1.2 Algebra & Equations

1.2 Algebra & Equations

Algebra is the foundation of the PMAEE Mathematics subtest. To succeed, candidates must solve equations quickly and accurately. This section reviews key algebraic structures: linear, quadratic, systems, fractions, exponents, and radicals.

Linear Equations

A linear equation is an algebraic equation of the first degree, meaning the highest exponent of the variable is 1. The standard form of a linear equation in one variable is:

$$ax + b = 0$$

where $a$ and $b$ are real numbers, and $a \neq 0$.

Solving Linear Equations

Solving a linear equation involves isolating the variable on one side. The golden rule is: whatever operation is performed on one side must also be performed on the other.

Steps to solve multi-step linear equations:

1. Simplify: Clear parentheses using the distributive property and combine like terms.
2. Isolate variable terms: Use addition or subtraction to move all terms containing the variable to one side, and constant terms to the other.
3. Solve: Use multiplication or division to isolate the variable (coefficient of 1).

Worked Example: Fractional Linear Equation

Solve for $x$:

$$\frac{2x - 3}{4} + \frac{x}{3} = 2$$

Solution: First, find the least common denominator (LCD) of 4 and 3, which is 12. Multiply the entire equation by 12 to eliminate fractions:

$$12 \cdot \left(\frac{2x - 3}{4}\right) + 12 \cdot \left(\frac{x}{3}\right) = 12 \cdot 2$$ $$3(2x - 3) + 4(x) = 24$$

Distribute:

$$6x - 9 + 4x = 24$$

Combine like terms:

$$10x - 9 = 24$$

Add 9 to both sides:

$$10x = 33$$

Divide by 10:

$$x = 3.3 \text{ or } \frac{33}{10}$$

Quadratic Equations

A quadratic equation is a polynomial equation of the second degree. The standard form is:

$$ax^2 + bx + c = 0$$

where $a, b, c$ are constants and $a \neq 0$.

Methods of Solving

1. Factoring: Find two numbers that multiply to $ac$ and add to $b$. Use the zero-product property: if $AB = 0$, then $A=0$ or $B=0$.
2. Quadratic Formula: Always works, even when factoring is difficult or impossible:

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

The Discriminant

The expression under the radical in the quadratic formula, $b^2 - 4ac$, is the discriminant ($D$). It determines the nature of the roots:

Worked Example: One Real Root

For what value(s) of $k$ does the equation $3x^2 + kx + 12 = 0$ have exactly one real root?

Solution: For exactly one real root, the discriminant must equal 0:

$$b^2 - 4ac = 0$$

Identify $a = 3$, $b = k$, $c = 12$:

$$k^2 - 4(3)(12) = 0$$ $$k^2 - 144 = 0$$ $$k^2 = 144 \Rightarrow k = \pm 12$$

Systems of Linear Equations

A system of linear equations consists of two or more equations with the same set of variables. PMAEE focuses on two-variable systems.

Methods of Solving

• Substitution: Solve one equation for one variable and substitute the result into the other equation.
• Elimination: Multiply one or both equations by constants so that adding or subtracting them eliminates one variable.

Solution Types

1. Consistent and Independent: One unique solution (lines intersect at a single point).
2. Consistent and Dependent: Infinitely many solutions (equations represent the same line).
3. Inconsistent: No solution (lines are parallel and never intersect).

Worked Example: Elimination Method

Solve the system:

$$3x - 2y = 8$$ $$x + 4y = 5$$

Solution: Multiply the second equation by 3 to match the $x$ coefficients:

$$3x + 12y = 15$$

Subtract the first equation from this new equation:

$$(3x + 12y) - (3x - 2y) = 15 - 8$$ $$14y = 7 \Rightarrow y = 0.5$$

Substitute $y = 0.5$ back into the second equation:

$$x + 4(0.5) = 5 \Rightarrow x + 2 = 5 \Rightarrow x = 3$$

The solution is $(3, 0.5)$.

Algebraic Fractions

An algebraic fraction is a fraction whose numerator and denominator are algebraic expressions. Simplifying and operating on these requires factoring and finding common multiples.

Key Rules

• Simplification: Factor both numerator and denominator completely and cancel out common factors.
• Addition/Subtraction: Find the least common denominator (LCD), rewrite each fraction with the LCD, and combine numerators.
• Multiplication: Multiply numerators together and denominators together, then simplify.
• Division: Multiply the first fraction by the reciprocal of the second (flip and multiply).

Worked Example: Simplification

Simplify the expression:

$$\frac{x^2 - 9}{x^2 + 5x + 6}$$

Solution: Factor the numerator (difference of squares) and denominator (quadratic trinomial):

$$\frac{(x - 3)(x + 3)}{(x + 2)(x + 3)}$$

Cancel the common factor $(x + 3)$:

$$\frac{x - 3}{x + 2} \quad (\text{where } x \neq -3, -2)$$

Exponents

An exponent indicates the number of times a base is multiplied by itself. Understanding exponent rules is essential for manipulating complex algebraic expressions.

Rules of Exponents

Let $a$ and $b$ be real numbers, and $m$ and $n$ be integers:

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

Radical Equations

A radical equation is an equation in which a variable is under a radical sign (root).

Steps to Solve Radical Equations

1. Isolate the radical term on one side of the equation.
2. Raise both sides of the equation to the power equal to the index of the radical (e.g., square for square roots, cube for cube roots).
3. Solve the resulting equation.
4. Check for extraneous solutions: This is the most common exam trap! Raising both sides of an equation to an even power can introduce solutions that do not satisfy the original equation.

Worked Example: Radical Equation with Extraneous Solution

Solve:

$$\sqrt{2x + 7} - x = 2$$

Solution: Isolate the radical:

$$\sqrt{2x + 7} = x + 2$$

Square both sides:

$$2x + 7 = (x + 2)^2$$ $$2x + 7 = x^2 + 4x + 4$$

Move all terms to one side to form a quadratic equation:

$$x^2 + 2x - 3 = 0$$

Factor the quadratic:

$$(x + 3)(x - 1) = 0$$

Thus, potential solutions are $x = -3$ and $x = 1$.

Check potential solutions in the original equation:

• For $x = 1$: $$\text{LHS} = \sqrt{2(1) + 7} - 1 = \sqrt{9} - 1 = 3 - 1 = 2 \quad (\text{LHS} = \text{RHS, Valid!})$$
• For $x = -3$: $$\text{LHS} = \sqrt{2(-3) + 7} - (-3) = \sqrt{1} + 3 = 4 \quad (\text{LHS} \neq \text{RHS, Extraneous!})$$

Thus, the only valid solution is $x = 1$.

Common Exam Traps

1. Extraneous Solutions: Always substitute your final answers back into the original radical equation.
2. Quadratic Signs: Forgetting the negative root, e.g., $x^2 = 25$ yields $x = 5$ and $x = -5$.
3. Fractional Denominators: Assuming solutions are valid even if they make a denominator in the original equation zero (which is undefined).

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