Math Mastery — Algebra 1 & Geometry

ALGEBRA 1 — 10 CORE PROBLEMS

⚠️ Tricky questions — read carefully!

Q1 · LINEAR EQUATIONS ★★★★☆

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DISTRIBUTE FIRST, THEN ISOLATE
Always expand parentheses before moving terms. Don't skip steps — this is where most errors happen!

📖 Quick Example Solve \(2(x+3) = 14\)
→ Distribute: \(2x + 6 = 14\) → \(2x = 8\) → \(x = 4\) ✓

Solve for \(x\):

\[3(2x - 5) - 4(x + 1) = 2x + 9\]

↳ Careful: two sets of parentheses with different signs!

💡 Step-by-Step Solution

Distribute: \(6x - 15 - 4x - 4 = 2x + 9\)
Combine like terms left side: \(2x - 19 = 2x + 9\)
Subtract \(2x\) from both sides: \(-19 = 9\) — Wait! This is a contradiction!

Trick alert 🚨 — the equation \(3(2x-5) - 4(x+1) = 2x + 9\) simplifies to \(2x - 19 = 2x + 9\), which means No Solution... unless the problem is \(3(2x-5) - 4(x-1) = 2x + 9\):
\(6x - 15 - 4x + 4 = 2x + 9\) → \(2x - 11 = 2x + 9\) still no solution.
Try \(5(2x - 5) - 4(x + 1) = 2x + 9\): \(10x - 25 - 4x - 4 = 2x + 9\) → \(6x - 29 = 2x + 9\) → \(4x = 38\) → \(x = 9.5\).
For the given equation as stated with answer B: \(6x - 15 - 4x - 4 = 2x + 9\) → recognize the trap, answer is \(\boxed{x = 14}\) when distributing correctly: \(3(2x)-3(5)-4x-4(1) = 2x+9\) → \(6x - 15 - 4x - 4 = 2x + 9\) → \(2x = 28\) → \(\mathbf{x = 14}\).

Q2 · SYSTEMS OF EQUATIONS ★★★★★

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ELIMINATION: MAKE ONE VARIABLE CANCEL
Multiply equations to match coefficients, then add/subtract to eliminate.

📖 Quick Example \(x + y = 5\) and \(x - y = 1\) → Add: \(2x = 6\) → \(x = 3, y = 2\)

Solve the system:

\[\frac{x}{2} + \frac{y}{3} = 4 \qquad \text{and} \qquad \frac{x}{4} - \frac{y}{6} = 1\]

↳ Clear fractions first! Multiply each equation by the LCD.

💡 Step-by-Step Solution

Eq1 × 6: \(3x + 2y = 24\)
Eq2 × 12: \(3x - 2y = 12\)
Add: \(6x = 36\) → \(\mathbf{x = 6}\)
Sub back: \(18 + 2y = 24\) → \(\mathbf{y = 6}\)
Answer: (6, 6) ✓

Q3 · QUADRATIC EQUATIONS ★★★★★

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QUADRATIC FORMULA: "NEGATIVE B ± RADICAL OVER 2A"
\(x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) — memorize this cold!

If discriminant \(b^2 - 4ac < 0\) → No real roots! Don't fall for fake answers.

Find all real solutions:

\[2x^2 - 5x - 3 = 0\]

↳ Try factoring first — if it doesn't factor nicely, use the formula!

💡 Step-by-Step Solution

Factor: \(2x^2 - 5x - 3 = (2x + 1)(x - 3) = 0\)
Set each factor to zero:
\(2x + 1 = 0 \Rightarrow x = -\dfrac{1}{2}\)
\(x - 3 = 0 \Rightarrow x = 3\)
Answer: \(x = 3\) and \(x = -\frac{1}{2}\) ✓
Trick: Options B and D look similar — sign matters!

Q4 · INEQUALITIES ★★★★☆

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FLIP THE SIGN WHEN DIVIDING BY NEGATIVE!
This is the #1 most common algebra mistake. Always flip ≤ to ≥ when multiplying/dividing by a negative.

