ACT Algebra and Functions

🔢 Solving Linear Equations

Part 1 of 7 — One-Step, Two-Step, Multi-Step & Variables on Both Sides

Linear equations are the backbone of ACT Algebra. You will see several on every test.

Golden Rule: Whatever you do to one side, do to the other.

Worked Examples

Example 1 — One-step: Solve $x - 9 = 4$.

$x - 9 = 4 \implies x = 13$

Example 2 — Two-step: Solve $3x + 7 = 22$.

$3x = 15 \implies x = 5$

Example 3 — Multi-step: Solve $2(x - 4) + 6 = 18$.

$2x - 8 + 6 = 18 \implies 2x - 2 = 18 \implies 2x = 20 \implies x = 10$

Example 4 — Variables on both sides: Solve $7x - 3 = 4x + 12$.

$3x = 15 \implies x = 5$

ACT Tip: Distribute first, combine like terms, then isolate $x$. Speed matters — practice until these steps are automatic.

Quick Check 🎯

Solve for x 🧮

1. $5x + 3 = 28$

2. $2(x - 6) = 10$

3. $9x - 4 = 5x + 16$

Identify the First Step 🔍

ACT-Style Practice

On the ACT you have roughly 1 minute per question. Try solving these without writing every step.

ACT Tip: When fractions appear, multiply every term by the LCD first — it clears the fractions instantly.

ACT-Style Questions 📋

📐 Systems of Equations

Part 2 of 7 — Substitution, Elimination & Word Problems

A system of equations is two (or more) equations with the same unknowns.

Goal: Find the $(x, y)$ pair that satisfies both equations simultaneously.

Substitution — Worked Example

Solve: $y = 2x + 1$ $3x + y = 16$

Substitute $y = 2x + 1$ into the second equation:

$3x + (2x + 1) = 16 \implies 5x + 1 = 16 \implies x = 3$

Back-substitute: $y = 2(3) + 1 = 7$.

Solution: $(3,\, 7)$

Elimination — Worked Example

Solve: $2x + 3y = 12$ $4x - 3y = 6$

Add the two equations:

$6x = 18 \implies x = 3$

Substitute back: $2(3) + 3y = 12 \implies y = 2$.

Solution: $(3,\, 2)$

ACT Tip: If the ACT asks only for $x$ or only for $y$, elimination is usually faster — you can skip the back-substitution step entirely.

Systems Practice 🎯

Find the Values 🧮

System: $3x + 2y = 19$ and $x - 2y = -3$.

1. What is $x$?

2. What is $y$?

3. What is $x + y$?

Word Problems → Systems

Example: A store sells pencils for $0.50 and pens for $1.25. Maria buys 14 items for $11.00. How many of each?

Let $p$ = pencils, $n$ = pens.

$p + n = 14$ $0.50p + 1.25n = 11$

Multiply the second equation by 4: $2p + 5n = 44$.

From the first: $p = 14 - n$ → substitute:

$2(14 - n) + 5n = 44 \implies 28 - 2n + 5n = 44 \implies 3n = 16$

Hmm — let's try $0.50 and $1.00:

$0.50p + 1.00n = 11 \implies p + 2n = 22$

Subtract from $p + n = 14$: $n = 8$, $p = 6$. ✓

ACT Tip: On the ACT, back-solve from the answer choices when the algebra gets messy — it's often faster.

Method Selection 🔍

ACT-Style Questions 📋

⚖️ Inequalities

Part 3 of 7 — Solving, Graphing, Compound & Absolute Value Inequalities

Inequalities work like equations with one critical rule:

Flip the inequality sign when you multiply or divide by a negative number.

Worked Examples

Example 1 — Basic inequality: Solve $3x - 7 > 8$.

$3x > 15 \implies x > 5$

Example 2 — Negative coefficient: Solve $-2x + 4 \leq 10$.

$-2x \leq 6 \implies x \geq -3$

Notice the inequality flipped when we divided by $-2$.

Example 3 — Compound inequality: Solve $-1 < 2x + 3 \leq 11$.

$-4 < 2x \leq 8 \implies -2 < x \leq 4$

Example 4 — Absolute value inequality: Solve $|x - 5| < 3$.

$-3 < x - 5 < 3 \implies 2 < x < 8$

ACT Tip: $|A| < k$ means $-k < A < k$ (AND). $|A| > k$ means $A < -k$ OR $A > k$.

