Algebra and Fundamentals - Detailed Explanation
Algebra forms the backbone of the Mathematics, Surveying, and Transportation Engineering (MSTE) component of the PRC Civil Engineer Licensure Examination. Board exam statistics consistently show that quadratic equations, progressions, logarithms, and word problems (work, mixture, age, motion) collectively account for roughly 30–40% of the algebra items. Mastery of these fundamentals is non-negotiable: they underpin structural analysis formulas, hydraulic flow equations, and cost-engineering computations you will encounter throughout your professional career. This chapter provides board-focused, step-by-step coverage of every major algebra topic — from the laws of exponents to the binomial theorem — with fully worked numerical examples modeled after actual PRC board exam question styles.
Laws of Exponents and Radicals
Examples
Simplify: (x^3 · y^(−2))^4 / (x^(−1) · y^3)
Solution
Step 1 — Apply power of a product to numerator:
(x^3)^4 · (y^(−2))^4 = x^12 · y^(−8)
Step 2 — Write the full expression:
x^12 · y^(−8) / (x^(−1) · y^3)
Step 3 — Apply quotient rule (subtract exponents):
x: 12 − (−1) = 13 → x^13
y: −8 − 3 = −11 → y^(−11)
Step 4 — Eliminate negative exponents:
Answer = x^13 / y^11
Work inside-out: power of a product first, then quotient rule. This three-step sequence avoids sign errors and is the recommended board-exam approach.
Simplify: √[3]{54x^5y^7} / √[3]{2xy}
Solution
Step 1 — Combine under a single cube root using quotient rule:
√[3]{54x^5y^7 / (2xy)} = √[3]{27x^4y^6}
Step 2 — Factor out perfect cubes:
27 = 3^3; x^4 = x^3 · x; y^6 = (y^2)^3
√[3]{3^3 · x^3 · y^6 · x} = √[3]{3^3} · √[3]{x^3} · √[3]{y^6} · √[3]{x}
= 3 · x · y^2 · √[3]{x}
Answer = 3xy^2 · ∛x
The key skill is identifying perfect cube factors. Group them, extract them outside the radical, and leave only the non-perfect-cube residue inside.
Exponents encode repeated multiplication in a compact form. Every algebra manipulation — simplifying polynomial expressions, solving exponential equations, working with scientific notation in load calculations — relies on the six fundamental exponent laws. Radicals are simply fractional exponents written in root notation: the nth root of a is a^(1/n). Understanding this equivalence lets you convert between radical and exponential form freely, which is essential when differentiating or integrating in later math courses.
The six laws:
- Product rule: a^m · a^n = a^(m+n)
- Quotient rule: a^m / a^n = a^(m−n)
- Power of a power: (a^m)^n = a^(mn)
- Power of a product: (ab)^n = a^n · b^n
- Zero exponent: a^0 = 1 (a ≠ 0)
- Negative exponent: a^(−n) = 1/a^n
Radical equivalences:
√[n]{a} = a^(1/n); a^(m/n) = √[n]{a^m} = (√[n]{a})^m; √[n]{ab} = √[n]{a} · √[n]{b}; √[n]{a/b} = √[n]{a} / √[n]{b}.
Applications
- Simplifying structural load formulas that contain power expressions (e.g., section modulus S = bd²/6 involves squaring).
- Scientific notation in geotechnical bearing capacity calculations (e.g., qu = 1.3cNc + γDNq — intermediate exponent arithmetic appears in Nc, Nq tables).
- Unit conversion in hydraulics: Q = CA√(2gh) requires square root manipulation.
Common misconceptions
- WRONG: a^m + a^n = a^(m+n). Exponent rules only apply to multiplication/division, not addition.
- WRONG: (a + b)^n = a^n + b^n. This is the binomial expansion mistake — always expand using the binomial theorem.
- WRONG: √(a² + b²) = a + b. The square root does NOT distribute over addition.
- WRONG: a^(−n) = −a^n. Negative exponent means reciprocal, not negation.
Related concepts
- Logarithms (inverse of exponentiation)
- Binomial Theorem (expansion of power expressions)
- Polynomial Equations (exponent rules used in factoring)
Common exam questions
Example
Which is equal to (16x^8)^(3/4)? — Answer: 8x^6 [16^(3/4) = (2^4)^(3/4) = 2^3 = 8; (x^8)^(3/4) = x^6]
Approach
Apply exponent laws in order: power of a power first, then product/quotient rule, then eliminate negative exponents. Write each step explicitly.
Example
Solve 2^(x+1) = 8^(x−1): rewrite 8 = 2^3 → 2^(x+1) = 2^(3x−3) → x+1 = 3x−3 → x = 2.
Approach
Express both sides as powers of the same base, then equate exponents. If bases cannot be made equal, take logarithms of both sides.
Key points to remember
- When multiplying like bases, ADD exponents; when dividing, SUBTRACT exponents.
- a^0 = 1 for any nonzero base — a common source of errors in polynomial evaluation.
- Negative exponent means reciprocal, NOT a negative number: 2^(−3) = 1/8, not −8.
