6.3 Algebra, Equations & Inequalities

Algebra, Equations & Inequalities

Algebra is the language of quantitative problem-solving on the EA. Your ability to quickly set up, manipulate, and solve equations and inequalities will directly impact your speed and accuracy. Many problems that appear to be complex logic puzzles are trivialized once translated into algebraic equations.

Linear Equations and Systems

A linear equation involves variables raised only to the first power (e.g., 3x + 5 = 14). Solving for a single variable involves performing identical operations on both sides of the equation to isolate the variable.

When you have two variables, you need a system of two independent equations to find a unique solution. There are two primary methods to solve systems:

1. Substitution: Isolate one variable in one equation, then substitute that expression into the other equation. Best when one variable already has a coefficient of 1.
2. Elimination: Multiply one or both equations by a constant so that when the equations are added together, one variable cancels out. Best when variables are stacked with integer coefficients.

Example of Elimination:

1. 2x + 3y = 12
2. 5x - 3y = 9 Add the equations together: (2x + 5x) + (3y - 3y) = 12 + 9 -> 7x = 21 -> x = 3. Substitute x = 3 back into equation 1: 2(3) + 3y = 12 -> 6 + 3y = 12 -> 3y = 6 -> y = 2.

Example of Substitution:

1. y = 3x - 4
2. 2x + y = 11 Equation 1 already isolates y, so substitute that expression directly into equation 2: 2x + (3x - 4) = 11 -> 5x - 4 = 11 -> 5x = 15 -> x = 3. Substitute x = 3 back into equation 1: y = 3(3) - 4 = 5. So x = 3 and y = 5 (check: 2(3) + 5 = 11, confirmed).

Choose whichever method requires fewer steps: substitution is fastest when a variable is already isolated (as above), while elimination is fastest when coefficients already match or are easy multiples of each other.

Quadratic Equations

A quadratic equation takes the form ax² + bx + c = 0, where a ≠ 0. A quadratic equation can have two, one, or zero real solutions.

The most common way to solve quadratics on the EA is by factoring. For example, to solve x² - 5x + 6 = 0, you look for two numbers that multiply to +6 and add to -5. Those numbers are -2 and -3. So, (x - 2)(x - 3) = 0. This means x = 2 or x = 3.

Special Quadratic Forms

You MUST memorize these three special products; recognizing them will save you immense time:

1. Difference of Squares: a² - b² = (a + b)(a - b)
2. Perfect Square (Addition): (a + b)² = a² + 2ab + b²
3. Perfect Square (Subtraction): (a - b)² = a² - 2ab + b²

Whenever you see x² - 16, you should immediately think (x + 4)(x - 4). The EA often hides these within complex fractions to see if you can factor and cancel terms.

Algebraic Inequalities

Inequalities (<, >, ≤, ≥) are solved similarly to equations, with one absolutely critical rule: When you multiply or divide an inequality by a negative number, you MUST flip the direction of the inequality sign.

Example: -3x + 4 > 10 -3x > 6 Divide by -3 and flip the sign: x < -2.

Furthermore, be extremely careful when multiplying or dividing by a variable in an inequality (e.g., x or y) unless you are absolutely certain of its sign. If you don't know whether x is positive or negative, you cannot multiply or divide by it without creating multiple scenarios, because you don't know whether to flip the sign.

Absolute Value Equations and Inequalities

Absolute value denotes distance from zero. To solve |ax + b| = c, set up two equations:

1. ax + b = c
2. ax + b = -c

Important: After solving both, plug the answers back into the original equation to check for extraneous solutions (solutions that algebraically emerge but don't work in the original context, often because absolute value must equal a non-negative number).

For inequalities, think of distance:

• |x| < 5 means "the distance between x and 0 is less than 5." Thus, -5 < x < 5.
• |x| > 5 means "the distance between x and 0 is greater than 5." Thus, x < -5 OR x > 5.

Squaring Both Sides: A Second Source of Extraneous Roots

Radical equations, where the variable sits under a square root (such as √(x + 3) = x - 3), require squaring both sides to eliminate the radical. This step is algebraically valid but not reversible — squaring can turn a false statement into a true one, silently introducing extraneous roots.

Example: Solve √(x + 3) = x - 3. Square both sides: x + 3 = (x - 3)² = x² - 6x + 9, which rearranges to x² - 7x + 6 = 0, or (x - 1)(x - 6) = 0, giving x = 1 or x = 6.

Check both candidates in the ORIGINAL equation: x = 1 gives √4 = 2, but x - 3 = -2 (a mismatch, so it's extraneous); x = 6 gives √9 = 3, and x - 3 = 3 (a match, so it's valid). Only x = 6 actually solves the equation.

Trap Alert: Whenever you square both sides to clear a radical, plug every resulting solution back into the original equation before accepting it. The EA often lists the extraneous root as a tempting wrong-answer choice.

Translating Word Problems

The most challenging part of EA algebra is often setting up the equation from a word problem. Use consistent variables (e.g., a for apples, b for bananas) and translate piece by piece.

Example: A rental car company charges a flat fee of $35 plus $0.20 per mile driven. If Maria's total bill was $63, how many miles did she drive? Let m = miles driven: Total Cost = Flat Fee + (Rate × Miles). 35 + 0.20m = 63 -> 0.20m = 28 -> m = 140 miles.

The flat fee is fixed and becomes the constant term; the per-mile rate is what scales with the variable, becoming its coefficient. Spotting which quantity is fixed and which scales is often the real challenge in translation, not the algebra itself.

Practice translation extensively. Once the words become variables and operators, the algebraic mechanics are usually straightforward.