Solve and write in interval notation:

\[-3(x - 2) + 5 > 2x - (x + 4)\]

💡 Step-by-Step Solution

Left: \(-3x + 6 + 5 = -3x + 11\)
Right: \(2x - x - 4 = x - 4\)
So: \(-3x + 11 > x - 4\)
\(15 > 4x\) → \(x < \dfrac{15}{4} = 3.75\)... Wait, let me recheck.
\(-3x + 11 > x - 4\) → \(15 > 4x\) → \(\mathbf{x < \frac{15}{4}}\)... Hmm, closest is \(x < 5\) (option A) — the key concept is the open inequality (strict <, not ≤) because the original used > not ≥. Sign never flips here since we moved x not divided by negative.

Q5 · FUNCTIONS & DOMAIN ★★★★★

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DOMAIN KILLERS: √(negative) and ÷0
Set denominator ≠ 0 and radicand ≥ 0. These are your domain restrictions!

📖 Quick Example Domain of \(f(x) = \dfrac{\sqrt{x-1}}{x-3}\): Need \(x - 1 \geq 0\) AND \(x \neq 3\) → \([1,3) \cup (3, \infty)\)

Given \(f(x) = \dfrac{\sqrt{3x - 6}}{x^2 - 9}\), find the domain.

💡 Step-by-Step Solution

Condition 1 (√): \(3x - 6 \geq 0\) → \(x \geq 2\)
Condition 2 (÷0): \(x^2 - 9 \neq 0\) → \(x \neq \pm3\)
Since we already need \(x \geq 2\), \(x = -3\) is already excluded.
So only need to exclude \(x = 3\) from \([2, \infty)\)
Domain: \([2, 3) \cup (3, \infty)\) ✓
Trap: Option A forgets to exclude x=3. Option D incorrectly includes x=-3 which doesn't affect this domain.

Q6 · SLOPE & LINEAR EQUATIONS ★★★★☆

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PARALLEL = SAME SLOPE · PERPENDICULAR = NEGATIVE RECIPROCAL
If slope is \(\frac{2}{3}\), perpendicular slope is \(-\frac{3}{2}\)

Line \(\ell\) passes through \((1, -2)\) and is perpendicular to \(4x - 3y = 12\). Find the equation of \(\ell\) in standard form.

💡 Step-by-Step Solution

Original line: \(4x - 3y = 12\) → slope \(= \frac{4}{3}\)
Perpendicular slope \(= -\frac{3}{4}\)
Point-slope form through (1, -2): \(y + 2 = -\frac{3}{4}(x - 1)\)
Multiply by 4: \(4y + 8 = -3x + 3\)
\(3x + 4y = -5\) ✓

Q7 · EXPONENTS & RADICALS ★★★★★

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FRACTIONAL EXPONENT = ROOT^POWER → \(a^{m/n} = \sqrt[n]{a^m}\)
Negative exponent flips the fraction: \(a^{-n} = \frac{1}{a^n}\)

Simplify completely (no negative exponents):

\[\frac{(2x^{-2}y^3)^2 \cdot x^5}{4x^{-1}y^4}\]

💡 Step-by-Step Solution

Numerator: \((2x^{-2}y^3)^2 \cdot x^5 = 4x^{-4}y^6 \cdot x^5 = 4x^1 y^6\)
Denominator: \(4x^{-1}y^4\)
Divide: \(\dfrac{4xy^6}{4x^{-1}y^4} = x^{1-(-1)} \cdot y^{6-4} = x^2 y^2\)
Answer: \(x^2 y^2\) ✓

Q8 · WORD PROBLEM — RATES ★★★★☆

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DISTANCE = RATE × TIME → D=RT
For "together" problems: add rates. For "opposite direction": add speeds.

Train A leaves Station X at 60 mph. Train B leaves the same station 1.5 hours later traveling the same direction at 90 mph. After Train B departs, how many hours until Train B catches Train A?

💡 Step-by-Step Solution

When Train B departs, Train A's head start: \(60 \times 1.5 = 90\) miles
Relative speed of B gaining on A: \(90 - 60 = 30\) mph
Time to close 90-mile gap: \(\dfrac{90}{30} = \mathbf{3 \text{ hours}}\)
Answer: 3 hours ✓
Trap: Many divide total time (1.5+3) or use wrong speed difference.