- Fractional exponent: numerator is the power, denominator is the root index.
- Simplify radical expressions by factoring out perfect nth powers before applying rules.
- √[n]{a^n} = |a| for even n (absolute value); equals a for odd n.
Logarithms
Examples
PRC-style: Given log 2 = 0.3010 and log 3 = 0.4771, find log 72.
Solution
Step 1 — Factor 72 into primes: 72 = 8 × 9 = 2^3 × 3^2
Step 2 — Apply log laws:
log 72 = log(2^3 × 3^2)
= log 2^3 + log 3^2
= 3 log 2 + 2 log 3
= 3(0.3010) + 2(0.4771)
= 0.9030 + 0.9542
= 1.8572
Verification: 10^1.8572 ≈ 72 ✓
Decompose the argument into prime factors, then apply the product and power rules. This is the standard approach for ALL 'given log 2 and log 3, find log N' problems on the board.
Solve for x: log₃(x² − 2x) = log₃(3).
Solution
Step 1 — Use one-to-one property (same base, same log = same argument):
x² − 2x = 3
Step 2 — Rearrange to standard quadratic form:
x² − 2x − 3 = 0
Step 3 — Factor:
(x − 3)(x + 1) = 0
x = 3 or x = −1
Step 4 — Check domain (argument must be positive):
For x = 3: x² − 2x = 9 − 6 = 3 > 0 ✓
For x = −1: x² − 2x = 1 + 2 = 3 > 0 ✓
Both solutions are valid: x = 3 or x = −1.
Always check that each solution keeps the log argument positive. Extraneous solutions are a classic board exam trap.
The logarithm log_b(x) = y means b^y = x: it answers the question 'to what power must b be raised to give x?' Common logarithm (base 10, written log) and natural logarithm (base e ≈ 2.71828, written ln) are the two forms you will encounter most on the board exam and in engineering computations (e.g., consolidation settlement uses ln; pH and sound level dB use log₁₀).
Five fundamental log laws:
- log_b(MN) = log_b M + log_b N (product → sum)
- log_b(M/N) = log_b M − log_b N (quotient → difference)
- log_b(M^p) = p · log_b M (power → coefficient)
- log_b(b) = 1 (base to itself = 1)
- log_b(1) = 0 (anything to power 0 = 1)
Change of base formula (critical for calculator use):
log_b x = log x / log b = ln x / ln b
The antilogarithm (antilog) is the inverse: if log x = y, then x = 10^y.
Applications
- Decibel scale in noise impact assessment: L = 10 log(I/I₀).
- pH calculation in water quality engineering: pH = −log[H⁺].
- Terzaghi consolidation: time factor uses natural log in primary consolidation equations.
- Compound interest and project payback period in Engineering Economics: uses log to solve for n.
Common misconceptions
- WRONG: log(M + N) = log M + log N. Addition inside the log has NO simple rule.
- WRONG: log(M − N) = log M − log N. The quotient rule applies to M/N, not M − N.
- WRONG: (log M)^n = n log M. The power rule applies to log(M^n), not to raising the entire log expression to a power.
- WRONG: log 0 = 0. log 0 is undefined (the domain of log is strictly x > 0).
Related concepts
- Laws of Exponents (logarithm is the inverse operation)
- Exponential Equations
- Engineering Economics — compound interest and annuities
Common exam questions
Example
Find log 12 given log 2 = 0.3010, log 3 = 0.4771: log 12 = log(4×3) = 2log2 + log3 = 0.6020 + 0.4771 = 1.0791.
Approach
Factor N into 2s and 3s, then apply product and power log laws with the given values.
Example
log(x + 2) + log(x − 1) = 1 → log[(x+2)(x−1)] = log 10 → x²+x−12=0 → x=3 (reject x=−4, domain violation).
Approach
Consolidate all log terms on one side using log laws, then convert to exponential form (or use one-to-one property if same base). Always check domain.
Key points to remember
- Log laws convert products and powers into sums — this is why engineers use them to simplify exponential growth/decay formulas.
- log(M + N) ≠ log M + log N — this is the single most common log error on board exams.
- ln x and log x differ by the factor ln 10 ≈ 2.3026: ln x = 2.3026 log x.
- The domain of log_b(x) is x > 0 only — logarithm of zero or negative number is undefined over the reals.
- Change of base: memorise log_b x = (log x)/(log b). Use this when your calculator only has log and ln keys.
- If log_b x = log_b y, then x = y (one-to-one property).
Quadratic and Polynomial Equations
Examples
PRC Board 2019-style: The sum of the roots of 3x² − 5x + k = 0 is 5/3. Find k if the roots are equal.
Solution
Step 1 — From Vieta's, sum of roots = −b/a = −(−5)/3 = 5/3 ✓ (already satisfied for any k).
Step 2 — For equal (repeated) roots, discriminant = 0:
b² − 4ac = 0
(−5)² − 4(3)(k) = 0
25 − 12k = 0
k = 25/12
Answer: k = 25/12
Verification: x = −b/2a = 5/6 (repeated root); product = (5/6)² = 25/36 = c/a = k/3 → k = 25/12 ✓
The problem gave us information to confirm the formula but the real question was finding k from the discriminant condition. Always identify WHICH condition is actually being used.
