Algebra: Finding the Unknown

Algebra sounds intimidating, but here is a secret that will change how you feel about it: algebra is just arithmetic with a blank space. Every time you were a child and saw "3 + ___ = 8" and figured out the blank was 5, you were already doing algebra. The only difference now is that we give that blank a name — usually a letter like x — so we can talk about it, move it around, and solve for it even in complicated situations. This chapter will take you from that simple idea all the way to the algebra questions on the Civil Service Exam, one careful step at a time.

1. Lesson Overview

Algebra is the branch of mathematics that uses letters to stand for unknown numbers. On the Civil Service Examination, algebra appears in two ways. Sometimes it is direct — "solve for x" — but far more often it is hidden inside a word problem about ages, money, mixtures, or quantities, where your real task is to translate ordinary Filipino-life sentences into an equation and then solve it.

You can expect algebra to appear throughout the Numerical Reasoning section, and the skill of "turning words into an equation" quietly powers many other topics too — profit and loss, ratio, work, and mixtures all lean on it. Realistically, algebra-flavored reasoning touches 6 to 10 items on a typical exam once you count the disguised word problems.

The difficulty is beginner to intermediate. The math itself is rarely hard; the challenge is setting up the problem correctly. Students who lose marks here almost never lose them because they can't compute — they lose them because they translated the words wrong or forgot to do the same thing to both sides. This lesson attacks exactly those weak points.

2. Learning Objectives

After completing this lesson you will be able to:

• Translate English (and everyday Filipino) sentences into algebraic equations.
• Solve linear equations in one variable by isolating the variable.
• Apply the rules of exponents and radicals confidently.
• Factor using the difference-of-squares and perfect-square patterns.
• Solve quadratic equations by factoring and by the quadratic formula.
• Solve systems of two equations in two unknowns.
• Set up and solve classic word problems: ages, consecutive integers, coins/tickets, and number relationships.
• Handle inequalities, including the sign-flip rule.
• Use back-solving and number-picking shortcuts to beat multiple-choice items faster.
• Avoid the specific setup and sign errors that cost examinees marks.

3. Prerequisites

Before starting, you should be comfortable with:

• The four basic operations on whole numbers, fractions, and decimals, including with negative numbers.
• Order of operations (PEMDAS/GEMDAS): do what's in parentheses first, then exponents, then multiplication/division left to right, then addition/subtraction.
• Negative numbers — adding, subtracting, and multiplying them (a negative times a negative is a positive).

Let's refresh the one prerequisite algebra depends on most: the idea of an equation as a balance. Picture a see-saw or a timbangan (weighing scale) that is perfectly level. The left side equals the right side. Algebra's golden rule follows directly from this picture: whatever you do to one side, you must do to the other, or the balance tips and the equality breaks. Keep the see-saw in mind for the entire chapter — it is the reason every solving step works.

4. Why This Topic Matters

Algebra is the math of "figure out the missing piece," and that situation is everywhere:

• Business. A sari-sari store owner knows total sales and the price per item and wants the number sold. A vendor knows cost and desired profit and wants the selling price.
• Personal finance. You have ₱5,000, you spend ₱1,200 on load and ₱2,300 on groceries — how much is left for savings? That's x = 5000 − 1200 − 2300.
• Cooking and scaling. A recipe for 4 servings needs to become 10 — algebra scales the ingredients.
• Construction and work. If 3 workers finish in 8 days, how long for 4 workers? A hidden equation.
• Government and salaries. Computing take-home pay after fixed deductions, or how many years of service are needed to reach a target pension.
• Travel. Knowing distance and speed to find travel time from Manila to Baguio.

Every one of these is "I know some things, I'm missing one thing, let me set up a relationship and solve." That is algebra, and that is why the exam tests it — you will use it for life.

5. Core Concepts

We build slowly. Each idea rests on the one before it.

What a variable really is

A variable is a letter that stands for a number we don't know yet. The most common is x, but any letter works. When we write x + 8 = 23, we are saying "some number, plus 8, gives 23." Solving means discovering what that number must be.

Analogy: A variable is like a wrapped gift. You don't know what's inside yet, but you can reason about it from clues ("it's the same weight as a book," "two of them equal a kilo"). Algebra is the set of tools for unwrapping the gift using the clues in the problem.

A coefficient is the number multiplied by a variable. In 3x, the 3 is the coefficient — it means "three of whatever x is." A constant is a plain number with no variable, like the 8 in x + 8. A term is a single piece: 3x, 8, and x² are each terms.

