6.1 Single-Variable Linear Equations & Algebraic Expressions

Single-Variable Linear Equations & Algebraic Expressions

In healthcare education and clinical nursing practice, algebraic fluency is not an abstract academic exercise. Nurses constantly solve single-variable linear equations to calculate individualized medication infusion rates, adjust dosages based on renal clearance or body surface area, determine syringe volumes, and evaluate fluid deficits across shifts. On the HESI A2 Mathematics subtest, algebraic questions assess your ability to manipulate symbolic expressions, apply inverse operations systematically, and maintain mathematical balance under time constraints. Mastering these core algebraic mechanics provides the foundation for error-free dosage calculations in clinical pharmacology.

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1. Expressions Versus Equations & The Mechanics of Substitution

Before manipulating algebraic statements, you must distinguish between an algebraic expression and an algebraic equation:

• Algebraic Expression: A mathematical phrase combining numbers (constants), letters (variables), and operation symbols ($+$, $-$, $\times$, $\div$). An expression does not contain an equals sign ($=$) and cannot be "solved" for a definitive numerical value on its own; it can only be simplified or evaluated when numerical values are assigned to its variables. Examples include $4x - 7$, $3a + 2b$, and $\frac{5y - 2}{3}$.
• Algebraic Equation: A mathematical statement asserting that two expressions are equal in value, joined by an equals sign ($=$). An equation can be solved to determine the specific numerical value (or values) of the variable that makes the statement true. Examples include $4x - 7 = 21$ and $3(x + 4) = 2x - 1$.

The Anatomy of an Algebraic Term

In an expression such as $-7x^2$:

• The number $-7$ is the coefficient (the numerical factor multiplying the variable).
• The letter $x$ is the variable (an unknown numerical quantity).
• The number $2$ is the exponent (the power to which the variable is raised).
• A standalone number, such as $+15$ in $3x + 15$, is a constant.

Evaluating Expressions by Substitution

To evaluate an algebraic expression, replace each variable with its given numerical value enclosed in parentheses, then compute the result following standard order of operations (PEMDAS: Parentheses, Exponents, Multiplication and Division from left to right, Addition and Subtraction from left to right).

Given Expression: 3x^2 - 4xy + 5y
Values: x = -2, y = 3

Step 1: Enclose substitutions in parentheses:
        3(-2)^2 - 4(-2)(3) + 5(3)

Step 2: Evaluate exponents first:
        (-2)^2 = +4
        3(4) - 4(-2)(3) + 5(3)

Step 3: Execute multiplications from left to right:
        3(4) = 12
        -4(-2)(3) = -4(-6) = +24
        5(3) = 15
        12 + 24 + 15

Step 4: Execute additions:
        12 + 24 = 36
        36 + 15 = 51

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2. The Balance Method & Inverse Operations

The central conceptual model for solving equations is the balance scale. An equation asserts that the expression on the left-hand side (LHS) has exactly the same quantitative value as the expression on the right-hand side (RHS). To preserve this equality, any operation performed on one side must be performed identically on the other side.

$\text{Left-Hand Side (LHS)} = \text{Right-Hand Side (RHS)}$

To isolate the unknown variable, you apply inverse operations in reverse operational order to "undo" the numbers attached to the variable.

Clinical One-Step Equation Examples

• Additive Balance: A patient's total fluid intake target is $2{,}200\text{ mL}$. They have consumed $1{,}450\text{ mL}$ thus far. How many milliliters $x$ remain?
  Equation: $1{,}450 + x = 2{,}200 \implies x = 2{,}200 - 1{,}450 = 750\text{ mL}$.
• Multiplicative Balance: An intravenous infusion delivers $125\text{ mL/hr}$. How many hours $t$ are required to deliver $1{,}000\text{ mL}$?
  Equation: $125t = 1{,}000 \implies t = \frac{1{,}000}{125} = 8\text{ hours}$.

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3. Solving Two-Step Linear Equations ($ax + b = c$)

A two-step linear equation contains two operations acting on the variable—typically multiplication or division combined with addition or subtraction. To isolate the variable, reverse the normal PEMDAS order of operations:

1. Step 1 (Undo Addition/Subtraction): Eliminate the constant term by adding or subtracting it across the equals sign.
2. Step 2 (Undo Multiplication/Division): Eliminate the variable's coefficient by dividing or multiplying both sides.

