2.1 Linear Equations in One Variable
Key Takeaways
- Multi-step linear equations are solved by applying inverse operations systematically: distributing constants, combining like terms, isolating variable terms on one side, and clearing coefficients.
- Clearing fractions by multiplying every term by the Least Common Denominator (LCD) and clearing decimals by multiplying by powers of 10 eliminates arithmetic errors on complex SAT questions.
- Absolute value equations of the form |ax + b| = c split into two distinct cases: ax + b = c or ax + b = -c (when c >= 0), and possess no real solutions when c < 0.
- Always verify solutions for absolute value equations to prevent accepting extraneous roots, especially when variable expressions appear outside the absolute value bars.
- The Digital SAT frequently tests algebraic efficiency by asking for the value of an expression (such as 3x - 4 or 2x + 7) rather than x alone; identify opportunities to scale equations directly.
Linear Equations in One Variable
Linear equations in one variable form the absolute bedrock of the Digital SAT Math section. More than 25% of the questions in the Algebra domain directly test your ability to construct, manipulate, and solve linear equations rapidly and without algebraic errors. Whether presented as abstract algebraic manipulations or embedded in real-world scenarios, single-variable linear equations reward systematic execution and mathematical precision.
1. Anatomy of Multi-Step Linear Equations
A linear equation in one variable is an equation that can be written in the standard form:
where $a$, $b$, $c$, and $d$ are real numbers, and $a \ne c$. The highest exponent on the variable $x$ is always 1, guaranteeing that the equation represents a relationship of degree one.
The Standard 5-Step Solving Algorithm
To solve any multi-step linear equation efficiently under timed test conditions, execute the following standardized sequence:
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| MULTI-STEP LINEAR EQUATION ALGORITHM |
| |
| [STEP 1: CLEAR FRACTIONS / DECIMALS] |
| Multiply every term on both sides by the Least Common Denominator (LCD). |
| | |
| v |
| [STEP 2: EXPAND & DISTRIBUTE] |
| Apply distributive property to eliminate all parentheses: a(bx + c). |
| | |
| v |
| [STEP 3: COMBINE LIKE TERMS] |
| Simplify variable terms and constant terms separately on each side. |
| | |
| v |
| [STEP 4: ISOLATE VARIABLE TERMS] |
| Use addition or subtraction to collect all variable terms on one side. |
| | |
| v |
| [STEP 5: ISOLATE THE VARIABLE] |
| Multiply or divide by the variable's coefficient to find x. |
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| Algebraic Operation | Property of Equality | Example Transformation |
|---|---|---|
| Addition | If $A = B$, then $A + c = B + c$ | $x - 7 = 12 \implies x = 19$ |
| Subtraction | If $A = B$, then $A - c = B - c$ | $3x + 5 = 20 \implies 3x = 15$ |
| Multiplication | If $A = B$, then $A \cdot c = B \cdot c$ ($c \ne 0$) | $\frac{x}{4} = 6 \implies x = 24$ |
| Division | If $A = B$, then $\frac{A}{c} = \frac{B}{c}$ ($c \ne 0$) | $5x = 35 \implies x = 7$ |
| Distributive | $a(b + c) = ab + ac$ | $-3(2x - 4) = -6x + 12$ |
2. Advanced Distributive Property & Negative Sign Traps
The most frequent source of lost points on linear algebra questions is improper distribution across subtraction or negative signs.
Negative Multipliers and Subtraction
When a negative sign or negative coefficient precedes parentheses, that negative sign must be distributed to every single term inside the parentheses:
[!WARNING] The Missing Negative Trap: In the expression $12 - 3(2x - 5)$, students frequently write $12 - 6x - 15$. The correct expansion is $12 - 6x + 15 = 27 - 6x$. Always circle or highlight the negative sign in front of the multiplier before distributing.
Step-by-Step Worked Example: Multi-Step Linear Equation
Problem: Solve the equation for $x$:
Step 1: Distribute on both sides
Step 2: Combine like terms on each side
Step 3: Collect variable terms on one side (add $12x$ to both sides)
Step 4: Collect constants on the opposite side (subtract 29 from both sides)
Step 5: Divide by the coefficient of $x$
3. Clearing Fractions and Decimals (The LCD Technique)
Fractions introduce unnecessary complexity and increase the risk of arithmetic error. You can eliminate all fractions in a single step by multiplying the entire equation by the Least Common Denominator (LCD) of all terms.
