11.4 DS Applied to Algebraic Equations, Systems, and Inequalities

11.4 DS Applied to Algebraic Equations, Systems, and Inequalities

Quick Summary: Algebraic Data Sufficiency questions assess whether mathematical constraints restrict variables to a unique solution or definitive relationship. The naive rule of "$N$ equations for $N$ variables" fails in DS due to linear dependency and combo shortcuts. In quadratics, single roots behave differently in Value versus Yes/No questions. In inequalities, multiplying by variables of unknown sign or invalidly subtracting inequalities are lethal errors.

---

The "N Equations for N Unknowns" Myth and Linear Dependency

In standard high-school algebra, students are taught a rule of thumb: "To solve for $N$ variables, you need $N$ distinct equations." On the GMAT, GMAC deliberately preys upon candidates who apply this heuristic mechanically.

Having two equations and two unknown variables does not guarantee a unique solution. Given a system of two linear equations: $\begin{cases} A_1 x + B_1 y = C_1 \\ A_2 x + B_2 y = C_2 \end{cases}$

Three distinct geometric relationships can occur:

1. Linearly Independent (Unique Solution): The lines intersect at exactly one point $(x, y)$. This occurs if and only if the coefficient ratios are unequal: $\frac{A_1}{A_2} \neq \frac{B_1}{B_2}$ Sufficiency Status: SUFFICIENT to find unique values for both $x$ and $y$.
2. Linearly Dependent / Coincident (Infinitely Many Solutions): Both equations represent the exact same line. This occurs when all ratios are identical: $\frac{A_1}{A_2} = \frac{B_1}{B_2} = \frac{C_1}{C_2}$ Example: $2x + 3y = 7$ and $4x + 6y = 14$. The second equation is simply the first equation multiplied by 2. It provides zero new information. Sufficiency Status: INSUFFICIENT to determine unique values for $x$ and $y$.
3. Inconsistent / Parallel (Zero Solutions): The lines are parallel and distinct ($\frac{A_1}{A_2} = \frac{B_1}{B_2} \neq \frac{C_1}{C_2}$). If equations derived from the two statements form an inconsistent system, recheck the stem constraints and derivation because a well-formed combined case must describe a coherent situation.

The Quick Ratio Check: Whenever you are presented with two linear equations in two variables on Data Sufficiency, immediately compute the ratio of the $x$-coefficients and the ratio of the $y$-coefficients. If $\frac{A_1}{A_2} \neq \frac{B_1}{B_2}$, you have a unique solution. Do not waste time actually solving for $x$ and $y$!

The Algebraic Combo Shortcut in Systems

The converse of the linear dependency trap is the Combo Shortcut. Frequently, a Data Sufficiency prompt does not ask for the individual values of $x$ or $y$; it asks for an algebraic combination such as $x + y$, $2x - y$, or $3x + 5y$.

In such cases, a single equation with two unknown variables can be completely SUFFICIENT.

• Consider the question: "What is the value of $x + y$?"
• Statement (1) gives: $4x + 4y = 28$.
• A candidate suffering from the "two variables require two equations" myth will look at Statement (1), see two unknowns with only one equation, and incorrectly declare it insufficient.
• The GMAT master simply divides the equation by 4: $4(x + y) = 28 \implies x + y = 7$.
• Statement (1) provides the exact target value uniquely and is 100% sufficient on its own!

---

Quadratic Equations: Single vs. Multiple Roots

Quadratic equations ($ax^2 + bx + c = 0$) appear constantly in Data Sufficiency. How they behave depends entirely on whether the question is a Value question or a Yes/No question.

1. Quadratics in Value Questions

In a Value question ("What is the value of $x$?"), a quadratic equation is insufficient if it yields two distinct real roots.

• Example: $x^2 - 5x + 6 = 0 \implies (x - 2)(x - 3) = 0 \implies x = 2 \text{ or } x = 3$. Two values $\implies$ INSUFFICIENT.

However, a quadratic statement is SUFFICIENT under two specific conditions:

1. The Perfect Square Binomial (Single Repeated Root): $x^2 - 6x + 9 = 0 \implies (x - 3)^2 = 0 \implies x = 3$ There is only one root. The value is unique $\implies$ SUFFICIENT.
2. External Constraints Eliminating One Root: If the question stem states that $x$ represents the width of a rectangle (or that $x > 0$), and Statement (1) yields $x^2 + 2x - 15 = 0 \implies (x + 5)(x - 3) = 0$, the roots are $-5$ and $3$. Because width cannot be negative, $x = 3$ uniquely $\implies$ SUFFICIENT.

2. Quadratics in Yes/No Questions

In a Yes/No question ("Is $x > 0$?"), having two distinct roots does not automatically mean insufficient! You must test both roots against the target question:

• If Statement (1) yields roots $x = 2$ and $x = 5$, both numbers are strictly greater than 0. The answer to "Is $x > 0$?" is a definitive YES in both cases! Thus, Statement (1) is SUFFICIENT.
• Only if one root produces a YES while the other root produces a NO (e.g., $x = -2$ and $x = 5$) is the statement insufficient.

---

Inequalities in DS: The Illegal Operations

Inequalities are among the most heavily tested and error-prone topics on the GMAT Focus Edition. When evaluating inequalities in Data Sufficiency, you must obey three strict operational laws:

Law 1: Never Multiply or Divide by an Unknown Sign

If you have the inequality $\frac{x}{y} > 2$, you cannot simply multiply both sides by $y$ to get $x > 2y$.