Inequality Skills 🎯

Solve the Inequality 🧮

Give the boundary value (the number $x$ is compared to).

1. $5x + 2 > 27$ → $x > \text{?}$

2. $-3x \geq 12$ → $x \leq \text{?}$

3. $|x - 1| < 4$ → lower bound of $x$ is?

Inequality Rules 🔍

Graphing Inequalities — Number Line Summary

ACT Tip: Open circle = strict ($<, >$). Closed circle = inclusive ($\leq, \geq$).

ACT-Style Questions 📋

📝 Linear Word Problems

Part 4 of 7 — Translating Words to Equations, Rate/Distance/Time & Mixtures

The hardest part of word problems is translation — turning English into algebra.

ACT Tip: Underline what they're asking for before you set up the equation.

Rate × Time = Distance

The classic formula: $d = rt$.

Example 1: A car travels at 55 mph for 3 hours. Distance?

$d = 55 \times 3 = 165 \text{ miles}$

Example 2: Two trains leave the same station in opposite directions. Train A travels at 60 mph and Train B at 80 mph. After how many hours are they 420 miles apart?

$60t + 80t = 420 \implies 140t = 420 \implies t = 3 \text{ hours}$

Example 3: You drive to work at 30 mph and return at 50 mph. The total trip is 40 miles each way. What is your average speed for the round trip?

Time there: $\frac{40}{30} = \frac{4}{3}$ hr. Time back: $\frac{40}{50} = \frac{4}{5}$ hr.

$\text{Avg speed} = \frac{\text{total distance}}{\text{total time}} = \frac{80}{\frac{4}{3} + \frac{4}{5}} = \frac{80}{\frac{32}{15}} = \frac{80 \times 15}{32} = 37.5 \text{ mph}$

ACT Tip: Average speed is NOT the average of the two speeds. Use total distance ÷ total time.

Translation Practice 🎯

Word Problem Workout 🧮

1. The sum of three consecutive integers is 48. What is the smallest?

2. A cyclist travels at 15 mph for $t$ hours covering 60 miles. What is $t$?

3. A phone plan charges $25/month plus $0.10 per text. If the bill is $40, how many texts were sent?

Mixture Problems

Example: A chemist mixes a 40% acid solution with a 70% acid solution to get 12 liters of 50% acid. How many liters of each?

Let $x$ = liters of 40%, so $12 - x$ = liters of 70%.

$0.40x + 0.70(12 - x) = 0.50(12)$

$0.40x + 8.4 - 0.70x = 6$

$-0.30x = -2.4 \implies x = 8$

Answer: 8 liters of 40% and 4 liters of 70%.

ACT Tip: For mixture problems, set up: $(amount_{1})(concentration_{1})$ + $(amount_{2})(concentration_{2})$ = (total)(target concentration).

Match the Formula 🔍

ACT-Style Questions 📋

📏 Absolute Value Equations

Part 5 of 7 — Solving $|x - a| = b$, Two Cases & Extraneous Solutions

The absolute value $|A|$ is the distance from $A$ to 0 on the number line. Because distance is always non-negative:

$|A| = b \implies A = b \quad \text{or} \quad A = -b \qquad (b \geq 0)$

If $b < 0$, there is no solution — absolute value can never be negative.

Key Insight: Always isolate the absolute value expression first, then split into two cases.

Worked Examples

Example 1: Solve $|x - 3| = 7$.

Case 1: $x - 3 = 7 \implies x = 10$

Case 2: $x - 3 = -7 \implies x = -4$

Solutions: $x = 10$ or $x = -4$

Example 2: Solve $|2x + 1| + 5 = 12$.

First isolate: $|2x + 1| = 7$.

Case 1: $2x + 1 = 7 \implies x = 3$

Case 2: $2x + 1 = -7 \implies x = -4$

Example 3 — No solution: Solve $|x + 4| = -2$.

No solution! Absolute value cannot equal a negative number.

Example 4 — Extraneous solutions: Solve $|3x - 6| = x + 2$.

Case 1: $3x - 6 = x + 2 \implies 2x = 8 \implies x = 4$

Check: $|12 - 6| = 6$ and $4 + 2 = 6$. ✓

Case 2: $3x - 6 = -(x + 2) \implies 3x - 6 = -x - 2 \implies 4x = 4 \implies x = 1$

Check: $|3 - 6| = 3$ and $1 + 2 = 3$. ✓

Both valid! But always check — sometimes one case is extraneous.

