ASVAB Mathematics Knowledge — Geometry and the Pythagorean Theorem

Linear equation solving: isolate the variable. Step 1: subtract 7 from both sides: 3x + 7 − 7 = 22 − 7, giving 3x = 15. Step 2: divide both sides by 3: 3x ÷ 3 = 15 ÷ 3, giving x = 5. Verify: 3(5) + 7 = 15 + 7 = 22 ✓. The two-step linear equation is the foundation of algebra. Strategy: undo operations in reverse order of operations — first reverse addition/subtraction, then reverse multiplication/division. Many ASVAB MK problems use this structure. Common errors: dividing before subtracting; sign errors; arithmetic errors. Always verify by substituting the answer back into the original equation. Linear equations in one variable always have exactly one solution unless the variable terms cancel.

Two approaches both work. Approach 1 — distribute first: 2(x − 4) = 14 becomes 2x − 8 = 14; add 8: 2x = 22; divide by 2: x = 11. Approach 2 — divide first: divide both sides by 2: x − 4 = 7; add 4: x = 11. Verify: 2(11 − 4) = 2(7) = 14 ✓. When the variable appears once inside parentheses with a coefficient, dividing first is often cleaner. When the equation involves multiple variable terms, distributing first is necessary. Order of operations matters during solving just as in evaluation. Parentheses with subtraction are a common source of sign errors — distribute carefully.

Distribute each coefficient through its parentheses, watching signs. 4(2x + 3) = 8x + 12. −2(x − 5) = −2x + 10 (note: −2 times −5 = +10). Combine: 8x + 12 − 2x + 10. Combine like terms: (8x − 2x) + (12 + 10) = 6x + 22. The most common error is sign mistakes when distributing a negative coefficient — −2(x − 5) is −2·x + (−2)·(−5) = −2x + 10, NOT −2x − 10. Always distribute the sign with the coefficient. Like terms (terms with the same variable raised to the same power) can be combined; constants are like terms with each other; x terms are like terms with each other.

System of two equations in two unknowns. Elimination method (often easiest for ASVAB): add the equations to eliminate y. (x + y) + (x − y) = 10 + 4, giving 2x = 14, so x = 7. Substitute back: 7 + y = 10, so y = 3. Verify both: 7 + 3 = 10 ✓ and 7 − 3 = 4 ✓. Substitution method: solve first equation for y: y = 10 − x; substitute into second: x − (10 − x) = 4; simplify: 2x − 10 = 4; solve: x = 7; then y = 3. Both methods give the same answer. Elimination works well when adding or subtracting eliminates a variable (coefficients are opposites or equal); substitution works when one equation is easily solved for a variable. On multiple-choice tests, you can also test answers by substituting — efficient for systems.

Area of a rectangle = length × width = 12 × 8 = 96 square units. Memorize basic geometry formulas: (1) Rectangle area = l × w, perimeter = 2(l + w); (2) Square area = s², perimeter = 4s; (3) Triangle area = ½ × base × height; (4) Circle area = πr², circumference = 2πr; (5) Parallelogram area = base × height; (6) Trapezoid area = ½(b₁ + b₂) × h. Perimeter sums the lengths around the boundary; area measures the surface inside. Volume formulas: rectangular prism = l × w × h; cylinder = πr²h; sphere = (4/3)πr³; cone = (1/3)πr²h. Units: area is in square units (sq ft, sq m, etc.); volume in cubic units; perimeter in linear units. The ASVAB MK includes several geometry questions; formula recognition is essential.

Area of a circle = πr², where r is the radius. With r = 5: A = π(5)² = 25π square units. Numerically: 25π ≈ 25 × 3.14159 ≈ 78.5 square units. Common ASVAB answer choices keep π symbolic to avoid arithmetic ambiguity. Related circle formulas: (1) Diameter = 2r (twice the radius); (2) Circumference = 2πr or πd (length around the circle); (3) Area = πr² (surface inside); (4) Arc length = (θ/360) × 2πr where θ is the central angle in degrees; (5) Sector area = (θ/360) × πr². Common error: confusing radius and diameter — read the problem carefully. The constant π ≈ 3.14 ≈ 22/7 (older approximation); for precise work, calculator π. Squaring the radius first, then multiplying by π, avoids confusion.