Solve 2x³ − 3x² − 11x + 6 = 0 by finding rational roots.
Solution
Step 1 — Rational Root Theorem: possible roots = ±{1, 2, 3, 6} / ±{1, 2}
= ±1, ±2, ±3, ±6, ±1/2, ±3/2
Step 2 — Test x = 3: P(3) = 2(27) − 3(9) − 11(3) + 6 = 54 − 27 − 33 + 6 = 0 ✓ → (x − 3) is a factor.
Step 3 — Synthetic division by (x − 3):
2 | −3 | −11 | 6 | 6 | 9 | −6 2 | 3 | −2 | 0
Quotient: 2x² + 3x − 2
Step 4 — Factor the quotient:
2x² + 3x − 2 = (2x − 1)(x + 2)
Roots: x = 3, x = 1/2, x = −2
Always test the simplest integer candidates first (±1, ±2, ±3 before fractions). Once one root is found, reduce the degree by synthetic division and repeat.
A quadratic equation has the standard form ax² + bx + c = 0 (a ≠ 0). It is the most frequently examined single-equation type on the MSTE paper. The quadratic formula gives both roots in one expression:
x = [−b ± √(b² − 4ac)] / (2a)
The discriminant D = b² − 4ac classifies the nature of roots:
- D > 0: two real, distinct roots
- D = 0: one real, repeated root (x = −b/2a)
- D < 0: two complex conjugate roots (no real solution)
Vieta's formulas (sum and product of roots) are shortcuts that appear directly in board exam questions:
- Sum of roots: r₁ + r₂ = −b/a
- Product of roots: r₁ · r₂ = c/a
For higher-degree polynomials:
- Rational Root Theorem: possible rational roots are ±(factors of constant term)/(factors of leading coefficient)
- Remainder Theorem: P(r) = remainder when P(x) is divided by (x − r)
- Factor Theorem: (x − r) is a factor of P(x) if and only if P(r) = 0
- Synthetic division: efficient algorithm for dividing by a linear factor (x − r)
Applications
- Structural analysis: computing reactions and internal forces often reduces to solving quadratic equations (e.g., location of zero shear in a beam under distributed load).
- Hydraulics: discharge through orifices and weirs, velocity head equations (V² = 2gh) are quadratic in V.
- Surveying: angle computation in triangulation can produce polynomial equations.
- NSCP 2015 Section 406 (beam design): the depth of equivalent rectangular stress block a = As·fy / (0.85f'c·b) is linear, but interaction diagram generation for columns involves quadratic relationships.
Common misconceptions
- WRONG: Every quadratic has two different real solutions. If D = 0, both roots are the same; if D < 0, roots are complex.
- WRONG: The product of roots c/a assumes the equation is in the form ax²+bx+c=0. Rearrange to standard form FIRST.
- WRONG: A polynomial of degree 3 'might' have only 2 roots. It always has exactly 3 roots (counting complex and repeated roots).
- WRONG: Synthetic division only works with integer divisors. It works for ANY linear divisor (x − r) where r can be any real or complex number.
Related concepts
- Quadratic Formula and Discriminant
- Vieta's Formulas
- Remainder and Factor Theorems
- Binomial Theorem (polynomial expansion)
- Systems of Equations
Common exam questions
Example
Roots are 2+3i and 2−3i: sum = 4, product = (2)²+(3)² = 13 → equation: x² − 4x + 13 = 0.
Approach
Use Vieta's in reverse: x² − (sum of roots)x + (product of roots) = 0. Multiply through by the leading coefficient if needed.
Example
x² − 6x + 9 = 0: D = 36 − 36 = 0 → one real, repeated root x = 3.
Approach
Compute discriminant D = b² − 4ac and classify: D > 0 real distinct; D = 0 real equal; D < 0 complex.
Example
Find remainder when x³ − 3x + 2 is divided by (x − 2): P(2) = 8 − 6 + 2 = 4.
Approach
Substitute x = r directly into the polynomial (Remainder Theorem). Set P(r) = 0 to find unknown coefficients (Factor Theorem).
Key points to remember
- Memorise the quadratic formula: x = [−b ± √(b² − 4ac)] / (2a). This is the single most useful formula in algebra.
- Vieta's formulas allow you to reconstruct a quadratic from its roots: x² − (sum)x + (product) = 0.
- The discriminant sign tells the nature of roots — board exams frequently ask this WITHOUT asking you to actually find the roots.
- For completing the square: (x + b/2a)² = (b² − 4ac)/4a² — this also gives the vertex of the parabola.
- Remainder Theorem: to find the remainder when P(x) is divided by (x − r), simply evaluate P(r).
- A polynomial of degree n has exactly n roots (counting multiplicity, including complex).
Arithmetic and Geometric Progressions
Examples
Find the sum of all even integers from 2 to 200.
Solution
Step 1 — Identify the AP: a₁ = 2, aₙ = 200, d = 2.
