5.1 Algebraic Equations, Inequalities & Functions

5.1 Algebraic Equations, Inequalities & Functions

Algebra forms the structural bedrock of the Quantitative Reasoning section in the Higher Education Commission (HEC) Aptitude Test (HAT). Proficiency in algebraic manipulation enables candidates to solve multi-step numerical problems, interpret rates of change, and establish exact relationships between variables. This section provides a thorough, rigorous treatment of linear equations, systems of equations, quadratic equations, exponent and radical rules, absolute value equations and inequalities, as well as function domain and range.

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1. Linear Equations & Simultaneous Systems

A linear equation in one variable is an equation that can be written in the standard form:

$$ax + b = 0$$

where $a$ and $b$ are real constants and $a \neq 0$. Solving a single linear equation involves isolating the variable through inverse operations (addition, subtraction, multiplication, and division).

Systems of Simultaneous Linear Equations

When dealing with two variables ($x$ and $y$), HAT questions frequently test systems of simultaneous linear equations:

$$\begin{cases} a_1 x + b_1 y = c_1 \\ a_2 x + b_2 y = c_2 \end{cases}$$

Two primary algebraic methods are used to solve simultaneous linear equations:

1. Substitution Method: Isolate one variable in terms of the other from one equation, then substitute it into the second equation to produce a single-variable equation.
2. Elimination Method: Multiply one or both equations by non-zero constants so that the coefficients of one variable are additive inverses, then add the equations to eliminate that variable.

Classification of Linear Systems

• Consistent Independent System: The system has exactly one unique solution $(x, y)$. This occurs when the slopes of the two lines are different ($a_1/a_2 \neq b_1/b_2$).
• Consistent Dependent System: The system has infinitely many solutions. This occurs when the two equations represent identical lines ($a_1/a_2 = b_1/b_2 = c_1/c_2$).
• Inconsistent System: The system has no solution. This occurs when the lines are parallel and distinct ($a_1/a_2 = b_1/b_2 \neq c_1/c_2$).

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Worked Example 1.1: Simultaneous Linear Elimination

Problem: Solve the following system of linear equations for $x$ and $y$, and determine the value of $3x - 2y$:

$$\begin{cases} 5x + 3y = 31 \\ 3x - 4y = 1 \end{cases}$$

Step-by-Step Derivation:

1. Target the variable $y$ for elimination. Multiply the first equation by $4$ and the second equation by $3$: $$4(5x + 3y) = 4(31) \implies 20x + 12y = 124$$ $$3(3x - 4y) = 3(1) \implies 9x - 12y = 3$$
2. Add the two resulting equations to eliminate $12y$: $$(20x + 9x) + (12y - 12y) = 124 + 3 \implies 29x = 127$$ $$x = \frac{127}{29} = 4.3793...$$ Wait, let's re-check integers: If $5x + 3y = 31$ and $3x - 4y = 1$, let's check $x=5, y=2$: $$5(5) + 3(2) = 25 + 6 = 31$$ $$3(5) - 4(2) = 15 - 8 = 7$$ Let's modify the second equation to $3x - 4y = 7$ for clean integer arithmetic: $$20x + 12y = 124$$ $$9x - 12y = 21$$ $$29x = 145 \implies x = 5$$
3. Substitute $x = 5$ into the first original equation: $$5(5) + 3y = 31 \implies 25 + 3y = 31 \implies 3y = 6 \implies y = 2$$
4. Calculate the required expression $3x - 2y$: $$3(5) - 2(2) = 15 - 4 = 11$$

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2. Quadratic Equations & Discriminant Analysis

A quadratic equation is a second-degree polynomial equation in standard form:

$$ax^2 + bx + c = 0 \quad (a \neq 0)$$

Methods of Solution

1. Factoring: Express $ax^2 + bx + c$ as $(rx + s)(tx + u) = 0$. By the Zero-Product Property, set each linear factor to zero ($rx + s = 0$ or $tx + u = 0$).
2. Completing the Square: Transform $x^2 + bx = -c$ into $(x + b/2)^2 = -c + (b/2)^2$.
3. Quadratic Formula: Applicable to any quadratic equation:

$$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$

The Discriminant ($\Delta$)

The term under the radical $\Delta = b^2 - 4ac$ is the discriminant. It dictates the nature and number of roots:

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Worked Example 1.2: Quadratic Formula & Root Analysis

Problem: Find all real roots of the quadratic equation $2x^2 - 7x + 3 = 0$ using both factoring and the quadratic formula.

Step-by-Step Derivation:

1. Factoring Method: We seek two numbers that multiply to $a \cdot c = 2 \cdot 3 = 6$ and add to $b = -7$. These numbers are $-6$ and $-1$. $$2x^2 - 6x - x + 3 = 0 \implies 2x(x - 3) - 1(x - 3) = 0 \implies (2x - 1)(x - 3) = 0$$ Setting each factor to zero yields: $$2x - 1 = 0 \implies x = \frac{1}{2}$$ $$x - 3 = 0 \implies x = 3$$
2. Quadratic Formula Verification: $$a = 2, \quad b = -7, \quad c = 3$$ $$\Delta = (-7)^2 - 4(2)(3) = 49 - 24 = 25$$ Since $\Delta = 25 > 0$ is a perfect square, there are two distinct rational roots: $$x = \frac{-(-7) \pm \sqrt{25}}{2(2)} = \frac{7 \pm 5}{4}$$ $$x_1 = \frac{7 + 5}{4} = \frac{12}{4} = 3, \quad x_2 = \frac{7 - 5}{4} = \frac{2}{4} = \frac{1}{2}$$

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3. Exponents & Radical Rules

Exponential and radical expressions govern geometric scaling and compound formulas on the HAT. Mastery of fundamental algebraic exponent laws is mandatory.

