Equations & Inequalities

🔢 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.

Special Cases — When the Variable Disappears

Sometimes the $x$-terms cancel completely. Look at what is left:

ACT Tip: "For what value of $k$ does $kx + 4 = 3x + 9$ have no solution?" Make the $x$-coefficients match ($k = 3$) while the constants differ ($4 \neq 9$). Same coefficients and same constants would give infinitely many solutions instead.

Looking ahead (quadratics): For $ax^2 + bx + c = 0$, the sum of the roots is $-\frac{b}{a}$ and the product of the roots is $\frac{c}{a}$. Example: $2x^2 - 10x + 12 = 0$ has roots $2$ and $3$; sum $= -\frac{-10}{2} = 5$, product $= \frac{12}{2} = 6$. You get the sum without solving.

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

The ACT Math test gives you 45 questions in 50 minutes — a little over 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 on the Coordinate Plane

The solution of a system is the point where the two lines intersect. Systems often show up in coordinate geometry, for example when finding a perpendicular bisector.

Perpendicular bisector of segment $\overline{AB}$: the line that passes through the midpoint of $\overline{AB}$ and is perpendicular to $\overline{AB}$ (its slope is the negative reciprocal of the segment's slope).

Example: $A(1, 2)$ and $B(5, 6)$. Midpoint $= (3, 4)$. Slope of $\overline{AB} = \frac{6 - 2}{5 - 1} = 1$, so the perpendicular slope is $-1$.

$y - 4 = -1(x - 3) \implies y = -x + 7$

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 $13.00. How many of each?

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

$p + n = 14$ $0.50p + 1.25n = 13$

Multiply the second equation by 4 to clear the decimals: $2p + 5n = 52$.

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

$2(14 - n) + 5n = 52 \implies 28 - 2n + 5n = 52 \implies 3n = 24 \implies n = 8$

Then $p = 14 - 8 = 6$.

Check: 6 pencils × $0.50 = $3.00 and 8 pens × $1.25 = $10.00, for a total of $13.00 ✓ — 6 pencils and 8 pens.

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.

Watch the reversal: "less than" names the starting quantity second in English but you write it first in algebra. "7 less than $3n$" means start at $3n$ and subtract 7. (Order doesn't matter for "more than," since $3n + 7 = 7 + 3n$.)

Consecutive integers:

• Consecutive integers: $n,\ n + 1,\ n + 2$
• Consecutive even integers: $n,\ n + 2,\ n + 4$ ($n$ even)
• Consecutive odd integers: $n,\ n + 2,\ n + 4$ ($n$ odd) — still $+2$, because odd numbers are also 2 apart

Example: three consecutive odd integers sum to 57 → $3n + 6 = 57 \implies n = 17$, so the integers are 17, 19, 21.

Geometry facts word problems assume: a circle's diameter = 2r (the radius is half the diameter); circumference $= 2\pi r = \pi \cdot$ diameter; area $= \pi r^2$. If a problem gives the diameter, halve it before using $\pi r^2$.

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.

Example 4 — Head start / catch-up: Car A leaves town at 40 mph. One hour later, Car B leaves the same place on the same road at 60 mph. How long after Car B leaves does it catch Car A?

Let $t$ = hours Car B has been driving. Car A has been driving 1 hour longer: $t + 1$. When B catches A, they have gone the same distance:

$60t = 40(t + 1) \implies 60t = 40t + 40 \implies 20t = 40 \implies t = 2 \text{ hours}$

Check: B drives $60 \times 2 = 120$ miles; A drives $40 \times 3 = 120$ miles ✓

Shortcut: Car A's head start is $40 \times 1 = 40$ miles. Car B closes the gap at $60 - 40 = 20$ mph (the difference of the speeds), so it takes $\frac{40}{20} = 2$ hours. Same direction → subtract speeds; opposite directions → add speeds.

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:

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

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 now consider $|x - 2| = 3x + 4$:

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

Check: $|-3 - 2| = 5$ but $3(-3) + 4 = -5$ ✗ — extraneous! An absolute value can't equal $-5$.

Case 2: $x - 2 = -(3x + 4) \implies x - 2 = -3x - 4 \implies 4x = -2 \implies x = -0.5$

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

Solution: $x = -0.5$ only.

ACT Tip: Solve both cases and check each one in the original equation. Neither case is "usually" the right one — and an answer choice listing both values is a common trap when one of them is extraneous.

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 shows up often in ACT factoring and simplifying questions. 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 — dropping a negative sign when distributing is one of the most common algebra mistakes.

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

The Enhanced ACT Math test gives you 50 minutes for 45 questions — about 1 minute 6 seconds per question. Every question has 4 answer choices, and there is no penalty for guessing, so never leave one blank.

Questions generally get harder as you go, so bank time early:

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!

Check: Problem 5 — Both leave together, so the train's lead grows at the difference of the speeds: $80 - 60 = 20$ mph. Set lead = 50: $80t - 60t = 50 \implies 20t = 50 \implies t = 2.5$ hours. (Train: $80 \times 2.5 = 200$ mi; car: $60 \times 2.5 = 150$ mi; lead $= 50$ mi ✓.) If one vehicle had a head start, add the head-start distance to that vehicle's side of the equation, as in Part 4, Example 4.

Final ACT-Style Questions 📋