3.2 Basic Algebra, Linear Equations, and Inequalities

Algebraic Foundations and Exponent Rules

Algebra involves representing numbers with symbols and manipulating expressions using properties of real numbers. Mastering the laws of exponents and algebraic expansion is critical for success on the LET.

Laws of Exponents

For all real numbers $x, y$ (where bases are non-zero when in denominators or raised to zero/negative powers) and integers $m, n$:

Polynomial Expansion and Factoring

Factoring simplifies complex expressions and enables solving higher-degree equations.

1. Greatest Common Monomial Factor (GCMF): $ab + ac = a(b + c)$
2. Difference of Two Squares: $a^2 - b^2 = (a - b)(a + b)$
3. Perfect Square Trinomials:
   $a^2 + 2ab + b^2 = (a + b)^2$ $a^2 - 2ab + b^2 = (a - b)^2$
4. Sum and Difference of Two Cubes:
   $a^3 + b^3 = (a + b)(a^2 - ab + b^2)$ $a^3 - b^3 = (a - b)(a^2 + ab + b^2)$

Worked Example 1: Simplifying Rational Algebraic Expressions

Problem: Simplify the expression: $\frac{x^2 - 9}{x^2 + 5x + 6}$.

Step 1: Factor the numerator using Difference of Squares.
$x^2 - 9 = (x - 3)(x + 3)$

Step 2: Factor the denominator trinomial.
Find two numbers that multiply to $6$ and add to $5 \implies 2$ and $3$.
$x^2 + 5x + 6 = (x + 2)(x + 3)$

Step 3: Cancel common factors.
$\frac{(x - 3)(x + 3)}{(x + 2)(x + 3)} = \frac{x - 3}{x + 2} \quad (\text{provided } x \neq -3, -2)$

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Linear Equations and Inequalities

A linear equation in one variable is an equation that can be written in the form $ax + b = 0$ ($a \neq 0$). A linear inequality replaces the equals sign with $<, \le, >, \text{ or } \ge$.

Solving Linear Inequalities and the Negative Rule Trap

When solving linear inequalities, follow standard algebraic properties of equality, with one critical exception:

CRITICAL RULE: When multiplying or dividing both sides of an inequality by a negative number, you MUST reverse the direction of the inequality symbol.

Worked Example 2: Linear Inequality with Sign Inversion

Problem: Solve for $x$: $5 - 3(x + 2) \ge 4x + 20$.

Step 1: Distribute and combine like terms on the left side.
$5 - 3x - 6 \ge 4x + 20$ $-3x - 1 \ge 4x + 20$

Step 2: Isolate variable terms on one side.
Subtract $4x$ from both sides:
$-7x - 1 \ge 20$ Add $1$ to both sides:
$-7x \ge 21$

Step 3: Divide by $-7$ and REVERSE the inequality sign.
$x \le \frac{21}{-7} \implies x \le -3$

Solution in Interval Notation: $(-\infty, -3]$

Standard LET Algebraic Word Problems

Word problems comprise a major share of the LET math subtest. Mastering systematic setups for age, work, motion, and mixture problems ensures accurate solutions.

1. Age Problems

Age problems involve establishing algebraic relationships across different time frames (past, present, future).

Worked Example 3: Age Problem Setup

Problem: A mother is currently four times as old as her daughter. In 12 years, the mother will be twice as old as her daughter will be then. How old is the daughter now?

Step 1: Set up a variable table.
Let $x$ = Daughter's present age.
Mother's present age = $4x$.

Step 2: Translate the future condition into an equation.
$\text{Mother's Future Age} = 2 \times (\text{Daughter's Future Age})$ $4x + 12 = 2(x + 12)$

Step 3: Solve for $x$.
$4x + 12 = 2x + 24$ $2x = 12 \implies x = 6$

Final Answer: The daughter is currently 6 years old (and the mother is 24 years old).

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2. Work Rate Problems

Work problems involve individual rates of completing a task.
$\text{Work Rate} = \frac{1}{\text{Time to complete job alone}}$ $\text{Combined Work Formula: } \frac{1}{T_{\text{together}}} = \frac{1}{t_1} + \frac{1}{t_2} + \dots$

Worked Example 4: Combined Work Rate

Problem: Pipe A can fill a water tank in 4 hours, while Pipe B can fill the same tank in 6 hours. If both pipes are opened simultaneously, how long will it take to fill the tank?

Step 1: Determine individual hourly work rates.
Pipe A rate = $\frac{1}{4}$ tank/hour; Pipe B rate = $\frac{1}{6}$ tank/hour.

Step 2: Sum the rates to get the combined rate.
$\text{Combined Rate} = \frac{1}{4} + \frac{1}{6} = \frac{3}{12} + \frac{2}{12} = \frac{5}{12}\text{ tank/hour}$

Step 3: Calculate total time $T$ (reciprocal of combined rate).
$T = \frac{12}{5} = 2.4\text{ hours} = 2\text{ hours and } 24\text{ minutes}$

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3. Quadratic Equations and the Discriminant

A quadratic equation is a second-degree polynomial equation: $ax^2 + bx + c = 0$ ($a \neq 0$).

$\text{Quadratic Formula: } x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ $\text{Discriminant: } \Delta = b^2 - 4ac$

Worked Example 5: Solving Quadratic Equations via Factoring

Problem: Solve for $x$: $2x^2 - 7x + 3 = 0$.

Step 1: Factor the quadratic expression.
Look for factors of $a \cdot c = 2 \cdot 3 = 6$ that sum to $b = -7 \implies -6$ and $-1$.
$2x^2 - 6x - x + 3 = 0$ $2x(x - 3) - 1(x - 3) = 0$ $(2x - 1)(x - 3) = 0$

Step 2: Set each factor to zero.
$2x - 1 = 0 \implies x = \frac{1}{2}$ $x - 3 = 0 \implies x = 3$

Final Answer: $x = \frac{1}{2}, 3$