Translating words into equations

This is the single most important algebra skill for the exam, because most questions are word problems. Certain words map directly to operations:

• "sum," "total," "increased by," "more than," "added to" → addition (+)
• "difference," "decreased by," "less than," "fewer than," "reduced by" → subtraction (−)
• "of," "times," "product," "twice," "double," "triple" → multiplication (×)
• "per," "divided by," "shared equally," "ratio" → division (÷)
• "is," "was," "equals," "will be," "results in" → the equals sign (=)

A number increased by 8 is 23. → x + 8 = 23 → x = 15.

Watch the order with subtraction phrases. "5 less than a number" translates to x − 5, not 5 − x. The number you are subtracting from comes first, even though it's mentioned last. Read "5 less than x" as "start with x, take away 5." This tiny reversal is one of the most common exam traps, so slow down every time you see "less than" or "fewer than."

Memory tip: "less than" and "subtracted from" both flip the order you read them in. "7 less than twice a number" = 2x − 7.

Isolating the variable — the see-saw in action

To solve a linear equation, get the variable alone on one side by undoing every operation attached to it, using the opposite operation, on both sides:

3x − 5 = 16 3x = 21 (added 5 to both sides — undoing the − 5) x = 7 (divided both sides by 3 — undoing the × 3)

Undo in reverse order of operations: deal with addition/subtraction first, then multiplication/division. Always check by substituting back: 3(7) − 5 = 21 − 5 = 16. ✓ It balances, so x = 7 is correct.

Combining like terms and clearing parentheses

Like terms have the same variable to the same power; you can add or subtract them: 3x + 5x = 8x, but 3x + 5x² cannot be combined (different powers). To clear parentheses, use the distributive property: a(b + c) = ab + ac.

2(x + 3) = 4x − 6 2x + 6 = 4x − 6 (distributed the 2) 6 + 6 = 4x − 2x (moved terms: subtract 2x from both sides, add 6 to both sides) 12 = 2x → x = 6.

Check: 2(6 + 3) = 18, and 4(6) − 6 = 18. ✓

Exponents and radicals

An exponent tells how many times to multiply a base by itself: x³ = x × x × x. The key rules:

• Multiplying same-base powers adds exponents: x³ × x² = x⁵.
• Dividing same-base powers subtracts exponents: x⁵ ÷ x² = x³.
• Raising a power to a power multiplies exponents: (x³)² = x⁶.
• Any nonzero number to the power 0 equals 1: 7⁰ = 1.
• A negative exponent means reciprocal: x⁻² = 1 ÷ x².
• A square root (√) asks "what number times itself gives this?" √49 = 7, because 7 × 7 = 49.

Common misconception: x³ × x² is NOT x⁶. You add the exponents (3 + 2 = 5) when multiplying, and multiply them (3 × 2 = 6) only when raising a power to a power. Keep these two rules apart — mixing them up is the #1 exponent error.

Two factoring patterns worth memorizing

Factoring means writing an expression as a product. Two patterns appear constantly:

• Difference of squares: a² − b² = (a + b)(a − b). Example: x² − 9 = (x + 3)(x − 3).
• Perfect-square trinomial: a² + 2ab + b² = (a + b)². Example: x² + 6x + 9 = (x + 3)².

Recognizing these lets you factor quadratics in seconds instead of guessing.

Quadratic equations

A quadratic has the form ax² + bx + c = 0 — the variable is squared. Two ways to solve:

By factoring: rewrite as a product equal to zero, then set each factor to zero (because if two things multiply to zero, at least one must be zero).

x² − 5x + 6 = 0 → (x − 2)(x − 3) = 0 → x = 2 or x = 3.

By the quadratic formula (works even when factoring is hard):

x = (−b ± √(b² − 4ac)) ÷ (2a)

For x² − 2x − 3 = 0, we have a = 1, b = −2, c = −3: x = (2 ± √(4 + 12)) ÷ 2 = (2 ± 4) ÷ 2 = 3 or −1.

Systems of two equations

When a problem has two unknowns, you need two equations. Solve by elimination (add or subtract the equations to cancel a variable) or substitution (solve one equation for a variable, plug into the other).

x + y = 10 and x − y = 4. Add the two equations: 2x = 14 → x = 7. Then y = 3.