ACT Tip: When the other side contains a variable, you must check both solutions in the original equation.

Absolute Value Check 🎯

Solve for x 🧮

For each, give the LARGER solution.

1. $|x - 4| = 9$

2. $|3x + 6| = 12$

3. $|x| + 3 = 10$

Absolute Value Concepts 🔍

Spotting Extraneous Solutions

Extraneous solutions most commonly appear when the right side contains a variable:

$|3x - 1| = 2x + 5$

Case 1: $3x - 1 = 2x + 5 \implies x = 6$ → Check: $|17| = 17$ ✓

Case 2: $3x - 1 = -(2x + 5) \implies 5x = -4 \implies x = -\frac{4}{5}$

Check: $|3(-0.8) - 1| = |-3.4| = 3.4$ vs. $2(-0.8) + 5 = 3.4$ ✓

Both work here, but consider $|x - 2| = 3x + 4$:

Case 2 gives $x - 2 = -3x - 4 \implies 4x = -2 \implies x = -0.5$

Check: $|-0.5 - 2| = 2.5$ but $3(-0.5) + 4 = 2.5$ ✓ — valid!

ACT Tip: If you only have time, solve Case 1 and check the answer choices — it's usually correct. Use Case 2 only if needed.

ACT-Style Questions 📋

🔧 Algebraic Manipulation

Part 6 of 7 — Factoring, Distributing, Combining Like Terms & Special Products

Strong algebraic manipulation skills save time on every section of the ACT Math. These are the building blocks.

Special Products — Memorise These!

Example 1: Factor $x^2 - 49$.

$x^2 - 49 = (x + 7)(x - 7)$

Example 2: Factor $4x^2 - 12x + 9$.

$4x^2 - 12x + 9 = (2x)^2 - 2(2x)(3) + 3^2 = (2x - 3)^2$

Example 3: Simplify $\frac{x^2 - 9}{x + 3}$.

$\frac{(x+3)(x-3)}{x+3} = x - 3 \quad (x \neq -3)$

ACT Tip: Difference of squares appears constantly on the ACT. Be ready to recognize it instantly.

Factoring Practice 🎯

Simplify 🧮

Give the numerical result.

1. Expand and simplify: $(x + 3)(x - 3)$ when $x = 5$. Answer?

2. If $x^2 + 6x + 9 = (x + a)^2$, what is $a$?

3. $2(3x + 4) - (x - 2) = ?x + ?$ — what is the coefficient of $x$?

Identify the Technique 🔍

Distribution & Combining Like Terms — Practice Table

ACT Tip: Don't skip sign distribution — the most common algebra mistake on the ACT is dropping a negative sign when distributing.

ACT-Style Questions 📋

🏆 Review & Mixed Practice

Part 7 of 7 — Cheat Sheet, Mixed ACT-Style Problems & Time Strategy

This final part pulls together everything from Parts 1–6.

Quick-Reference Cheat Sheet

ACT Math Time Strategy

You have 60 minutes for 60 questions — exactly 1 minute per question.

Top 5 Algebra Speed Tips:

1. Back-solve from answer choices on tough questions
2. Plug in numbers when variables make things abstract
3. Eliminate obviously wrong answers first
4. Memorise special products — never FOIL $a^2 - b^2$
5. Clear fractions immediately by multiplying by the LCD

Mixed Practice — Set 1 🎯

Mixed Skills 🧮

1. Solve: $4(x - 3) = 2(x + 5)$. What is $x$?

2. Factor: $x^2 - 16 = (x + a)(x - a)$. What is $a$?

3. System: $x + y = 9$ and $x - y = 3$. What is $x$?

Strategy Selection 🔍

Final Mixed Problems

Try these under timed conditions — 6 minutes for 6 questions.

Check: Problem 3 — Case 1: $x + 3 = 2x - 1 \implies x = 4$. Case 2: $x + 3 = -(2x-1) \implies 3x = -2 \implies x = -\frac{2}{3}$. Check Case 2: $|{-\frac{2}{3}}+3| = \frac{7}{3}$ but $2(-\frac{2}{3})-1 = -\frac{7}{3} < 0$ ✗ Extraneous!

Final ACT-Style Questions 📋