6.2 Multi-Step Equations, Variables on Both Sides, & Special Solution Sets

6.2 Multi-Step Equations, Variables on Both Sides, & Special Solution Sets

As you advance to TABE 13&14 Mathematics Levels D and A, algebraic equations expand in complexity to include multiple grouping symbols, rational fractions, decimal rates, and variable terms distributed across both sides of the equal sign. Solving these problems requires an unshakeable, systematic procedure.

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1. The 5-Step Master Algorithm for Multi-Step Equations

Regardless of how complicated a linear equation appears, applying the following five steps in order will reduce it to a simple solution:

Step 1: CLEAR FRACTIONS / DECIMALS (Multiply every term by LCD or 10^n)
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Step 2: DISTRIBUTE (Eliminate all parentheses: a(bx + c) = abx + ac)
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Step 3: COMBINE LIKE TERMS (Simplify left side, then simplify right side)
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Step 4: COLLECT VARIABLES ON ONE SIDE (Add/subtract variable terms)
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Step 5: ISOLATE THE VARIABLE (Undo addition/subtraction, then multiplication/division)

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2. Step-by-Step Worked Demonstrations

Example 1: Multi-Step Equation with Distributive Property & Like Terms

Problem: Solve $5(2x - 3) - 3(x + 4) = 25$.

$$\begin{aligned} \text{Step 2 (Distribute):} &\quad 10x - 15 - 3x - 12 = 25 \quad (\text{Note: } -3 \cdot +4 = -12) \\ \text{Step 3 (Combine Like Terms):} &\quad (10x - 3x) + (-15 - 12) = 25 \\ &\quad 7x - 27 = 25 \\ \text{Step 5 (Undo Subtraction):} &\quad 7x = 25 + 27 \\ &\quad 7x = 52 \\ \text{Step 5 (Undo Multiplication):} &\quad x = \frac{52}{7} = 7\frac{3}{7} \end{aligned}$$

Example 2: Clearing Fractions using the Least Common Denominator (LCD)

Problem: Solve $\frac{2x - 1}{3} - \frac{x + 2}{4} = \frac{7}{12}$.

1. Identify the LCD: The denominators are $3, 4,$ and $12$. Their $\text{LCD} = 12$.
2. Multiply Every Term by the LCD ($12$): $$12 \cdot \left(\frac{2x - 1}{3}\right) - 12 \cdot \left(\frac{x + 2}{4}\right) = 12 \cdot \left(\frac{7}{12}\right)$$
3. Simplify Denominators: $$4(2x - 1) - 3(x + 2) = 7$$
4. Distribute: $$8x - 4 - 3x - 6 = 7$$
5. Combine Like Terms: $$5x - 10 = 7$$
6. Isolate Variable: $$5x = 17 \implies x = \frac{17}{5} = 3.4$$

Example 3: Clearing Decimals

Problem: Solve $0.08x + 0.12(500 - x) = 46$.

1. Clear Decimals: Multiply all terms by $100$: $$8x + 12(500 - x) = 4600$$
2. Distribute: $$8x + 6000 - 12x = 4600$$
3. Combine Like Terms: $$-4x + 6000 = 4600$$
4. Subtract 6000: $$-4x = -1400$$
5. Divide by $-4$: $$x = \frac{-1400}{-4} = 350$$

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3. Equations with Variables on Both Sides ($ax + b = cx + d$)

When variable terms appear on both the left and right sides of the equation, eliminate one variable term using the Addition or Subtraction Property of Equality.

Example: Solve $7x - 18 = 2x + 32$.

$$\begin{aligned} \text{Subtract } 2x \text{ from both sides:} &\quad 7x - 2x - 18 = 2x - 2x + 32 \\ &\quad 5x - 18 = 32 \\ \text{Add } 18 \text{ to both sides:} &\quad 5x = 32 + 18 \\ &\quad 5x = 50 \\ \text{Divide by } 5: &\quad x = 10 \end{aligned}$$

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4. Special Solution Sets: Conditional, Inconsistent, & Identity Equations

Not every linear equation simplifies to a single numerical value. In algebra, linear equations are categorized into three distinct solution types:

Linear Equation Solution Classification:
├── 1. Conditional Equation ──────> One Unique Solution (x = k)
├── 2. Inconsistent Equation ─────> No Solution (False statement: 3 = 8 or 0x = k)
└── 3. Identity Equation ─────────> Infinitely Many Solutions (True statement: 5 = 5 or 0x = 0)

Clinical Analysis of Special Cases

Special Case A (Inconsistent / No Solution): Solve $3(2x - 4) = 6x + 7$.

$$\begin{aligned} 6x - 12 &= 6x + 7 \\ 6x - 6x - 12 &= 6x - 6x + 7 \quad (\text{Subtract } 6x \text{ from both sides}) \\ -12 &= 7 \quad \mathbf{\text{Contradiction (False!)}} \end{aligned}$$

Conclusion: No Solution (there is no value of $x$ in existence that can make $-12 = 7$).

