Course Cheatsheet

Section 02: Equations & Problem-Solving Strategies

The IDEA Method

A systematic approach for solving any mathematical problem:

Identify: What type of problem is this? What are the unknowns?
Develop: Choose the appropriate method and create a plan
Execute: Carry out the solution step by step
Assess: Check your answer – does it make sense? Does it satisfy the original equation?

Using the IDEA Method Effectively

Always start by defining your variables clearly. Write down what each variable represents, including units. This single habit prevents most translation errors in word problems and makes it much easier to verify your final answer.

Translating Words to Mathematics

English Phrase	Symbol	Example
“is”, “equals”, “is equal to”	\(=\)	“The cost is 50” \(\to\) \(C = 50\)
“less than”, “fewer than”	\(<\)	“x is less than 10” \(\to\) \(x < 10\)
“greater than”, “more than”	\(>\)	“x exceeds 5” \(\to\) \(x > 5\)
“at least”, “no less than”	\(\geq\)	“at least 5 units” \(\to\) \(x \geq 5\)
“at most”, “no more than”	\(\leq\)	“at most 100” \(\to\) \(x \leq 100\)
“increased by”, “plus”	\(+\)	“price increased by 5” \(\to\) \(p + 5\)
“decreased by”, “minus”	\(-\)	“reduced by 20%” \(\to\) \(x - 0.2x\)
“of”, “times”	\(\times\)	“30% of sales” \(\to\) \(0.3S\)
“per”	\(\div\)	“cost per unit” \(\to\) \(\frac{C}{n}\)

Business Word Problem Strategy

For business problems, always set up these relationships first:

Revenue \(= \text{Price} \times \text{Quantity}\)
Cost \(= \text{Fixed Costs} + \text{Variable Cost} \times \text{Quantity}\)
Profit \(= \text{Revenue} - \text{Cost}\)
Break-even occurs when \(\text{Revenue} = \text{Cost}\) (i.e., Profit \(= 0\))

Identify which of these relationships the problem involves before writing any equations.

Business Vocabulary

Term	Definition	Formula
Revenue (\(R\))	Total income from sales	\(R = P \times Q\)
Cost (\(C\))	Total expenses	\(C = F + vQ\) (\(F\) = fixed, \(v\) = variable per unit)
Profit (\(\Pi\))	Revenue minus cost	\(\Pi = R - C\)
Break-even	Point where profit = 0	Solve \(R = C\)
Margin	Profit as % of revenue	\(\frac{\Pi}{R} \times 100\%\)
Markup	Increase from cost to price	\(\frac{P - v}{v} \times 100\%\)

Linear Equations

Standard Form: \(ax + b = c\)

Solving Multi-Step Equations:

Clear fractions by multiplying all terms by the LCD
Expand parentheses using the distributive property
Collect like terms: variables on one side, constants on the other
Isolate the variable by dividing by its coefficient
Verify by substituting back into the original equation

Example with Fractions:

\[\frac{2x - 1}{3} + \frac{x + 2}{4} = 5\]

LCD \(= 12\)
Multiply through: \(4(2x - 1) + 3(x + 2) = 60\)
Expand: \(8x - 4 + 3x + 6 = 60\)
Combine: \(11x + 2 = 60\)
Solve: \(x = \frac{58}{11}\)

Equation Method Selection

Choose your approach based on what you see:

Simple coefficients and one variable isolated: Direct substitution
Fractions: Multiply by LCD first to clear them
Parentheses: Distribute before combining terms
Multiple equations: Use substitution or elimination (see Systems section)
Higher powers: Check if factoring, quadratic formula, or substitution applies

Inequalities

Key Rules:

All rules for equations apply, except: when multiplying or dividing by a negative number, you must flip the inequality sign
Example: \(-2x > 6\) becomes \(x < -3\) (sign flipped!)