Step 2 — Find number of terms:
n = (aₙ − a₁)/d + 1 = (200 − 2)/2 + 1 = 99 + 1 = 100
Step 3 — Apply sum formula:
S₁₀₀ = n/2 · (a₁ + aₙ)
= 100/2 · (2 + 200)
= 50 · 202
= 10,100
Answer: S = 10,100
The formula Sₙ = n/2 · (a₁ + aₙ) is the most efficient when both endpoints are known. Finding n first is the critical first step.
A geometric series has first term 24 and sum to infinity 32. Find the common ratio and the 4th term.
Solution
Step 1 — Use S∞ formula:
S∞ = a₁/(1 − r)
32 = 24/(1 − r)
1 − r = 24/32 = 3/4
r = 1/4
Step 2 — Check |r| = 1/4 < 1 ✓ (convergent)
Step 3 — Find the 4th term:
a₄ = a₁ · r³ = 24 · (1/4)³ = 24/64 = 3/8
Answer: r = 1/4, a₄ = 3/8
S∞ problems always give you a₁ and S∞ to find r. Once r is found, use the nth term formula. Always verify |r| < 1 for infinite GP validity.
Insert 3 arithmetic means between 5 and 25.
Solution
Step 1 — The sequence is: 5, ___, ___, ___, 25 → 5 terms total.
a₁ = 5, a₅ = 25, n = 5
Step 2 — Find d:
aₙ = a₁ + (n−1)d
25 = 5 + 4d
d = 5
Step 3 — Write the means:
a₂ = 5 + 5 = 10
a₃ = 15
a₄ = 20
Answer: the three arithmetic means are 10, 15, 20.
Inserting k arithmetic means between two numbers creates a sequence of k+2 terms total. Set a₁ = first number, aₙ = last number, n = k+2.
A sequence is an ordered list of numbers following a specific pattern. The two most important types for the board exam are the arithmetic progression (AP) and the geometric progression (GP).
ARITHMETIC PROGRESSION (AP):
Each term is obtained by adding a constant common difference d.
- nth term: aₙ = a₁ + (n−1)d
- Sum of n terms: Sₙ = n/2 · (2a₁ + (n−1)d) = n/2 · (a₁ + aₙ)
- Arithmetic mean of two numbers: AM = (a + b)/2
- If a, b, c are in AP: b − a = c − b → 2b = a + c
GEOMETRIC PROGRESSION (GP):
Each term is obtained by multiplying by a constant common ratio r.
- nth term: aₙ = a₁ · r^(n−1)
- Sum of n terms: Sₙ = a₁(r^n − 1)/(r − 1) for r ≠ 1
- Sum to infinity (|r| < 1): S∞ = a₁/(1 − r)
- Geometric mean of two numbers: GM = √(ab)
- If a, b, c are in GP: b/a = c/b → b² = ac
HARMONIC PROGRESSION (HP):
A sequence is HP if the reciprocals of its terms form an AP.
- Harmonic mean: HM = 2ab/(a+b)
- Relationship: AM ≥ GM ≥ HM (always true for positive numbers)
Applications
- Earthworks and grading: earthwork volumes in prismatoid formula involve sums of cross-sections (AP structure).
- Structural engineering: reinforced concrete stirrup spacing is sometimes designed in arithmetic progression for economy.
- Engineering Economics: uniform gradient series and geometric gradient series in cash flow analysis are direct applications of AP and GP formulas.
- Population growth modeling (GP) used in water supply and sewerage design to project future demand.
Common misconceptions
- WRONG: Applying S∞ = a₁/(1−r) when |r| ≥ 1. This formula only works for |r| < 1 (convergent series).
- WRONG: Confusing AP sum formula Sₙ=n/2(2a₁+(n−1)d) with the nth term formula aₙ=a₁+(n−1)d. Sum includes the factor n/2; nth term does not.
- WRONG: In 'insert k means' problems, using n = k instead of n = k + 2. The two given endpoints ARE counted as terms.
- WRONG: Checking if a sequence is GP by subtracting consecutive terms instead of dividing.
Related concepts
- Harmonic Progression and Harmonic Mean
- Binomial Theorem (involves summation)
- Engineering Economics — Uniform and Geometric Gradient Series
- Series and Sequences in Calculus
Common exam questions
Example
Sum of 8 terms of GP: a₁ = 3, r = 2 → S₈ = 3(2⁸−1)/(2−1) = 3(255) = 765.
Approach
For AP: Sₙ = n/2[2a₁ + (n−1)d]. For GP: Sₙ = a₁(r^n − 1)/(r−1). Identify whether sequence is AP or GP first by checking differences vs. ratios.
Example
AP: a₁=5, aₙ=95, Sₙ=1000 → n = 2(1000)/(5+95) = 20 terms.
Approach
Use Sₙ = n/2(a₁ + aₙ) for AP and solve for n. This is a linear equation in n.
Example
S∞=16, r=1/2 → a₁ = 16(1−1/2) = 8.
Approach
Apply S∞ = a₁/(1−r). If two of the three values (S∞, a₁, r) are given, solve for the third. Verify |r| < 1.