• If $y > 0$, multiplying by $y$ yields $x > 2y$.
• If $y < 0$, multiplying by $y$ reverses the inequality, yielding $x < 2y$.
• If the sign of $y$ is unknown, clearing the denominator is an illegal operation that invalidates your analysis.

The Legal Workaround: If you must clear a variable denominator whose sign is unknown, multiply both sides by $y^2$ (which is guaranteed to be positive since $y \neq 0$): $\frac{x}{y} > 2 \implies \frac{x}{y} \cdot y^2 > 2 \cdot y^2 \implies xy > 2y^2$

Law 2: Squaring Inequalities Requires Positive Verification

You can only square both sides of an inequality ($a > b \implies a^2 > b^2$) if you are 100% certain that both sides are non-negative ($a > b \ge 0$).

• If both sides are positive: $5 > 3 \implies 25 > 9$ (Valid).
• If negative numbers are involved: $2 > -5$, but $2^2 < (-5)^2$ since $4 < 25$ (Inequality flips!).
• If $-4 > -7$, then $(-4)^2 < (-7)^2$ since $16 < 49$ (Inequality flips!). Unless both quantities are confirmed positive, squaring an inequality is invalid.

Law 3: Combining Inequalities (Adding vs. Subtracting)

• ADDING inequalities facing the SAME direction is ALWAYS LEGAL: $\begin{cases} a > b \\ c > d \end{cases} \implies a + c > b + d \quad (\text{Guaranteed True})$
• SUBTRACTING inequalities facing the same direction is NEVER LEGAL: Consider $10 > 2$ and $8 > 1$. If you subtract them: $10 - 8 = 2$, and $2 - 1 = 1$. Here $2 > 1$ happens to hold. But now consider $10 > 2$ and $9 > 1$. Subtracting gives $10 - 9 = 1$ on the left and $2 - 1 = 1$ on the right, which gives $1 > 1$, which is false! Never subtract inequalities.

---

Step-by-Step Worked DS Examples

Worked Example 1: Linear Independence vs. Combo Shortcut

Problem: What is the value of $x + y$?

• Statement (1): $2x - y = 8$
• Statement (2): $3x + 3y = 21$

Step-by-Step Solution:

1. Analyze Target: We need the unique value of the composite quantity $(x + y)$.
2. Evaluate Statement (1) Alone: $2x - y = 8$. Can we manipulate $2x - y$ into $x + y$ without another equation? No, the coefficients of $x$ and $y$ are in the ratio $2 : -1$, whereas $x + y$ requires a ratio of $1 : 1$. Statement (1) allows infinitely many pairs of $(x, y)$, each yielding a different value for $x + y$ (e.g., $x=4, y=0 \implies x+y=4$; $x=5, y=2 \implies x+y=7$). Statement (1) is insufficient.
   • Scratchpad: Eliminate AD. Retain BCE.
3. Evaluate Statement (2) Alone (Sterile Isolation): $3x + 3y = 21$. Notice the common factor of 3 across all terms! Factor out 3: $3(x + y) = 21 \implies x + y = \frac{21}{3} = 7$ Statement (2) directly isolates the requested combination $x + y = 7$ uniquely! Statement (2) ALONE is sufficient.
4. Final Verdict: Statement (2) alone is sufficient, but Statement (1) alone is not sufficient. The correct answer is Choice B.

---

Worked Example 2: Quadratic Equation in a Yes/No Setting

Problem: Is $x > 0$?

• Statement (1): $x^2 - 7x + 10 = 0$
• Statement (2): $x^2 - 3x - 10 = 0$

Step-by-Step Solution:

1. Analyze Target: This is a Yes/No question: "Is $x > 0$?" A definitive YES or a definitive NO is sufficient.
2. Evaluate Statement (1) Alone: Factor the quadratic: $(x - 2)(x - 5) = 0 \implies x = 2 \text{ or } x = 5$.
   • If $x = 2$: Is $2 > 0$? YES.
   • If $x = 5$: Is $5 > 0$? YES. Even though Statement (1) yields two distinct numerical roots, both roots produce a definitive YES to the target question! Statement (1) ALONE is sufficient.
   • Scratchpad: Retain AD. Eliminate BCE.
3. Evaluate Statement (2) Alone: Wipe Statement 1 completely. Factor Statement (2): $x^2 - 3x - 10 = 0 \implies (x - 5)(x + 2) = 0 \implies x = 5 \text{ or } x = -2$
   • If $x = 5$: Is $5 > 0$? YES.
   • If $x = -2$: Is $-2 > 0$? NO. Because Statement (2) yields both YES and NO, it produces a MAYBE. Statement (2) is insufficient.
4. Final Verdict: Statement (1) alone is sufficient, but Statement (2) alone is not sufficient. The correct answer is Choice A.

---

High-Frequency Algebraic Traps

• Trap 1: The Equation-Counting Fallacy
  Assuming two equations with two variables are sufficient without testing coefficient ratios. If the equations are linearly dependent (coincident lines), they yield infinitely many solutions.

• Trap 2: The Variable in Denominator Trap
  Multiplying both sides of an inequality by an algebraic expression with an unknown sign (such as $k$ or $x - 3$), forgetting that negative values flip the inequality direction.

• Trap 3: The Inequality Subtraction Fallacy
  Subtracting one inequality from another. Adding inequalities facing the same direction is mathematically valid; subtracting them is completely invalid.