Solution Notation (Intervals):

Inequality	Interval
\(x < a\)	\((-\infty, a)\)
\(x \leq a\)	\((-\infty, a]\)
\(x > a\)	\((a, \infty)\)
\(x \geq a\)	\([a, \infty)\)
\(a < x < b\)	\((a, b)\)
\(a \leq x \leq b\)	\([a, b]\)

Flipping the Inequality Sign

The most common error with inequalities is forgetting to flip the sign when multiplying or dividing by a negative number. This happens because multiplying by a negative reverses the order on the number line. For example, \(2 < 5\) but \(-2 > -5\).

Business Example: A company has costs \(C = 5000 + 20x\) and revenue \(R = 50x\). To make at least 4000 profit:

\(50x - (5000 + 20x) \geq 4000\)
\(30x \geq 9000\), so \(x \geq 300\) units

Systems of Linear Equations

Two Methods for 2x2 Systems

1. Substitution Method – best when one variable is already isolated or has coefficient 1:

Isolate one variable in one equation
Substitute into the other equation
Solve for the remaining variable
Back-substitute to find the first variable
Verify in both original equations

2. Elimination Method – best for symmetric systems or integer coefficients:

Align equations vertically
Multiply equations to create opposite coefficients for one variable
Add or subtract equations to eliminate that variable
Solve for the remaining variable
Back-substitute and verify

Three Possible Outcomes

Outcome	Geometric Meaning	How to Detect
Unique solution	Lines intersect once	\(\frac{a_1}{a_2} \neq \frac{b_1}{b_2}\)
No solution	Parallel lines	\(\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}\)
Infinite solutions	Same line	\(\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}\)

Business Example – Market Equilibrium:

Given demand \(Q_d = 100 - 2P\) and supply \(Q_s = 20 + 3P\), find equilibrium:

Set \(Q_d = Q_s\): \(100 - 2P = 20 + 3P\)
Solve: \(5P = 80\), so \(P = 16\) and \(Q = 68\)

3x3 Systems

For three equations with three unknowns:

Use one equation to eliminate a variable from the other two
Solve the resulting 2x2 system
Back-substitute to find the eliminated variable

Quadratic Equations

Standard Form: \(ax^2 + bx + c = 0\)

The Discriminant

The discriminant \(\Delta = b^2 - 4ac\) determines the nature of solutions:

\(\Delta\) Value	Solutions	Graph	Factorability
\(\Delta > 0\), perfect square	Two rational solutions	Crosses x-axis twice	Factorable over integers
\(\Delta > 0\), not perfect square	Two irrational solutions	Crosses x-axis twice	Not factorable over integers
\(\Delta = 0\)	One repeated solution	Touches x-axis	Perfect square trinomial
\(\Delta < 0\)	No real solutions	Above or below x-axis	Not factorable over reals

Three Solution Methods

1. Factoring (when \(\Delta\) is a perfect square):

Factor the expression and apply the Zero Product Property: if \(AB = 0\), then \(A = 0\) or \(B = 0\)
Example: \(x^2 - 5x + 6 = 0 \to (x-2)(x-3) = 0 \to x = 2\) or \(x = 3\)

2. Quadratic Formula (always works):

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

3. Completing the Square:

Move constant to right side
Add \(\left(\frac{b}{2a}\right)^2\) to both sides
Factor left side as a perfect square
Take square root of both sides

Method Selection Guide

Calculate Δ = b² - 4ac
│
├─ Δ < 0 → No real solutions (stop here)
│
├─ Δ = 0 → One solution: x = -b/(2a)
│
└─ Δ > 0 → Two solutions
           └─ Is Δ a perfect square?
              ├─ YES → Try factoring first (fastest)
              └─ NO → Use quadratic formula

Vieta’s Formulas (Sum and Product of Roots)

For \(ax^2 + bx + c = 0\) with roots \(x_1\) and \(x_2\):

Sum of roots: \(x_1 + x_2 = -\frac{b}{a}\)
Product of roots: \(x_1 \cdot x_2 = \frac{c}{a}\)

This is useful for checking your answers quickly. For \(x^2 - 5x + 6 = 0\), the roots should sum to \(5\) and multiply to \(6\). Indeed \(2 + 3 = 5\) and \(2 \times 3 = 6\).