Key points to remember
- In AP, the common difference d = aₙ − aₙ₋₁ (any term minus the preceding term).
- In GP, the common ratio r = aₙ/aₙ₋₁ (any term divided by the preceding term).
- S∞ for GP only exists (converges) when |r| < 1. For |r| ≥ 1, the infinite sum diverges (no finite answer).
- Board exams often give you two conditions (e.g., a₁ and S∞) and ask you to find r or individual terms — set up TWO equations.
- For n terms in AP with first and last term known, use Sₙ = n/2 · (a₁ + aₙ) — this is faster than the other sum formula.
- The number of terms n = [(aₙ − a₁)/d] + 1 for AP — a formula you MUST have for 'how many terms' problems.
Binomial Theorem
Examples
Find the 6th term of (2x − 1)^9.
Solution
Step 1 — Identify: a = 2x, b = −1, n = 9, and we want the 6th term → r = 5 (since r+1=6).
Step 2 — Apply general term formula:
T₆ = C(9,5) · (2x)^(9−5) · (−1)^5
Step 3 — Compute each part:
C(9,5) = 9!/(5!·4!) = 126
(2x)^4 = 16x^4
(−1)^5 = −1
Step 4 — Multiply:
T₆ = 126 · 16x^4 · (−1) = −2016x^4
Answer: T₆ = −2016x^4
This is a direct application of the general term formula — a staple on every MSTE board exam. Identify a, b, n, and r (= term number minus 1), then compute C(n,r), powers, and multiply.
Find the term independent of x (the constant term) in (x² − 2/x)^9.
Solution
Step 1 — Write general term:
T(r+1) = C(9,r) · (x²)^(9−r) · (−2/x)^r
= C(9,r) · x^(18−2r) · (−2)^r · x^(−r)
= C(9,r) · (−2)^r · x^(18−2r−r)
= C(9,r) · (−2)^r · x^(18−3r)
Step 2 — For constant term, set exponent of x = 0:
18 − 3r = 0
r = 6
Step 3 — Compute the constant term:
T₇ = C(9,6) · (−2)^6 · x^0
= 84 · 64 · 1
= 5376
Answer: Constant term = 5376
The 'constant term' or 'term independent of x' question requires setting the net power of x equal to zero and solving for r. This is the highest-difficulty binomial theorem question on the board exam.
The Binomial Theorem provides a formula for expanding (a + b)^n without repeated multiplication. It is particularly useful in the board exam for finding a specific term in the expansion — a task that would be impractical by direct multiplication for large n.
General expansion formula:
(a + b)^n = Σ[k=0 to n] C(n,k) · a^(n−k) · b^k
Where C(n,k) = n!/[k!(n−k)!] is the binomial coefficient, also written as ⁿCₖ or (n choose k).
Key formulas:
- The (r+1)th term [also called the general term or Tᵣ₊₁]:
T(r+1) = C(n,r) · a^(n−r) · b^r
- The MIDDLE term occurs at r = n/2 if n is even.
- Expanded sum of coefficients: set a = b = 1 → 2^n
- Sum of coefficients of expanded (1+x)^n: set x = 1 → 2^n
Properties:
- Coefficients are symmetric: C(n,k) = C(n, n−k)
- Pascal's triangle gives coefficients for small n
- The largest coefficient for even n is the middle one: C(n, n/2)
Applications
- Approximation: (1 + x)^n ≈ 1 + nx for small x — used in structural engineering to simplify small-strain expressions.
- Probability: binomial distribution (the probability mass function uses binomial coefficients) — applied in quality control and reliability analysis.
- Numerical methods: Taylor/Maclaurin series expansion, which underpins finite element and finite difference methods.
Common misconceptions
- WRONG: T(r+1) = C(n,r+1)·a^(n−r−1)·b^(r+1). The index in C(n,k) equals the exponent of b, which is r (not r+1).
- WRONG: Forgetting to include the sign of b when b is negative. (−1)^r alternates sign.
- WRONG: The middle term of (a+b)^n when n is odd — for odd n, there are TWO middle terms at positions (n+1)/2 and (n+3)/2.
- WRONG: Sum of coefficients by setting a=1, b=0 gives 1, not 2^n. Set BOTH a=1 AND b=1 to get 2^n.
Related concepts
- Combinatorics — Permutations and Combinations
- Pascal's Triangle
- Polynomial Equations (binomial expansion produces polynomials)
- Probability — Binomial Distribution
Common exam questions
Example
3rd term of (3a − b)^7: r=2, T₃ = C(7,2)(3a)^5(−b)^2 = 21·243a^5·b^2 = 5103a^5b^2.
Approach
Use T(r+1) = C(n,r)·a^(n−r)·b^r directly. Identify r = (term number − 1). Be careful with the sign of b.
Example
Term with x^4 in (x+2)^10: exponent = 10−r = 4 → r=6 → T₇ = C(10,6)·x^4·2^6 = 210·64·x^4 = 13440x^4.
Approach
Write the general term T(r+1), collect the net exponent of the variable as a function of r, set equal to the target power, solve for r, then substitute.