Pythagorean theorem: in a right triangle, a² + b² = c², where a and b are the legs and c is the hypotenuse (the longest side, opposite the right angle). Here: 3² + 4² = 9 + 16 = 25, so c² = 25, and c = √25 = 5. The 3-4-5 right triangle is a famous Pythagorean triple — memorize it and its multiples (6-8-10, 9-12-15, etc.). Other common Pythagorean triples to memorize: 5-12-13, 8-15-17, 7-24-25. Note: the hypotenuse is always the longest side. The Pythagorean theorem only applies to RIGHT triangles. Common applications: finding distances, ladder problems, navigation, construction. The theorem extends to coordinate geometry as the distance formula: distance = √((x₂-x₁)² + (y₂-y₁)²), which is just the Pythagorean theorem in disguise.

Volume of a rectangular prism (box) = length × width × height = 5 × 4 × 3 = 60 cubic units. Multiply in any order: 5 × 4 = 20, then 20 × 3 = 60. Volume formulas to know: (1) Rectangular prism: V = lwh; (2) Cube: V = s³ (side cubed); (3) Cylinder: V = πr²h (area of circular base times height); (4) Cone: V = (1/3)πr²h (one-third the cylinder); (5) Sphere: V = (4/3)πr³; (6) Pyramid: V = (1/3) × base area × height. Surface area is different — sum of all face areas. For a rectangular box, surface area = 2(lw + lh + wh). Units: volume is always cubic. Common error: confusing volume and surface area — read the question. Real-world applications: container capacity, concrete volume, water flow, gas volumes.

2⁵ means 2 multiplied by itself 5 times: 2 × 2 × 2 × 2 × 2 = 4 × 2 × 2 × 2 = 8 × 2 × 2 = 16 × 2 = 32. Memorize common powers: 2⁰ = 1, 2¹ = 2, 2² = 4, 2³ = 8, 2⁴ = 16, 2⁵ = 32, 2⁶ = 64, 2⁷ = 128, 2⁸ = 256, 2⁹ = 512, 2¹⁰ = 1024. Also: 3² = 9, 3³ = 27, 3⁴ = 81; 4² = 16, 4³ = 64; 5² = 25, 5³ = 125. Exponent rules: (1) a^m · a^n = a^(m+n); (2) a^m / a^n = a^(m−n); (3) (a^m)^n = a^(m·n); (4) a⁰ = 1 (for a ≠ 0); (5) a^(−n) = 1/a^n; (6) a^(1/n) = ⁿ√a (the n-th root); (7) (ab)^n = a^n · b^n; (8) (a/b)^n = a^n / b^n. Common error: 2⁵ ≠ 2 × 5 = 10. Exponents are repeated multiplication, not multiplication.

Simplify radicals by finding perfect-square factors. Factor 48: 48 = 16 × 3, where 16 is a perfect square. So √48 = √(16 × 3) = √16 × √3 = 4√3. The perfect-square factor comes out of the radical; the remaining factor stays inside. Steps: (1) Find the largest perfect-square factor; (2) Rewrite as a product; (3) Apply √(ab) = √a × √b; (4) Simplify the perfect-square root. Perfect squares to recognize: 1, 4, 9, 16, 25, 36, 49, 64, 81, 100, 121, 144. Example: √72 = √(36 × 2) = 6√2. Example: √200 = √(100 × 2) = 10√2. If no perfect-square factor exists beyond 1, the radical is already simplified (like √7, √15, √23). Common error: stopping with non-largest factor — √48 = √(4 × 12) = 2√12, which is not fully simplified because 12 has a perfect-square factor of 4 itself.