge subtest. Converting verbal descriptions into precise algebraic notation allows you to resolve complex relationships efficiently.

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Systems of Linear Equations in Two Variables

A system of linear equations consists of two or more linear equations containing the same set of variables: $\begin{cases} a_1 x + b_1 y = c_1 \\ a_2 x + b_2 y = c_2 \end{cases}$

Solving a $2 \times 2$ system means finding the ordered pair $(x, y)$ that satisfies both equations simultaneously.

The Three Possible Solution Types

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Algebraic Solution Methods

Method 1: The Substitution Method

Best used when one variable already has a coefficient of $+1$ or $-1$.

1. Isolate one variable in whichever equation is simplest.
2. Substitute the resulting expression into the other equation.
3. Solve the resulting single-variable equation.
4. Back-substitute the numerical value into the isolated expression to find the second variable.

Substitution Walkthrough

Solve the system: $\begin{cases} 2x + 3y = 12 \quad \text{--- (Eq. 1)} \\ x - y = 1 \qquad \quad \text{--- (Eq. 2)} \end{cases}$

1. Isolate $x$ in Eq. 2: $x = y + 1$
2. Substitute $(y + 1)$ for $x$ in Eq. 1: $2(y + 1) + 3y = 12$
3. Solve for $y$: $2y + 2 + 3y = 12 \implies 5y + 2 = 12 \implies 5y = 10 \implies y = 2$
4. Back-substitute $y = 2$ into $x = y + 1$: $x = 2 + 1 = 3$ Solution = $(3, 2)$

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Method 2: The Elimination Method (Addition Method)

Best used when equations are both written in standard form ($ax + by = c$).

1. Write both equations in standard form $ax + by = c$.
2. Multiply one or both equations by non-zero constants so that coefficients of one variable become exact additive opposites (e.g., $+6y$ and $-6y$).
3. Add the two equations vertically to eliminate that variable.
4. Solve the remaining single-variable equation.
5. Back-substitute to find the other variable.

Elimination Walkthrough

Solve the system: $\begin{cases} 3x + 4y = 18 \quad \text{--- (Eq. 1)} \\ 2x - 3y = -5 \quad \text{--- (Eq. 2)} \end{cases}$

1. To eliminate $y$, multiply Eq. 1 by 3 and Eq. 2 by 4: $3(3x + 4y = 18) \implies 9x + 12y = 54$ $4(2x - 3y = -5) \implies 8x - 12y = -20$
2. Add the equations vertically: $(9x + 8x) + (12y - 12y) = 54 + (-20)$ $17x = 34 \implies x = 2$
3. Back-substitute $x = 2$ into Eq. 1: $3(2) + 4y = 18 \implies 6 + 4y = 18 \implies 4y = 12 \implies y = 3$ Solution = $(2, 3)$

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Word-to-Algebra Translation Dictionary

Translating English sentences into algebraic expressions requires recognizing key operational signal phrases.

⚠️ THE REVERSAL TRAP IN SUBTRACTION: Pay extreme attention to phrases like "less than" or "subtracted from".

• "5 less than $x$" means $x - 5$ (NOT $5 - x$).
• "7 subtracted from $y$" means $y - 7$ (NOT $7 - y$).

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AFQT Word Problem Solving Framework

Follow these four steps to convert AFQT narrative word problems into solvable linear models:

1. Define Variables: Assign clear letter variables to unknown quantities (e.g., $n = \text{number}$, $c = \text{cost}$).
2. Set Up Equation(s): Translate the narrative relationships into one or two linear equations.
3. Solve Algebraically: Use inverse operations, substitution, or elimination.
4. Verify Context: Ensure the numerical answer makes physical sense (e.g., counts of objects cannot be negative).

Worked Real-World AFQT Problem: Mixture / Supply Order

A military supply depot orders 6 tactical vests and 4 helmets for a total cost of $640. A second order containing 3 tactical vests and 5 helmets costs $530. What is the individual cost of one helmet?

1. Define Variables: Let $v = \text{cost of one tactical vest}$
   Let $h = \text{cost of one helmet}$

2. Build System of Equations: $\begin{cases} 6v + 4h = 640 \quad \text{--- (Order 1)} \\ 3v + 5h = 530 \quad \text{--- (Order 2)} \end{cases}$

3. Solve by Elimination: To find $h$, eliminate $v$ by multiplying Order 2 by $-2$: $-2(3v + 5h = 530) \implies -6v - 10h = -1060$ Add this modified equation to Order 1: $(6v - 6v) + (4h - 10h) = 640 - 1060$ $-6h = -420$ $h = \frac{-420}{-6} = 70$

4. Conclusion: The cost of one helmet is $70.