Inequalities and the sign-flip rule

An inequality uses <, >, ≤, or ≥ instead of =. You solve it just like an equation, with one crucial exception: if you multiply or divide both sides by a negative number, you must flip the inequality sign.

−2x < 6 → x > −3 (divided by −2, so < became >).

Why the flip? Multiplying by a negative reverses order on the number line: 2 < 3, but −2 > −3. So the inequality symbol must turn around to stay true. Addition and subtraction never require a flip — only multiplying or dividing by a negative does.

Solving equations that contain fractions

When variables sit inside fractions, the cleanest first move is to clear the denominators — multiply every term by the least common denominator (LCD). This turns a messy fraction equation into a plain linear one.

x/3 + x/4 = 7. The LCD of 3 and 4 is 12. Multiply every term by 12: 12·(x/3) + 12·(x/4) = 12·7 → 4x + 3x = 84 → 7x = 84 → x = 12.

Check: 12/3 + 12/4 = 4 + 3 = 7. ✓ Clearing denominators up front prevents almost all fraction errors — do it before anything else.

Simplifying algebraic expressions with exponents

Sometimes the task isn't to solve but to simplify. Apply the exponent rules term by term, and remember that an exponent outside a parenthesis hits every factor inside.

(2x²y)³ = 2³ · (x²)³ · y³ = 8x⁶y³.

Notice the 2 gets cubed too (→ 8), the x² becomes x⁶ (power of a power: 2 × 3), and y becomes y³. Forgetting to raise the coefficient (leaving it as 2 instead of 8) is a frequent slip.

Percentages living inside algebra

Many word problems blend percentages with a variable. Convert the percent to a decimal and treat it as an ordinary coefficient.

30% of a number is 45. → 0.30x = 45 → x = 150.

This is exactly the "find the whole" idea from the Percentage lesson, now written as an equation. When you see "of a number," reach for a variable and a multiplication.

Literal equations — solving for one letter among many

A literal equation is a formula with several letters, and you're asked to rearrange it to isolate one of them. The method is identical to solving for x — undo operations on both sides — you just carry the other letters along as if they were numbers.

Solve A = ½bh for h (the area formula for a triangle). Multiply both sides by 2: 2A = bh. Divide both sides by b: h = 2A ÷ b.

This skill matters because many exam problems hand you a formula (for perimeter, area, distance, or interest) and ask you to solve for a quantity that isn't already alone. Treat every other letter as a stand-in for a number and the fear disappears.

Now that every core idea is in place, let's reinforce them with pictures, formulas, and lots of examples.

6. Visual Learning Suggestions

• [Illustration Suggestion] The balance scale. Draw a level see-saw with "3x − 5" on the left pan and "16" on the right. Show each solving step as removing/adding the same weight from both pans so it stays level. This makes "do the same to both sides" unforgettable.
• [Illustration Suggestion] The translation table. A two-column chart: left column English phrases ("5 more than a number," "twice a number," "3 less than a number"), right column their algebra (x + 5, 2x, x − 3). Build your own and add to it as you practice.
• [Illustration Suggestion] Area model for factoring. Draw a square of side (x + 3). Its area is (x + 3)² = x² + 6x + 9, shown as one x-by-x square, two 3-by-x rectangles, and one 3-by-3 square. Seeing the pieces explains where the "middle term" 6x comes from.
• [Illustration Suggestion] Number line for inequalities. Draw a number line, mark −3 with an open circle (for >), and shade to the right to picture "x > −3" — every number bigger than −3 is a solution.

7. Formula Library

Why the quadratic formula works (briefly): it is what you get by "completing the square" on the general form ax² + bx + c = 0 — a guaranteed method that finds the solutions of any quadratic, even one that won't factor with whole numbers. You don't need to derive it; just memorize it and know it always works.

8. Pattern Recognition

Train yourself to spot the setup instantly:

9. Problem-Solving Framework

Step 1 — Read and identify the unknown. What exactly is the question asking for? Name it x. ↓ Step 2 — Express everything in terms of that variable. If Ben's age is x, write Ana's age as 2x, and so on. ↓ Step 3 — Find the equality. Locate the sentence that says two things are equal ("is," "will be," "totals") and write the equation. ↓ Step 4 — Solve by isolating the variable, doing the same to both sides. ↓ Step 5 — Answer the actual question. If x is Ben's age but the question asks for Ana's, don't stop at x — compute 2x. ↓ Step 6 — Check by substituting back into the original wording, not just the equation.