Worked Example 1: Standard Two-Step Model

Solve for $x$: $5x + 23 = 68$

1. Subtract $23$ from both sides to isolate the variable term: $5x + 23 - 23 = 68 - 23$ $5x = 45$
2. Divide both sides by $5$ to isolate $x$: $\frac{5x}{5} = \frac{45}{5}$ $x = 9$

Worked Example 2: Negative Coefficients and Subtraction

Solve for $y$: $-4y - 17 = 31$

1. Add $17$ to both sides: $-4y - 17 + 17 = 31 + 17$ $-4y = 48$
2. Divide both sides by the negative coefficient $-4$: $\frac{-4y}{-4} = \frac{48}{-4}$ $y = -12$

Worked Example 3: Fractional Variable Term

Solve for $m$: $\frac{m}{3} - 8 = 14$

1. Add $8$ to both sides: $\frac{m}{3} = 22$
2. Multiply both sides by the denominator $3$: $3 \times \left(\frac{m}{3}\right) = 3 \times 22$ $m = 66$

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4. Multi-Step Equations: Distribution & Variables on Both Sides

More complex equations require organizing terms before inverse operations can be applied. This multi-step algorithm handles the linear equations you will meet in HESI A2 practice:

Multi-Step Equation Solving Hierarchy:
  1. Distribute: Clear all parentheses by multiplying through: a(bx + c) = abx + ac
  2. Combine Like Terms: Simplify expressions on each side of the '=' sign independently
  3. Move Variable Terms: Add or subtract variable terms so all variables reside on one side
  4. Move Constant Terms: Add or subtract constants so all numerical terms reside on the opposite side
  5. Isolate Variable: Divide or multiply by the variable's coefficient
  6. Back-Substitute: Verify the computed root in the original equation

The Distributive Property and Sign Traps

The distributive property states that multiplying a sum by a number gives the same result as multiplying each addend by that number and summing the products:

$a(bx + c) = abx + ac$

When the factor outside the parentheses is negative, distribute the negative sign to every term inside:

$-3(2x - 5) = (-3)(2x) + (-3)(-5) = -6x + 15$

A common mistake is failing to multiply the second term or mishandling double negatives (e.g., writing $-6x - 15$).

Worked Multi-Step Example: Variables on Both Sides

Solve for $x$: $4(2x - 3) - 3 = 2(x + 6) + 3$

• Step 1: Apply the distributive property to both sides: $8x - 12 - 3 = 2x + 12 + 3$
• Step 2: Combine like terms on each side: $8x - 15 = 2x + 15$
• Step 3: Transpose the variable term to the left (subtract $2x$ from both sides): $8x - 2x - 15 = 2x - 2x + 15$ $6x - 15 = 15$
• Step 4: Transpose the constant term to the right (add $15$ to both sides): $6x - 15 + 15 = 15 + 15$ $6x = 30$
• Step 5: Isolate the variable (divide by $6$): $\frac{6x}{6} = \frac{30}{6}$ $x = 5$

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5. Clearing Fractions and Decimals

Working with fractional and decimal coefficients increases cognitive load and heightens the risk of calculation errors. You can clear them in the first step by multiplying the entire equation by an appropriate scalar factor.

Clearing Fractions with the Least Common Denominator (LCD)

Identify all denominators in the equation, determine their least common denominator (LCD), and multiply every single term on both sides of the equation by that LCD.

Worked Example: Clearing Fractions

Solve for $x$: $\frac{3}{4}x - \frac{1}{2} = \frac{1}{3}x + 2$

1. Identify the LCD: Denominators are $4, 2,$ and $3$. The least common multiple of $4$ and $3$ is $12$.
2. Multiply every term by $12$: $12 \times \left(\frac{3}{4}x\right) - 12 \times \left(\frac{1}{2}\right) = 12 \times \left(\frac{1}{3}x\right) + 12 \times (2)$
3. Simplify each term: $\left(\frac{12 \times 3}{4}\right)x - 6 = \left(\frac{12 \times 1}{3}\right)x + 24$ $9x - 6 = 4x + 24$
4. Solve the resulting integer equation:
   • Subtract $4x$ from both sides: $5x - 6 = 24$
   • Add $6$ to both sides: $5x = 30$
   • Divide by $5$: $x = 6$

Clearing Decimals by Multiplying by Powers of 10

If an equation contains decimals, count the maximum number of decimal places across all terms. Multiply every term by $10$ (for tenths), $100$ (for hundredths), or $1{,}000$ (for thousandths).

Worked Example: Clearing Decimals

Solve for $p$: $0.4p + 1.8 = 0.15p + 3.3$

1. Count decimal places: $0.4$, $1.8$, and $3.3$ have one place; $0.15$ has two places. Multiply every term by $100$: $100(0.4p) + 100(1.8) = 100(0.15p) + 100(3.3)$ $40p + 180 = 15p + 330$
2. Transpose and solve:
   • Subtract $15p$ from both sides: $25p + 180 = 330$
   • Subtract $180$ from both sides: $25p = 150$
   • Divide by $25$: $p = 6$

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6. Systematic Verification Checklist (Back-Substitution)

In high-stakes clinical scenarios, dosage calculations undergo independent double-checks prior to drug administration. Similarly, you can catch most algebra slips by verifying your solution through back-substitution.