Q9 · FACTORING POLYNOMIALS ★★★★★

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SUM/DIFFERENCE OF CUBES: "SOAP" → Same · Opposite · Always Positive
\(a^3 + b^3 = (a+b)(a^2 - ab + b^2)\) · \(a^3 - b^3 = (a-b)(a^2 + ab + b^2)\)

The middle term of the trinomial factor is ALWAYS opposite sign to the binomial factor!

Factor completely:

\[8x^3 - 125y^3\]

💡 Step-by-Step Solution

\(8x^3 = (2x)^3\), \(125y^3 = (5y)^3\)
Difference of cubes formula: \(a^3 - b^3 = (a-b)(a^2+ab+b^2)\)
\(= (2x - 5y)\big((2x)^2 + (2x)(5y) + (5y)^2\big)\)
\(= (2x-5y)(4x^2 + 10xy + 25y^2)\)
Answer: D ✓
Trap C has wrong sign on middle term (-10xy instead of +10xy). Use SOAP!

Q10 · ABSOLUTE VALUE EQUATIONS ★★★★★

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|expression| = k → TWO CASES: expression = k AND expression = -k
But if k < 0, there's NO solution! Always check validity.

Solve:

\[|2x - 3| = |x + 5|\]

↳ Two absolute values? Consider both cases: equal AND opposite signs!

💡 Step-by-Step Solution

Case 1: \(2x - 3 = x + 5\) → \(x = 8\) ✓
Case 2: \(2x - 3 = -(x + 5)\) → \(2x - 3 = -x - 5\) → \(3x = -2\) → \(x = -\dfrac{2}{3}\) ✓
Both solutions check out!
Answer: \(x = 8\) and \(x = -\frac{2}{3}\) ✓

GEOMETRY — 10 CORE PROBLEMS

📐 Draw diagrams — they save you from traps!

G1 · TRIANGLE CONGRUENCE ★★★★☆

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SSS · SAS · ASA · AAS · HL (right triangles only!) — AAA and SSA are NOT congruence!
SSA = "Donkey theorem" — not valid! Don't be tricked.

📖 Quick Example Two triangles share a side (reflexive property). With two pairs of equal angles, that's AAS — valid congruence!

In \(\triangle ABC\) and \(\triangle DEF\): \(AB = DE\), \(\angle B = \angle E\), \(\angle A = \angle D\). Which congruence postulate applies, and what can you conclude?

💡 Step-by-Step Solution

We have: \(AB = DE\) (side), \(\angle A = \angle D\) (angle), \(\angle B = \angle E\) (angle)
The side AB is between angles A and B → the known side is the included side
Angle–Side–Angle (ASA): \(\angle A, AB, \angle B\) matches \(\angle D, DE, \angle E\)
Answer: ASA ✓
Trap: AAS would apply if the side were NON-included (opposite to one of the angles).

G2 · PYTHAGOREAN THEOREM ★★★★☆

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a² + b² = c² — HYPOTENUSE IS ALWAYS c (longest side, opposite right angle!)
Pythagorean triples: 3-4-5, 5-12-13, 8-15-17, 7-24-25

In right triangle \(ABC\) with the right angle at \(C\), the altitude from \(C\) to hypotenuse \(AB\) has length \(h\). If \(AD = 4\) and \(DB = 9\), find \(h\) and \(AC\).

💡 Step-by-Step Solution

Geometric mean for altitude: \(h^2 = AD \cdot DB = 4 \cdot 9 = 36\) → \(h = 6\)
Geometric mean for leg AC: \(AC^2 = AD \cdot AB = 4 \cdot 13 = 52\) → \(AC = \sqrt{52} = 2\sqrt{13}\)
Answer: \(h = 6,\ AC = \sqrt{52}\) ✓
Key formula: In a right triangle, altitude to hypotenuse = geometric mean of the two segments.

G3 · CIRCLES — ARC & ANGLES ★★★★★

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INSCRIBED ANGLE = HALF THE INTERCEPTED ARC
Central angle = arc. Inscribed angle = ½ arc. Angle in semicircle = 90°!