Why Step 5 matters: exam writers love to make x an intermediate value, not the final answer. Solving for x and stopping is a classic way to pick a wrong choice that's deliberately listed.

10. Worked Examples

Beginner

Example 1. Solve x + 8 = 23. Solution: Subtract 8 from both sides: x = 15. Check: 15 + 8 = 23. ✓ Difficulty: ★☆☆☆☆

Example 2. Solve 3x − 5 = 16. Solution: Add 5: 3x = 21. Divide by 3: x = 7. Check: 3(7) − 5 = 16. ✓ Difficulty: ★☆☆☆☆

Intermediate

Example 3. Solve 2(x + 3) = 4x − 6. Thinking: Distribute first, then gather variables on one side. Solution: 2x + 6 = 4x − 6 → 12 = 2x → x = 6. Check: both sides equal 18. ✓ Difficulty: ★★☆☆☆

Example 4. Three consecutive integers add up to 72. Find them. Thinking: "Consecutive integers" → x, x + 1, x + 2. Solution: x + (x + 1) + (x + 2) = 72 → 3x + 3 = 72 → 3x = 69 → x = 23. The integers are 23, 24, 25. Check: 23 + 24 + 25 = 72. ✓ Shortcut: for three consecutive integers, the middle one is the average: 72 ÷ 3 = 24, so the set is 23, 24, 25. Difficulty: ★★☆☆☆

Example 5 (ages). Ana is twice as old as Ben. In 5 years, the sum of their ages will be 40. How old is Ben now? Thinking: Let Ben = x now, so Ana = 2x. In 5 years each is 5 older. Solution: (x + 5) + (2x + 5) = 40 → 3x + 10 = 40 → 3x = 30 → x = 10. Ben is 10. Check: Ana is 20 now; in 5 years they're 15 and 25, which sum to 40. ✓ Difficulty: ★★☆☆☆

Advanced

Example 6 (quadratic by factoring). Solve x² − 5x + 6 = 0. Thinking: Find two numbers that multiply to +6 and add to −5: those are −2 and −3. Solution: (x − 2)(x − 3) = 0 → x = 2 or x = 3. Check: 2² − 5(2) + 6 = 0 ✓ and 3² − 5(3) + 6 = 0 ✓. Difficulty: ★★★☆☆

Example 7 (quadratic formula). Solve 2x² − 4x − 6 = 0. Thinking: Divide the whole equation by 2 first to simplify: x² − 2x − 3 = 0. Solution: a = 1, b = −2, c = −3. x = (2 ± √(4 + 12)) ÷ 2 = (2 ± 4) ÷ 2 = 3 or −1. Alternative: factor x² − 2x − 3 = (x − 3)(x + 1) → same answers. Difficulty: ★★★☆☆

Example 8 (system of equations). The sum of two numbers is 25 and their difference is 9. Find the numbers. Solution: x + y = 25, x − y = 9. Add: 2x = 34 → x = 17, so y = 8. Check: 17 + 8 = 25 and 17 − 8 = 9. ✓ Difficulty: ★★★☆☆

Civil Service Exam Level

Example 9 (tickets — two variables). A barangay fun-run sold 200 tickets for ₱21,600 total. Adult tickets cost ₱120 and child tickets ₱80. How many adult tickets were sold? Thinking: Let a = adult tickets, c = child tickets. Two facts → two equations. Solution: a + c = 200 and 120a + 80c = 21,600. Substitute c = 200 − a: 120a + 80(200 − a) = 21,600 → 120a + 16,000 − 80a = 21,600 → 40a = 5,600 → a = 140 (and 60 child tickets). Check: 140 × 120 = 16,800; 60 × 80 = 4,800; total = 21,600. ✓ Difficulty: ★★★★☆

Example 10 (ages, past tense). A father is 4 times as old as his son. Five years ago, he was 7 times as old. Find their present ages. Thinking: Let son = x now, father = 4x. Five years ago each was 5 younger. Solution: 4x − 5 = 7(x − 5) → 4x − 5 = 7x − 35 → 30 = 3x → x = 10. Son is 10, father is 40. Check: five years ago they were 5 and 35, and 35 = 7 × 5. ✓ Difficulty: ★★★★☆

Example 11 (number relationship). Three times a number, decreased by 7, equals the number increased by 9. Find the number. Solution: 3x − 7 = x + 9 → 2x = 16 → x = 8. Check: 3(8) − 7 = 17, and 8 + 9 = 17. ✓ Difficulty: ★★★☆☆