Break-Even Example: Profit \(P = -2x^2 + 120x - 1600\)

Set \(P = 0\): \(-2x^2 + 120x - 1600 = 0\)
Divide by \(-2\): \(x^2 - 60x + 800 = 0\)
\(x = \frac{60 \pm \sqrt{3600 - 3200}}{2} = \frac{60 \pm 20}{2}\)
Break-even at \(x = 20\) or \(x = 40\)

Biquadratic Equations

Form: \(ax^4 + bx^2 + c = 0\)

Strategy: Use substitution \(u = x^2\):

Replace to get \(au^2 + bu + c = 0\) (a standard quadratic)
Solve for \(u\) using any quadratic method
Back-substitute: if \(u = k\) and \(k \geq 0\), then \(x = \pm\sqrt{k}\)
If \(k < 0\), that value of \(u\) gives no real solutions for \(x\)

Example: \(x^4 - 5x^2 + 4 = 0\)

Let \(u = x^2\): \(u^2 - 5u + 4 = 0 \to (u-1)(u-4) = 0\)
\(u = 1 \to x = \pm 1\); \(u = 4 \to x = \pm 2\)
Four solutions: \(x \in \{-2, -1, 1, 2\}\)

Fractional (Rational) Equations

Key Steps:

Identify domain restrictions – find all values that make any denominator zero
Find the LCD and multiply all terms by it to clear fractions
Solve the resulting polynomial equation
Check every solution against domain restrictions

Always Check Domain Restrictions

Before solving any fractional equation, write down all values that make denominators zero. After solving, reject any solution that equals a restricted value. This step is not optional – skipping it is one of the most common exam errors.

Example: Solve \(\frac{2}{x} + \frac{3}{x-1} = 1\)

Domain: \(x \neq 0\), \(x \neq 1\)
LCD: \(x(x-1)\)
Clear fractions: \(2(x-1) + 3x = x(x-1)\)
Expand: \(2x - 2 + 3x = x^2 - x\)
Rearrange: \(x^2 - 6x + 2 = 0\)
\(x = 3 \pm \sqrt{7}\) (both valid since neither equals 0 or 1)

Cross Multiplication: For \(\frac{a}{b} = \frac{c}{d}\), cross multiply to get \(ad = bc\) (faster than finding LCD).

Work Rate Problems

When two workers or machines combine efforts:

\[\frac{1}{\text{time}_A} + \frac{1}{\text{time}_B} = \frac{1}{\text{time}_{\text{together}}}\]

Example: Machine A takes \(x\) hours alone, Machine B takes \(x + 3\) hours. Together they finish in 2 hours:

\(\frac{1}{x} + \frac{1}{x+3} = \frac{1}{2}\)
Solve: \(x^2 - x - 6 = 0 \to (x-3)(x+2) = 0\)
Since \(x > 0\): Machine A takes 3 hours, Machine B takes 6 hours

Radical Equations

Strategy: Isolate, Power, Check

Isolate the radical term on one side
Raise both sides to the appropriate power (square for \(\sqrt{\ }\), cube for \(\sqrt[3]{\ }\))
Solve the resulting equation
Check ALL solutions in the original equation

Extraneous Solutions in Radical Equations

Squaring both sides of an equation can introduce extraneous solutions – values that satisfy the squared equation but not the original. You must substitute every candidate solution back into the original equation to verify it works. This is especially important on exams.

Example: Solve \(\sqrt{x + 3} = x - 1\)

Square: \(x + 3 = (x - 1)^2 = x^2 - 2x + 1\)
Rearrange: \(x^2 - 3x - 2 = 0\)
\(x = \frac{3 \pm \sqrt{17}}{2}\)
Check \(x_1 \approx 3.56\): \(\sqrt{6.56} \approx 2.56\) and \(3.56 - 1 = 2.56\) – valid
Check \(x_2 \approx -0.56\): \(\sqrt{2.44} \approx 1.56\) and \(-0.56 - 1 = -1.56\) – invalid (signs differ)

Multiple Radicals: Isolate one radical at a time, then square.