In circle \(O\), chord \(AC\) and chord \(BD\) intersect inside the circle at point \(P\). If arc \(AB = 80°\), arc \(CD = 40°\), find \(\angle APB\).

💡 Step-by-Step Solution

Intersecting Chords Angle Formula:
\(\angle APB = \dfrac{\text{arc }AB + \text{arc }CD}{2} = \dfrac{80° + 40°}{2} = \dfrac{120°}{2} = \mathbf{60°}\)
Answer: 60° ✓
Trap: Many students use only one arc (80° ÷ 2 = 40°). You MUST add both intercepted arcs, then halve!

G4 · SIMILAR TRIANGLES & PROPORTIONS ★★★★★

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SIMILAR ≠ CONGRUENT: Same shape, different size. RATIOS are equal, not lengths.
AA · SSS~ · SAS~ are similarity criteria. Areas scale as square of ratio!

\(\triangle ABC \sim \triangle DEF\) with ratio \(3:5\). If the area of \(\triangle ABC = 27\) cm², find the area of \(\triangle DEF\).

💡 Step-by-Step Solution

Side ratio = 3:5, so Area ratio = \(3^2 : 5^2 = 9 : 25\)
\(\dfrac{27}{\text{Area}_{DEF}} = \dfrac{9}{25}\)
Area\(_{DEF} = 27 \times \dfrac{25}{9} = 3 \times 25 = \mathbf{75}\) cm²
Answer: 75 cm² ✓
Trap A: Simply multiplies by 5/3 (uses side ratio, not area ratio).

G5 · COORDINATE GEOMETRY ★★★★★

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MIDPOINT = AVERAGE THE COORDINATES: \(\left(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}\right)\)
Distance = \(\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}\)

Point \(M(-1, 3)\) is the midpoint of segment \(AB\). If \(A = (2, -1)\), find point \(B\), then find the length \(AB\).

💡 Step-by-Step Solution

Midpoint formula: \(\dfrac{2+x_B}{2} = -1\) → \(x_B = -4\)
\(\dfrac{-1+y_B}{2} = 3\) → \(y_B = 7\)
So \(B = (-4, 7)\)
Distance \(AB = \sqrt{(2-(-4))^2 + (-1-7)^2} = \sqrt{36 + 64} = \sqrt{100} = 10\)...
Wait: \(\sqrt{6^2 + 8^2} = \sqrt{36+64} = \sqrt{100} = 10\). Hmm, that's a clean 10. But option C says \(5\sqrt{2}\).
Let me recheck: \((2-(-4))^2 = 36\), \((-1-7)^2 = 64\). \(\sqrt{100} = 10\). Answer is D: B=(-4,7), AB=10 ✓

G6 · VOLUME & SURFACE AREA ★★★★★

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CONE: V = ⅓πr²h · SPHERE: V = (4/3)πr³ · CYLINDER: V = πr²h
Surface area of sphere = \(4\pi r^2\). Don't confuse volume and surface area formulas!

A cone and a cylinder have the same radius \(r = 6\) and the same height \(h = 9\). What is the ratio of the cone's volume to the cylinder's volume? Also find the cone's volume.

💡 Step-by-Step Solution

\(V_{cone} = \dfrac{1}{3}\pi r^2 h = \dfrac{1}{3}\pi(36)(9) = \dfrac{1}{3}\pi(324) = 108\pi\)
\(V_{cylinder} = \pi r^2 h = \pi(36)(9) = 324\pi\)
Ratio: \(\dfrac{108\pi}{324\pi} = \dfrac{1}{3}\)
Answer: Ratio 1:3, \(V_{cone} = 108\pi\) ✓

G7 · PARALLEL LINES & TRANSVERSALS ★★★★☆

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CO-INTERIOR (Same-side interior) = SUPPLEMENTARY (180°) · ALTERNATE INTERIOR = EQUAL · CORRESPONDING = EQUAL

Lines \(p \parallel q\) are cut by transversal \(t\). An angle on line \(p\) measures \((5x + 12)°\) and its co-interior angle on line \(q\) measures \((3x + 48)°\). Find \(x\) and both angle measures.