Example 12 (inequality). A student needs an average of at least 80 over three tests. She scored 75 and 82 on the first two. What must she score on the third to reach the target? Thinking: "at least 80 average" → (75 + 82 + x) ÷ 3 ≥ 80. Solution: 157 + x ≥ 240 → x ≥ 83. She needs at least 83. Check: with 83, the average is (75 + 82 + 83) ÷ 3 = 240 ÷ 3 = 80. ✓ Difficulty: ★★★★☆

Example 13 (fraction equation). Solve x/2 + x/5 = 14. Thinking: Clear denominators — the LCD of 2 and 5 is 10. Solution: Multiply every term by 10: 5x + 2x = 140 → 7x = 140 → x = 20. Check: 20/2 + 20/5 = 10 + 4 = 14. ✓ Difficulty: ★★★☆☆

Example 14 (exponent simplification). Simplify (3a³b)². Solution: 3² · (a³)² · b² = 9a⁶b². Common mistake: leaving the coefficient as 3 instead of squaring it to 9. Difficulty: ★★★☆☆

Example 15 (hidden distance–rate). A driver covers a route in 4 hours. Driving 20 kph faster, he could cover the same route in 3 hours. Find his original speed. Thinking: Distance is the same both ways, and distance = speed × time. Let original speed = x. Solution: 4x = 3(x + 20) → 4x = 3x + 60 → x = 60. His original speed is 60 kph (route length = 240 km). Check: 4 × 60 = 240; 3 × 80 = 240. ✓ Difficulty: ★★★★☆

Example 16 (product / quadratic word problem). Two positive numbers differ by 4, and their product is 45. Find them. Thinking: Let the smaller be x, the larger x + 4. Solution: x(x + 4) = 45 → x² + 4x − 45 = 0 → (x + 9)(x − 5) = 0 → x = 5 (reject −9, since the numbers are positive). The numbers are 5 and 9. Check: 9 − 5 = 4 and 5 × 9 = 45. ✓ Difficulty: ★★★★☆

Example 17 (literal equation). The perimeter of a rectangle is P = 2(l + w). Solve for w. Solution: Divide both sides by 2: P ÷ 2 = l + w. Subtract l: w = (P ÷ 2) − l, or equivalently w = (P − 2l) ÷ 2. Difficulty: ★★★☆☆

Example 18 (perimeter system). A rectangle's perimeter is 46 cm and its length is 5 cm more than its width. Find the length and width. Thinking: Perimeter 2(l + w) = 46, so l + w = 23; and l = w + 5. Solution: Substitute: (w + 5) + w = 23 → 2w + 5 = 23 → 2w = 18 → w = 9. So width = 9 cm, length = 14 cm. Check: 2(14 + 9) = 2(23) = 46. ✓ Difficulty: ★★★★☆

Example 19 (digit problem). In a two-digit number, the tens digit is twice the units digit. The number is 27 more than the number formed by reversing its digits. Find the number. Thinking: Let the units digit be u, so the tens digit is 2u. The value of the number is 10(2u) + u = 21u; the reversed number is 10u + 2u = 12u. Solution: 21u = 12u + 27 → 9u = 27 → u = 3, so tens digit = 6. The number is 63. Check: reversed is 36, and 63 − 36 = 27. ✓ Difficulty: ★★★★★

Example 20 (investment mix). Aling Nena splits ₱10,000 into two placements: part at 5% yearly interest and the rest at 8%. After one year she earns ₱680 in total interest. How much did she put at each rate? Thinking: Let x = amount at 5%, so (10,000 − x) is at 8%. Total interest is the sum. Solution: 0.05x + 0.08(10,000 − x) = 680 → 0.05x + 800 − 0.08x = 680 → −0.03x = −120 → x = 4,000. So ₱4,000 at 5% and ₱6,000 at 8%. Check: 0.05(4,000) + 0.08(6,000) = 200 + 480 = 680. ✓ Difficulty: ★★★★★

11. Exam Tricks

• The "less than" reversal. "8 less than a number" is x − 8. The exam includes 8 − x as a wrong choice specifically to catch fast readers. Slow down on every "less than."
• Solve for the right thing. If x is the son's age but the question asks the father's, the son's age will be sitting there as a tempting wrong answer. Re-read the question after solving.
• The forgotten second solution. Quadratics usually have two answers. If a question asks "which of the following is a solution," check whether more than one choice works — or whether the real-world context (like an age or a length) rules out a negative solution.
• Back-solving beats algebra on multiple choice. If solving looks messy, plug the answer choices into the original equation and see which one fits. On a four-choice item this is often faster and error-proof.
• Distribute the negative. −(x − 3) equals −x + 3, not −x − 3. A dropped sign after a minus outside parentheses is a classic silent error.