Key points to remember
- The general term formula: T(r+1) = C(n,r) · a^(n−r) · b^r. Memorise and apply directly — do NOT expand the whole expression.
- The INDEX r starts at 0. The 1st term has r=0, the 2nd has r=1, the kth term has r=k−1.
- To find the term containing a specific power of x: set the exponent equal to the target power and solve for r.
- Sum of all binomial coefficients in (a+b)^n equals 2^n (set a=b=1).
- Alternating sum of coefficients in (1−1)^n = 0 → C(n,0) − C(n,1) + C(n,2) − ... = 0 for n > 0.
- The coefficient of the middle term is the largest for any given n.
Word Problems — Work, Mixture, Age, and Motion
Examples
Pipe A fills a tank in 3 hours, Pipe B fills it in 5 hours, and Drain C empties it in 4 hours. How long to fill the tank with all three open?
Solution
Step 1 — Assign rates (fraction of tank per hour):
Rate of A = 1/3; Rate of B = 1/5; Rate of C = −1/4 (drain)
Step 2 — Net rate = sum of all rates:
Net rate = 1/3 + 1/5 − 1/4
Step 3 — Find common denominator (LCD = 60):
= 20/60 + 12/60 − 15/60
= 17/60 tanks/hour
Step 4 — Time to fill one tank:
t = 1 / (17/60) = 60/17 ≈ 3.53 hours
Answer: t = 60/17 ≈ 3.53 hours
The drain pipe gets a NEGATIVE rate because it removes fluid. The net rate is 17/60 of a tank per hour, so invert to get the fill time. This is a classic three-pipe board exam problem.
PRC-style mixture: How many liters of a 60% alcohol solution must be added to 8 L of a 20% alcohol solution to produce a 50% solution?
Solution
Step 1 — Let x = liters of 60% solution to be added.
Step 2 — Solute balance (alcohol):
(0.20)(8) + (0.60)(x) = (0.50)(8 + x)
Step 3 — Expand:
1.6 + 0.6x = 4 + 0.5x
Step 4 — Solve:
0.6x − 0.5x = 4 − 1.6
0.1x = 2.4
x = 24 liters
Answer: Add 24 liters of the 60% solution.
Verification: Final volume = 32 L; alcohol = 1.6 + 14.4 = 16 L; concentration = 16/32 = 50% ✓
Set up the solute balance equation: (conc₁)(vol₁) + (conc₂)(vol₂) = (final conc)(total vol). Convert percentages to decimals immediately to avoid arithmetic errors.
A is 3 times as old as B. In 4 years, A will be twice as old as B. Find their present ages.
Solution
Step 1 — Assign variables: let B's present age = x; then A's present age = 3x.
Step 2 — In 4 years:
A: 3x + 4; B: x + 4
Step 3 — Set up the condition 'A will be twice B':
3x + 4 = 2(x + 4)
3x + 4 = 2x + 8
x = 4
Step 4 — Find ages:
B = 4 years; A = 12 years.
Verification: In 4 years: A = 16, B = 8; 16 = 2(8) ✓
Age problems almost always need two conditions to produce two equations (or one equation in one unknown, as here). Always verify by substituting back into ALL original conditions.
Word problems translate physical situations into algebraic equations. They consistently appear on the MSTE paper, accounting for a significant portion of the algebra section. The key skill is recognising WHICH mathematical model applies to each scenario.
WORK PROBLEMS:
The fundamental principle is: add RATES, not times.
- If A finishes a job in a days, A's rate = 1/a (jobs per unit time).
- If A, B, C work together: combined rate = 1/a + 1/b + 1/c = 1/t (where t is the time together).
- If one pipe fills and another drains: net rate = (fill rate) − (drain rate).
MIXTURE PROBLEMS:
- Amount of solute = (concentration) × (volume).
- Before and after mixing: solute₁ + solute₂ = solute_final.
- Percent problems: express percentages as decimals.
AGE PROBLEMS:
- Assign present ages as variables.
- Use tense carefully: 'was x years ago' → subtract x; 'will be in x years' → add x.
- Write two equations from two different time references.
MOTION PROBLEMS:
- Fundamental relation: Distance = Rate × Time (D = RT).
- Opposite directions: D_total = D₁ + D₂ = R₁t + R₂t (for same time)
- Same direction: D_difference = (R₁ − R₂)t
- Round trip: total time = d/R₁ + d/R₂
Applications
- Construction scheduling: work problems model crew productivity — if crew A builds a wall in 5 days and crew B in 8 days, scheduling together applies the rate-addition principle.
- Water supply: mixture problems model blending reservoir water of different turbidities or chemical concentrations (DPWH design standards).
- Traffic engineering: motion problems model vehicle headways, speed studies, and travel time surveys per DPWH RAAM guidelines.
- Project management (CPM/PERT): work rates relate to activity durations and resource allocation.
Common misconceptions
- WRONG: Adding times in work problems. If A takes 4 hr and B takes 6 hr, they do NOT take 10 hr together — always add RATES.
- WRONG: Using the final concentration instead of the TARGET concentration as the right-hand side of a mixture equation.