Example: \(\sqrt{x + 5} + \sqrt{x} = 5\)

Isolate: \(\sqrt{x + 5} = 5 - \sqrt{x}\)
Square: \(x + 5 = 25 - 10\sqrt{x} + x\)
Simplify: \(10\sqrt{x} = 20\), so \(\sqrt{x} = 2\), thus \(x = 4\)
Check: \(\sqrt{9} + \sqrt{4} = 3 + 2 = 5\) – valid

Cubic Equations

Form: \(ax^3 + bx^2 + cx + d = 0\)

Solution Strategies:

1. Special Forms (from Section 01):

Sum of cubes: \(a^3 + b^3 = (a + b)(a^2 - ab + b^2)\)
Difference of cubes: \(a^3 - b^3 = (a - b)(a^2 + ab + b^2)\)

2. Rational Root Theorem: Test possible roots \(x = \pm\frac{\text{factors of } d}{\text{factors of } a}\)

If \(x = r\) works, then \((x - r)\) is a factor
Divide the cubic by \((x - r)\) to get a quadratic
Solve the quadratic with any standard method

3. Factor by Grouping: Group terms in pairs and look for common factors.

Example: \(x^3 - 6x^2 + 11x - 6 = 0\)

Test \(x = 1\): \(1 - 6 + 11 - 6 = 0\) – it works
Factor out \((x - 1)\): \((x - 1)(x^2 - 5x + 6) = 0\)
Factor further: \((x - 1)(x - 2)(x - 3) = 0\)
Solutions: \(x = 1, 2, 3\)

Exponential Equations

Strategy 1 – Same Base: If \(a^{f(x)} = a^{g(x)}\), then \(f(x) = g(x)\)

Example: \(3^{2x-1} = 81 = 3^4\), so \(2x - 1 = 4\) and \(x = 2.5\)

Strategy 2 – Logarithm Method: For \(a^x = b\) with different bases:

Take \(\log\) of both sides: \(x \log(a) = \log(b)\)
Solve: \(x = \frac{\log(b)}{\log(a)}\)

Strategy 3 – Substitution: For equations like \(4^x - 3 \cdot 2^x + 2 = 0\):

Note \(4^x = (2^x)^2\), let \(u = 2^x\)
Get \(u^2 - 3u + 2 = 0 \to (u-1)(u-2) = 0\)
\(u = 1 \to 2^x = 1 \to x = 0\); \(u = 2 \to 2^x = 2 \to x = 1\)

Handling Mixed Bases

When you see different bases in an exponential equation, look for ways to express them as powers of a common base. For example, \(8 = 2^3\), \(27 = 3^3\), \(4 = 2^2\), \(9 = 3^2\), \(16 = 2^4\). If no common base exists, use the logarithm method.

Exponential Inequalities: For base \(a > 1\), the function is increasing, so the inequality direction is preserved: \(a^{f(x)} > a^{g(x)} \Rightarrow f(x) > g(x)\).

Logarithmic Equations

Strategy 1 – Combine and Convert:

Use log properties to combine into a single logarithm
Convert to exponential form: \(\log_a(X) = Y\) becomes \(a^Y = X\)

Strategy 2 – Change of Base:

\[\log_a(x) = \frac{\log_b(x)}{\log_b(a)}\]

This is especially useful when an equation has logarithms with different bases.

Domain Checking for Logarithmic Equations

The argument of every logarithm must be strictly positive. When solving logarithmic equations, always check that your solutions make all log arguments positive. Reject any solution that causes a negative or zero argument.

Example: Solve \(\log(x) + \log(x - 3) = 1\)

Domain: \(x > 3\) (both arguments must be positive)
Combine: \(\log(x(x - 3)) = 1\)
Convert: \(x(x - 3) = 10^1 = 10\)
Solve: \(x^2 - 3x - 10 = 0 \to (x - 5)(x + 2) = 0\)
\(x = 5\) (valid, since \(5 > 3\)) or \(x = -2\) (rejected, since \(-2 < 3\))

Logarithmic Systems:

\(\log(x) + \log(y) = 2\) and \(\log(x) - \log(y) = 1\)

Add: \(2\log(x) = 3 \to x = 10^{1.5} = 10\sqrt{10}\)
Subtract: \(2\log(y) = 1 \to y = \sqrt{10}\)

Business Application Examples

Break-Even Analysis

Setup: Find where Revenue = Cost.