12. Common Mistakes

• Not doing the same thing to both sides. Adding 5 to the left only breaks the balance. Every operation must hit both sides. This is the root of most "I don't know where I went wrong" moments — each line of your solution should keep the see-saw level.
• Mis-translating "less than." Writing 5 − x instead of x − 5. The order flips. The same reversal trap hides in "subtracted from" and "fewer than." Reading the phrase aloud as "start with the number, then take away" fixes it.
• Confusing exponent rules. Adding exponents when you should multiply them (or vice versa): multiply same-base powers by adding exponents; raise a power to a power by multiplying them. And when a coefficient is inside a power — (3a)² — remember to raise the coefficient too: 9a², not 3a².
• Sign errors with negatives. Especially forgetting to flip the inequality sign when dividing by a negative, and mishandling −(x − 3), which is −x + 3, not −x − 3. Distribute the minus sign carefully every time.
• Stopping at x. Forgetting Step 5 — answering the intermediate variable instead of the quantity the question actually asked for. If you let x be the son's age but the question wants the father's, the wrong value is sitting there as a listed choice.
• Skipping the check. Not substituting the answer back into the original words to confirm it makes sense. Ten seconds of checking saves an entire mark.
• Rejecting nothing in quadratics. When a word problem gives two solutions, forgetting to discard the one that can't be real (a negative age, a negative length, a fractional person).

13. Shortcuts

• Back-solving: substitute each answer choice into the equation; pick the one that balances. Great when the algebra is messy.
• Number-picking: for "which expression equals…" questions with variables in the answers, plug in an easy number (like x = 2) for the variable everywhere, compute, and match.
• Consecutive-integer trick: the sum of an odd count of consecutive integers is that count times the middle number. Three consecutive integers summing to 72 → middle is 72 ÷ 3 = 24.
• Elimination first: for a system, if adding or subtracting the two equations immediately cancels a variable, do that — it's faster than substitution.
• Factor before formula: always try quick factoring on a quadratic before reaching for the quadratic formula; it's faster when it works.

14. Memory Techniques

• "Do unto both sides." The golden rule of equations — whatever you do to one side, do to the other.
• "Less than flips." Any phrase with "less than" or "subtracted from" reverses the order you read it in.
• "MAdd, PMultiply." Multiplying same-base powers → Add exponents; Power of a power → Multiply exponents.
• "Zero product": if two things multiply to zero, one of them is zero — the whole reason factoring solves quadratics.
• "Flip for negative": dividing or multiplying an inequality by a negative flips the sign.

15. Real Civil Service Exam Strategy

• Budget your time. A direct "solve for x" should take under 45 seconds. A word problem might take 60–90 seconds — most of it spent on the setup, which is time well spent.
• Set up before you compute. Resist the urge to start calculating. A correct equation almost solves itself; a wrong equation guarantees a wrong answer no matter how clean your arithmetic.
• Use the choices. On multiple choice, back-solving and number-picking are legitimate, fast, and reliable. There are no bonus points for doing "real" algebra.
• When stuck, name the unknown. If you freeze, just write "Let x = the thing I'm asked for" and express the next fact in terms of x. Momentum usually follows.
• Check with reality. An age can't be negative; a count of people must be a whole number. Use common sense to eliminate impossible choices.

16. Practice Questions

Easy

1. Solve x − 6 = 10.
2. Solve 5x = 45.
3. Simplify x⁴ × x³.

Medium

4. Solve 4x + 3 = 2x + 15.
5. The sum of two consecutive integers is 47. Find them.
6. Solve for x: 3(x − 2) = 2x + 4.

Hard

7. Solve x² − 7x + 12 = 0.
8. A number added to twice itself gives 39. Find the number.

Challenge

9. Mother is 3 times as old as her daughter. In 12 years, she will be twice as old. Find their present ages.
10. Adult movie tickets cost ₱150 and student tickets ₱90. A group of 12 paid ₱1,440. How many were students?
11. Solve x/4 + x/6 = 5.
12. A rectangle's length is 3 more than its width, and its area is 40. Find its dimensions.
13. A number decreased by 20% equals 48. Find the number.
14. Simplify (2m³n²)³.