- WRONG: In age problems, applying the time shift to only one person. Both persons' ages change by the same number of years.
- WRONG: In motion problems, using total distance divided by total time to find average speed (this only works when times are equal, not distances).
Related concepts
- Systems of Linear Equations
- Ratio and Proportion
- Percentage and Fraction Arithmetic
- Engineering Economics — productivity and rate analysis
Common exam questions
Example
A can paint in 6 hours; B joins after 2 hours; total time = t. A's fraction + B's fraction = 1 → t/6 + (t−2)/4 = 1 → t = 18/5 = 3.6 hr.
Approach
Assign variable t as total time from start. Person/pipe A works for t hours; person/pipe B works for (t − start delay) hours. Set the sum of completed fractions equal to 1.
Example
Two cars leave the same point opposite directions at 60 km/h and 80 km/h; time to be 350 km apart: 60t + 80t = 350 → t = 2.5 hr.
Approach
Write D = RT for each. If moving in opposite directions, D₁ + D₂ = total distance. If same direction, D₁ − D₂ = gap.
Key points to remember
- Work: add RATES. Rate = 1/(time to complete job alone). NEVER add times directly.
- Mixture: set up solute balance (amount in = amount out × concentration). Volume changes when adding pure substance or solution.
- Age: define ONE variable (usually present age of one person), express everything else in terms of it.
- Motion: D = RT. For problems with TWO vehicles, write a SEPARATE D=RT for each, then relate them (same distance, same time, sum of distances, etc.).
- Upstream/downstream boat problems: effective speed downstream = (boat speed + current); upstream = (boat speed − current).
- Coin/number problems: set up equations for TOTAL COUNT and TOTAL VALUE separately.
Practice problems
Solve 2x² + 3x − 5 = 0 using the quadratic formula. Verify using Vieta's formulas.
Solution
Step 1 — Identify coefficients: a = 2, b = 3, c = −5.
Step 2 — Apply quadratic formula:
x = [−3 ± √(9 − 4(2)(−5))] / (2·2)
= [−3 ± √(9 + 40)] / 4
= [−3 ± √49] / 4
= [−3 ± 7] / 4
Step 3 — Two solutions:
x₁ = (−3 + 7)/4 = 4/4 = 1
x₂ = (−3 − 7)/4 = −10/4 = −5/2
Step 4 — Vieta's check:
Sum: 1 + (−5/2) = −3/2 = −b/a = −3/2 ✓
Product: 1 × (−5/2) = −5/2 = c/a = −5/2 ✓
Answer: x = 1 or x = −5/2
The discriminant D = 49 > 0, confirming two real distinct roots. Vieta's formulas provide a fast self-check — always use them on the exam when time permits.
Find the sum to infinity S∞ of the GP: 8, 4, 2, 1, ...
Solution
Step 1 — Find common ratio: r = 4/8 = 1/2.
Step 2 — Check |r| = 1/2 < 1 ✓ (series converges).
Step 3 — Apply S∞ formula:
S∞ = a₁/(1 − r) = 8/(1 − 1/2) = 8/(1/2) = 16
Answer: S∞ = 16
This is a standard GP convergence problem. The sum approaches 16 but never quite reaches it — each term adds half of the remaining gap to 16. The formula gives the exact limit.
Pipe A fills a tank in 3 hours, Pipe B fills it in 5 hours, and Drain C empties it in 4 hours. How long to fill the tank with all three open?
Solution
Step 1 — Rates per hour: A = 1/3; B = 1/5; C = −1/4.
Step 2 — Net rate:
R_net = 1/3 + 1/5 − 1/4 = 20/60 + 12/60 − 15/60 = 17/60
Step 3 — Time to fill:
t = 1/(17/60) = 60/17 ≈ 3.53 hours
Answer: t = 60/17 h ≈ 3 h 32 min
The drain reduces the net rate. The LCD of 3, 5, 4 is 60 — working in sixtieths of a tank prevents fraction arithmetic errors.
Find the 6th term of (2x − 1)^9.
Solution
Step 1 — Identify: a = 2x, b = −1, n = 9; for 6th term, r = 5.
Step 2 — General term: T(r+1) = C(n,r)·a^(n−r)·b^r
T₆ = C(9,5)·(2x)^(9−5)·(−1)^5
Step 3 — Evaluate:
C(9,5) = 126
(2x)^4 = 16x^4
(−1)^5 = −1
T₆ = 126 × 16x^4 × (−1) = −2016x^4
Answer: T₆ = −2016x^4
The negative sign from b = −1 raised to an odd power (r = 5) makes the term negative. Always compute the sign of b^r separately to avoid errors.
Given log 2 = 0.3010 and log 3 = 0.4771, find log 36.
Solution
Step 1 — Factor: 36 = 4 × 9 = 2² × 3².
Step 2 — Apply log laws:
log 36 = log(2² × 3²)
= 2 log 2 + 2 log 3
= 2(0.3010) + 2(0.4771)
= 0.6020 + 0.9542
= 1.5562
Answer: log 36 = 1.5562
Always factorise the argument into the given base values (2 and 3 here). The product rule converts multiplication into addition; the power rule converts exponents into coefficients.