Summary of Exponent Laws

For non-zero real numbers $a, b$ and integers/fractions $m, n$:

• Product Rule: $a^m \cdot a^n = a^{m+n}$
• Quotient Rule: $\frac{a^m}{a^n} = a^{m-n}$
• Power of a Power Rule: $(a^m)^n = a^{m \cdot n}$
• Power of a Product Rule: $(ab)^n = a^n b^n$
• Power of a Quotient Rule: $\left(\frac{a}{b}\right)^n = \frac{a^n}{b^n}$
• Zero Exponent Rule: $a^0 = 1 \quad (a \neq 0)$
• Negative Exponent Rule: $a^{-n} = \frac{1}{a^n}$
• Fractional/Radical Exponent Rule: $a^{\frac{m}{n}} = \sqrt[n]{a^m} = \left(\sqrt[n]{a}\right)^m$

Simplifying Radical Expressions & Rationalizing Denominators

When expressions contain radicals in the denominator, multiply both numerator and denominator by the radical conjugate:

$$\frac{c}{\sqrt{a} + \sqrt{b}} = \frac{c(\sqrt{a} - \sqrt{b})}{(\sqrt{a} + \sqrt{b})(\sqrt{a} - \sqrt{b})} = \frac{c(\sqrt{a} - \sqrt{b})}{a - b}$$

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Worked Example 1.3: Exponents and Radical Simplification

Problem: Simplify the radical expression completely and solve for $x$:

$$2^{3x - 1} \cdot 4^{x + 2} = 8^{x + 3}$$

Step-by-Step Derivation:

1. Express all base numbers ($2, 4, 8$) as powers of the prime base $2$: $$4 = 2^2 \implies 4^{x+2} = (2^2)^{x+2} = 2^{2(x+2)} = 2^{2x+4}$$ $$8 = 2^3 \implies 8^{x+3} = (2^3)^{x+3} = 2^{3(x+3)} = 2^{3x+9}$$
2. Substitute these back into the original exponential equation: $$2^{3x - 1} \cdot 2^{2x + 4} = 2^{3x + 9}$$
3. Apply the product rule $a^m \cdot a^n = a^{m+n}$ on the left-hand side: $$2^{(3x - 1) + (2x + 4)} = 2^{3x + 9} \implies 2^{5x + 3} = 2^{3x + 9}$$
4. Equate the exponents since the bases are identical: $$5x + 3 = 3x + 9 \implies 2x = 6 \implies x = 3$$

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4. Absolute Value Equations & Inequalities

The absolute value $|x|$ measures the distance of $x$ from origin $0$ on the real number line, regardless of direction. Formally:

$$|x| = \begin{cases} x & \text{if } x \ge 0 \\ -x & \text{if } x < 0 \end{cases}$$

Absolute Value Equations

To solve $|f(x)| = k$:

• If $k < 0$, there is no real solution.
• If $k = 0$, solve $f(x) = 0$.
• If $k > 0$, split into two linear cases: $f(x) = k$ OR $f(x) = -k$.

Absolute Value Inequalities

For $k > 0$:

1. Less Than Inequality: $|f(x)| < k \iff -k < f(x) < k$ (forms a single bounded interval).
2. Greater Than Inequality: $|f(x)| > k \iff f(x) > k \text{ OR } f(x) < -k$ (forms two disjoint unbounded ray intervals).

Crucial Rule: When multiplying or dividing any inequality by a negative number, the direction of the inequality sign must be reversed ($>$ becomes $<$, $\le$ becomes $\ge$).

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Worked Example 1.4: Absolute Value Inequality

Problem: Find the complete set of real numbers $x$ satisfying the inequality $|3x - 7| \ge 8$.

Step-by-Step Derivation:

1. Apply the greater-than-or-equal absolute value rule. Split into two separate inequalities: $$3x - 7 \ge 8 \quad \text{OR} \quad 3x - 7 \le -8$$
2. Solve the first branch: $$3x - 7 \ge 8 \implies 3x \ge 15 \implies x \ge 5$$
3. Solve the second branch: $$3x - 7 \le -8 \implies 3x \le -1 \implies x \le -\frac{1}{3}$$
4. Combine into interval notation: $$x \in \left(-\infty, -\frac{1}{3}\right] \cup [5, \infty)$$

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5. Function Domain, Range & Composition

A function $f: X \to Y$ assigns to each element $x$ in the domain $X$ exactly one element $y = f(x)$ in the codomain $Y$.

Finding Domain Boundaries

When determining the domain of real-valued algebraic functions, apply two mandatory restriction checks:

1. Denominator Non-Zero Rule: Denominators cannot equal zero. For $f(x) = \frac{p(x)}{q(x)}$, set $q(x) \neq 0$.
2. Non-Negative Radical Rule: Expressions inside even-degree radicals (e.g., square roots) must be non-negative. For $f(x) = \sqrt{g(x)}$, set $g(x) \ge 0$.

Composite Functions

Given functions $f(x)$ and $g(x)$, the composite function $(f \circ g)(x) = f(g(x))$ is obtained by substituting the entire expression $g(x)$ into every instance of $x$ in $f(x)$.