Answers and Explanations

1. x = 16. Add 6 to both sides.
2. x = 9. Divide both sides by 5.
3. x⁷. Multiplying same-base powers adds exponents (4 + 3).
4. x = 6. 4x − 2x = 15 − 3 → 2x = 12 → x = 6.
5. 23 and 24. x + (x + 1) = 47 → 2x + 1 = 47 → x = 23.
6. x = 10. 3x − 6 = 2x + 4 → x = 10.
7. x = 3 or x = 4. (x − 3)(x − 4) = 0.
8. 13. x + 2x = 39 → 3x = 39 → x = 13.
9. Daughter 12, mother 36. Let daughter = x, mother = 3x. In 12 years: 3x + 12 = 2(x + 12) → 3x + 12 = 2x + 24 → x = 12. Check: in 12 years, 48 = 2 × 24. ✓
10. 6 students (and 6 adults). Let s = students, a = adults, a + s = 12 and 150a + 90s = 1,440. Substitute a = 12 − s: 150(12 − s) + 90s = 1,440 → 1,800 − 60s = 1,440 → −60s = −360 → s = 6. Check: 6 × 150 + 6 × 90 = 900 + 540 = 1,440. ✓
11. x = 12. LCD is 12: 3x + 2x = 60 → 5x = 60 → x = 12.
12. Width 5, length 8. w(w + 3) = 40 → w² + 3w − 40 = 0 → (w + 8)(w − 5) = 0 → w = 5 (reject −8), length = 8. Check: 5 × 8 = 40. ✓
13. 60. "Decreased by 20%" means × 0.80: 0.80x = 48 → x = 60.
14. 8m⁹n⁶. 2³ · (m³)³ · (n²)³ = 8m⁹n⁶.

17. Summary

Algebra is arithmetic with a named blank. Its golden rule is the balance: do the same operation to both sides, and the equality holds. Most exam algebra is really translation — turning a sentence into an equation — so master the word-to-symbol map, and mind the "less than" reversal. Isolate the variable by undoing operations in reverse; keep the exponent rules straight (add when multiplying powers, multiply when raising a power to a power); recognize the two factoring patterns; and know both routes to a quadratic — factoring and the formula. For two unknowns, build two equations. Above all, set up carefully, solve for what's actually asked, and check against reality — those three habits protect nearly every mark in this topic.

18. Cheat Sheet

Translation words: sum/more than = +, less than/difference = − (flips order), of/times/twice = ×, per/divided = ÷, is/equals/will be = =.

19. Frequently Asked Questions

How do I know which letter to use for the variable? It doesn't matter mathematically — x is customary. Choose a letter that reminds you of the quantity if it helps (a for adults, s for students).

When should I factor versus use the quadratic formula? Try factoring first; it's faster. If you can't find whole-number factors within a few seconds, switch to the formula, which always works.

Why does a negative times a negative give a positive? Think of it as "the opposite of the opposite." Taking away a debt (a negative) repeatedly acts like a gain. For the exam, just remember the rule: (−)(−) = (+).

A quadratic gave me two answers but the problem is about age. Which do I keep? Keep only the answer that makes real-world sense. Ages, lengths, and counts can't be negative, so discard a negative solution.

Is back-solving "cheating"? No. On a multiple-choice exam, any correct method that gets the right answer earns the full mark. Back-solving is a smart, legitimate tool.

20. Mastery Checklist

• ☐ I can translate a sentence into an equation, minding the "less than" order.
• ☐ I can isolate a variable by doing the same operation to both sides.
• ☐ I can combine like terms and clear parentheses with the distributive property.
• ☐ I know the exponent rules and don't confuse adding vs. multiplying exponents.
• ☐ I can factor a difference of squares and a perfect-square trinomial.
• ☐ I can solve a quadratic by factoring and by the formula.
• ☐ I can solve a system of two equations in two unknowns.
• ☐ I can set up and solve age, consecutive-integer, and ticket/coin problems.
• ☐ I remember to flip the inequality sign when dividing by a negative.
• ☐ I answer the quantity actually asked for, and I check against reality.

When these are all ticked, you've mastered exam algebra — and you've built the translation skill that powers profit-and-loss, ratio, work, and mixture problems too.