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| FRACTION-CLEARING WORKFLOW |
| |
| Given: (3x - 1)/4 - (2x + 5)/6 = (x + 7)/12 |
| |
| 1. Identify Denominators: 4, 6, 12 ---> LCD = 12 |
| 2. Multiply EVERY term by 12: |
| 12 * [(3x - 1)/4] - 12 * [(2x + 5)/6] = 12 * [(x + 7)/12] |
| 3. Simplify before multiplying: |
| 3(3x - 1) - 2(2x + 5) = 1(x + 7) |
| 4. Expand: |
| 9x - 3 - 4x - 10 = x + 7 |
| 5. Solve clean integers: |
| 5x - 13 = x + 7 ===> 4x = 20 ===> x = 5 |
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Clearing Decimals
Similarly, when an equation contains decimals, multiply every term by $10$, $100$, or $1,000$ depending on the maximum number of decimal places:
Multiply the entire equation by $100$:
4. Absolute Value Linear Equations: $|ax + b| = c$
The absolute value of a real number $u$, denoted $|u|$, represents its geometric distance from 0 on the number line. Because distance is always non-negative, $|u| \ge 0$ for all real numbers.
The Two-Case Fundamental Rule
For any algebraic expression $u$ and constant $c$:
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| ABSOLUTE VALUE EQUATION BRANCHING |
| |
| |ax + b| = c |
| | |
| +-----------------+-----------------+ |
| | | |
| c >= 0 c < 0 |
| | | |
| +---------+---------+ v |
| | | [NO SOLUTION] |
| v v |
| Case 1: Case 2: |
| ax + b = c ax + b = -c |
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Worked Example: Absolute Value with Variable on Right Side
Problem: Solve $|2x - 5| = 3x + 1$ for all real solutions.
Case 1 (Positive branch):
Case 2 (Negative branch):
Check for Extraneous Solutions:
- Test $x = -6$: $|2(-6) - 5| = |-17| = 17$. Right side: $3(-6) + 1 = -17$. Since $17 \ne -17$, $x = -6$ is extraneous.
- Test $x = \frac{4}{5}$: $|2(0.8) - 5| = |1.6 - 5| = |-3.4| = 3.4$. Right side: $3(0.8) + 1 = 2.4 + 1 = 3.4$. Valid!
- Final Answer: $x = \frac{4}{5}$.
[!IMPORTANT] Extraneous Solutions in Absolute Value: When the right-hand side contains variables, you must verify your solutions in the original equation. Any algebraic solution that causes the non-absolute value side to be negative is extraneous and must be discarded.
5. The "Expression Value" Trap on the Digital SAT
The College Board frequently structures questions that ask for the value of an algebraic expression rather than the variable $x$ alone. For example, a question might state:
The Direct Scaling Strategy vs. Traditional Solving
| Method | Steps Required | Time Spent |
|---|---|---|
| Traditional Method | $6x = 36 \implies x = 6 \implies 2(6) - 5 = 12 - 5 = 7$ | ~25 seconds |
| Direct Scaling | Notice $(6x - 15) = 3(2x - 5)$. Divide both sides by 3: $2x - 5 = \frac{21}{3} = 7$ | ~5 seconds |
Recognizing proportional relationships between the given equation and the requested expression saves critical seconds and reduces algebraic calculation errors.
Desmos Calculator Verification Workflow
On the Digital SAT, the built-in Desmos graphing calculator can instantly verify single-variable linear equations:
- Type the left-hand side as
y = LHS(e.g.,y = -4(3x - 5) + 6). - Type the right-hand side as
y = RHS(e.g.,y = 2(x + 8) - 5x + 13). - Click the point of intersection. The $x$-coordinate of the intersection is the exact solution.
If 5(2x - 3) - 3(x + 4) = 4(x - 2) + 7, what is the value of 6x - 5?
What is the solution to the equation (2x + 3)/5 - (x - 1)/2 = 3/10?
What is the sum of all real solutions to the absolute value equation |3x - 6| = x + 2?