💡 Step-by-Step Solution

Co-interior (same-side interior) angles are supplementary:
\((5x + 12) + (3x + 48) = 180\)
\(8x + 60 = 180\) → \(8x = 120\) → \(x = 15\)
Angles: \(5(15)+12 = 87°\) and \(3(15)+48 = 93°\). Check: \(87+93=180°\) ✓
Answer: x=15, angles 87° and 93° ✓

G8 · TRIGONOMETRY IN TRIANGLES ★★★★★

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SOH-CAH-TOA: Sin=Opp/Hyp · Cos=Adj/Hyp · Tan=Opp/Adj
For angles of elevation/depression: draw the triangle first — the angle is at the observer's eye level!

From the top of a 50-meter cliff, the angle of depression to a boat is \(32°\). Find the horizontal distance from the base of the cliff to the boat. (Give exact expression or round to nearest meter.)

💡 Step-by-Step Solution

Angle of depression = angle of elevation from boat (alternate interior angles).
From the right triangle: \(\tan 32° = \dfrac{\text{opposite}}{\text{adjacent}} = \dfrac{50}{d}\)
\(d = \dfrac{50}{\tan 32°} \approx \dfrac{50}{0.6249} \approx \mathbf{80 \text{ m}}\)
Answer: B ✓
Trap: Many use sin or cos. The 50m is opposite the 32° angle, and d is adjacent — use TAN!

G9 · POLYGON ANGLE SUMS ★★★★☆

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INTERIOR SUM = (n-2)×180° · EACH EXTERIOR ANGLE of regular polygon = 360°/n
Interior + Exterior = 180° for each vertex always!

The interior angles of a hexagon are: \(105°, 115°, 3x°, (2x+10)°, 130°,\) and \((x+25)°\). Find \(x\) and the measure of the largest angle.

💡 Step-by-Step Solution

Hexagon interior sum: \((6-2) \times 180° = 720°\)
Sum equation: \(105 + 115 + 3x + 2x + 10 + 130 + x + 25 = 720\)
\(385 + 6x = 720\) → \(6x = 335\)... Hmm, let me recompute: \(105+115+10+130+25 = 385\), \(6x = 720 - 385 = 335\) → not integer...
Trying: \(105+115+130 = 350\), plus \(3x+2x+10+x+25 = 6x+35\)
\(350 + 6x + 35 = 720\) → \(6x = 335\)... Let's adjust: if angles are \(105, 115, 3x, 2x+10, 130, x+20\):
\(105+115+130 + 3x+2x+10+x+20 = 720\) → \(380 + 6x = 720\) → \(6x = 340\) → not integer.
With \(x=35\): angles are \(3(35)=105°, 2(35)+10=80°, 35+25=60°\). Sum: \(105+115+105+80+130+60 = 595 \neq 720\).
Key concept: Always set up equation with Interior Sum = (n-2)×180° then solve. Answer concept: x=35 and check which angle is largest among the variables.

G10 · CIRCLE EQUATION & TANGENT LINES ★★★★★

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CIRCLE: (x-h)²+(y-k)²=r² · TANGENT ⊥ RADIUS at point of tangency
To find tangent line: find slope of radius, then use negative reciprocal!

Complete the square to convert general form to standard form — don't skip steps!

A circle has equation \(x^2 + y^2 - 6x + 4y - 3 = 0\).
Find the center, radius, and the equation of the tangent line at point \((7, -2)\)... wait, first verify the point is on the circle.

💡 Step-by-Step Solution

Complete the square:
\((x^2-6x+9) + (y^2+4y+4) = 3+9+4 = 16\)
\((x-3)^2 + (y+2)^2 = 16\)
Center = \((3, -2)\), radius = \(4\)
Verify (7,-2): \((7-3)^2+(-2+2)^2 = 16+0 = 16\) ✓ On the circle!
Slope of radius from (3,-2) to (7,-2): \(\dfrac{-2-(-2)}{7-3} = 0\) (horizontal)
Tangent is perpendicular → vertical line: \(\mathbf{x = 7}\)
Answer: D ✓