Special Case B (Identity / Infinitely Many Solutions): Solve $4(3x + 2) - 5 = 12x + 3$.

$$\begin{aligned} 12x + 8 - 5 &= 12x + 3 \\ 12x + 3 &= 12x + 3 \\ 12x - 12x + 3 &= 12x - 12x + 3 \\ 3 &= 3 \quad \mathbf{\text{Identity (Always True!)}} \end{aligned}$$

Conclusion: Infinitely Many Solutions (any real number substituted for $x$ will satisfy the equation).

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5. Simple Rational & Radical Equations and Extraneous Solutions

The Level A specification includes solving simple rational and radical equations in one variable and giving examples showing how extraneous solutions may arise. Both types are solved by the same balance principle, with one mandatory extra step: every solution must be checked in the original equation.

Rational equations (a variable in a denominator)

Multiply both sides by the least common denominator to clear the fractions, then solve normally.

Worked example. Solve $\dfrac{12}{x} + 3 = 7$.

$$\frac{12}{x} = 4 \;\Rightarrow\; 12 = 4x \;\Rightarrow\; x = 3$$

Check: $\dfrac{12}{3} + 3 = 4 + 3 = 7$. ✓

Worked example with an excluded value. Solve $\dfrac{x}{x-4} = \dfrac{4}{x-4} + 3$.

Multiply through by $(x - 4)$:

$$x = 4 + 3(x - 4) \;\Rightarrow\; x = 4 + 3x - 12 \;\Rightarrow\; -2x = -8 \;\Rightarrow\; x = 4$$

Check: substituting $x = 4$ makes the denominator $x - 4$ equal to zero, which is undefined. So $x = 4$ is an extraneous solution and the equation has no solution.

Why extraneous solutions appear. Multiplying both sides by an expression containing the variable is not a reversible move when that expression can equal zero. The algebra produces a candidate; only the check confirms it. Before you begin, note the excluded values — the values that make any denominator zero — and reject any candidate that matches one.

Radical equations (a variable under a root)

Isolate the radical, then square both sides.

Worked example. Solve $\sqrt{2x + 7} = 5$.

$$2x + 7 = 25 \;\Rightarrow\; 2x = 18 \;\Rightarrow\; x = 9$$

Check: $\sqrt{2(9) + 7} = \sqrt{25} = 5$. ✓

Worked example producing an extraneous root. Solve $\sqrt{x + 6} = x$.

$$x + 6 = x^2 \;\Rightarrow\; x^2 - x - 6 = 0 \;\Rightarrow\; (x - 3)(x + 2) = 0$$

Candidates: $x = 3$ and $x = -2$.

The only solution is $x = 3$. Squaring both sides created the second candidate because squaring destroys sign information: $(-2)^2$ and $2^2$ are both 4, so the squared equation cannot tell them apart.

The rule to carry into the test: whenever you multiply by a variable expression or square both sides, substitute every answer back into the original equation. A candidate that fails the check is discarded, and an equation can legitimately end up with no solution at all.

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6. Real-World Applications: Break-Even & Comparative Cost Analysis

A major TABE application is determining the exact point at which two competing pricing structures, rental plans, or production methods cost the exact same amount (break-even analysis).

Practical Workplace Example: Equipment Rental Comparison

Problem: A general contractor needs to rent an industrial generator.

• Supplier A charges a $\text{\textdollar}60$ pickup fee plus $\text{\textdollar}35$ per day.
• Supplier B charges a $\text{\textdollar}120$ pickup fee plus $\text{\textdollar}25$ per day. For how many rental days $d$ will the total cost from both suppliers be identical?

1. Formulate Equations: $$\text{Cost}_A = 60 + 35d \qquad \text{Cost}_B = 120 + 25d$$
2. Equate the Costs: $$60 + 35d = 120 + 25d$$
3. Solve for $d$: $$\begin{aligned} 35d - 25d &= 120 - 60 \\ 10d &= 60 \\ d &= 6 \text{ days} \end{aligned}$$
4. Financial Interpretation: At exactly $6$ days, both suppliers charge $60 + 35(6) = \text{\textdollar}270$. For rentals under 6 days, Supplier A is more economical; for long-term rentals exceeding 6 days, Supplier B's lower daily rate makes it cheaper.