A coffee shop: fixed costs 2000/month, variable cost 1.50/coffee, selling price 3.50/coffee.

Cost: \(C = 2000 + 1.50x\)
Revenue: \(R = 3.50x\)
Break-even: \(3.50x = 2000 + 1.50x \to 2x = 2000 \to x = 1000\) coffees

Mixture and Investment Problems

Investment Split: Allocating 10,000 between bonds (4%) and stocks (9%) to earn 650/year:

Let \(x\) = amount in bonds, \(10000 - x\) = amount in stocks
\(0.04x + 0.09(10000 - x) = 650\)
\(0.04x + 900 - 0.09x = 650\)
\(-0.05x = -250 \to x = 5000\) in each

Compound Interest Time Problems

How long to double at 8% annual interest?

\(2P = P(1.08)^t\)
\(2 = 1.08^t\)
\(t = \frac{\ln(2)}{\ln(1.08)} \approx 9.01\) years
Quick check: Rule of 72 gives \(\frac{72}{8} = 9\) years

Growth and Decay Models

Many real-world problems use these standard forms:

Exponential growth: \(A(t) = A_0 \cdot e^{kt}\) where \(k > 0\)
Exponential decay: \(A(t) = A_0 \cdot e^{-kt}\) where \(k > 0\)
Logistic growth: \(P(t) = \frac{L}{1 + Ce^{-kt}}\) (bounded by carrying capacity \(L\))
Saturation model: \(R(t) = L(1 - e^{-kt})\) (approaches limit \(L\))

To find the rate constant \(k\), substitute a known data point and solve using logarithms.

Quick Reference

Equation Type	Standard Form	Method
Linear	\(ax + b = c\)	Isolate \(x\)
Quadratic	\(ax^2 + bx + c = 0\)	Factor, formula, or complete square
Biquadratic	\(ax^4 + bx^2 + c = 0\)	Substitute \(u = x^2\)
Fractional	\(\frac{a}{f(x)} + \frac{b}{g(x)} = c\)	LCD, clear fractions, check domain
Radical	\(\sqrt{f(x)} = g(x)\)	Square both sides, check solutions
Cubic	\(ax^3 + bx^2 + cx + d = 0\)	Find one root, factor, solve quadratic
Exponential	\(a^{f(x)} = b\)	Same base or take logarithms
Logarithmic	\(\log_a(f(x)) = c\)	Convert to \(f(x) = a^c\)
System (2x2)	Two equations, two unknowns	Substitution or elimination

Problem-Solving Strategies

Identify the equation type before choosing a method
Define variables with clear descriptions and units for word problems
Check the discriminant before trying to factor quadratics
State domain restrictions immediately for fractional and logarithmic equations
Verify every solution – especially for radical, fractional, and logarithmic equations
Translate systematically for word problems: define variables, write relationships, set up equation, solve, interpret
Use substitution when you see repeated sub-expressions or higher powers that are multiples of lower powers
Break complex problems into steps – solve for one thing at a time

Common Mistakes to Avoid

Inequalities: Forgetting to flip the sign when multiplying or dividing by a negative number
Fractions: Not checking domain restrictions after solving (solutions that make denominators zero must be rejected)
Radicals: Not checking for extraneous solutions after squaring both sides
Logarithms: Forgetting that \(\log\) arguments must be strictly positive
Word problems: Mixing up revenue and profit, or forgetting units in the final answer
Systems: Not verifying the solution in all original equations
Quadratics: Using the quadratic formula with wrong signs (double-check \(a\), \(b\), \(c\) from standard form)
Break-even: Setting Revenue \(= 0\) instead of Revenue \(=\) Cost