A man is 4 times as old as his son. In 5 years, he will be 3 times as old as his son. Find their present ages.
Solution
Step 1 — Let son's present age = x; father's = 4x.
Step 2 — In 5 years: father = 4x+5, son = x+5.
Condition: 4x + 5 = 3(x + 5)
4x + 5 = 3x + 15
x = 10
Step 3 — Present ages:
Son = 10 years; Father = 40 years.
Verification: In 5 years: father 45, son 15; 45 = 3(15) ✓
Answer: Son is 10; Father is 40.
Define the younger person's age as the single variable — this minimises algebraic complexity. Apply the future-age condition to generate the equation.
Two trains start from cities A and B, 360 km apart, at the same time, traveling toward each other. Train A travels at 80 km/h and Train B at 100 km/h. When and where do they meet?
Solution
Step 1 — Combined closing speed = 80 + 100 = 180 km/h.
Step 2 — Time to meet:
t = 360/180 = 2 hours
Step 3 — Distance from A:
D_A = 80 × 2 = 160 km from city A
D_B = 100 × 2 = 200 km from city B
Check: 160 + 200 = 360 ✓
Answer: They meet after 2 hours, 160 km from city A.
When two objects move toward each other, add their speeds to get the closing rate. The time is then total distance divided by combined speed — a one-step solution once you set it up correctly.
Find the constant term in the expansion of (x² − 2/x)^9.
Solution
Step 1 — General term:
T(r+1) = C(9,r)·(x²)^(9−r)·(−2/x)^r
= C(9,r)·x^(18−2r)·(−2)^r·x^(−r)
= C(9,r)·(−2)^r·x^(18−3r)
Step 2 — Set exponent of x = 0:
18 − 3r = 0 → r = 6
Step 3 — Compute:
T₇ = C(9,6)·(−2)^6·x^0
= 84 × 64 × 1
= 5376
Answer: Constant term = 5376
The term independent of x has exponent 0. Find r by setting the net exponent equal to zero, then evaluate the coefficient. C(9,6) = C(9,3) = 84 by symmetry.
Exam preparation tips
- MEMORISE the quadratic formula, AP/GP sum formulas, S∞ = a₁/(1−r), and the binomial general term T(r+1) = C(n,r)·a^(n−r)·b^r. These five formulas alone account for 60–70% of algebra board exam problems.
- For 'given log 2 and log 3' problems, IMMEDIATELY factorise the target number into 2s and 3s. Do not attempt to use the calculator — the exam tests whether you know the log product and power rules.
- In work problems, write down rates (1/a, 1/b, 1/c) on scratch paper BEFORE setting up the equation. Assign a negative rate to drain/emptying elements. This mechanical first step prevents the 'add times' error.
- Use VIETA'S FORMULAS as a 10-second self-check after solving any quadratic. Sum the roots and verify = −b/a; multiply roots and verify = c/a. This catches arithmetic errors without re-solving.
- For the binomial theorem, the two most common board question types are: (1) find the rth term, and (2) find the term independent of x or containing a specific power. Practice the 'set exponent = target, solve for r' technique until it is automatic.
- In progression problems, FIRST determine whether the sequence is AP (constant difference) or GP (constant ratio) by testing two consecutive terms. Misidentifying the type is the most costly mistake.
- For mixture problems, ALWAYS write the solute balance equation in the form: (c₁)(V₁) + (c₂)(V₂) = (c_final)(V_final). Converting percentages to decimals immediately avoids a very common arithmetic error.
- Time management on the MSTE: aim to spend no more than 2 minutes per algebra item. If a direct formula applies, use it immediately rather than deriving from first principles. The board exam rewards formula recall and clean arithmetic over derivation.
- Practice SYNTHETIC DIVISION until it is mechanical — it appears in polynomial factoring, remainder theorem, and partial fraction problems across all areas of the exam.
- Age problems: ALWAYS verify your answer by substituting back into all given conditions (present AND future/past). A solution that satisfies only one condition is incomplete.
- For S∞ (infinite GP sum), always CHECK that |r| < 1 BEFORE writing the formula. Writing S∞ = a₁/(1−r) when |r| ≥ 1 is an automatic wrong answer.
- In PRC board exams from 2015–2023, quadratic equation problems (using the formula or Vieta's) appeared in EVERY administration. Treat this as a guaranteed item — perfect your technique.
Diagrams
Algebra and Fundamentals — Master Concept Map
A hierarchical mind map showing all major topic clusters in Algebra and Fundamentals. Use this as a study checklist to ensure no subtopic is overlooked before the board exam.
Quadratic Equation Problem-Solving Decision Tree
A complete decision flowchart for solving any quadratic equation on the board exam. Follow the branches: first check standard form, then attempt factoring; if factoring fails, apply the quadratic formula. Discriminant analysis classifies the nature of roots. Always verify with Vieta's formulas as a final check.
Work Problem Setup Flowchart
Step-by-step procedural flowchart for setting up and solving work problems. The critical branch identifies drain or emptying elements (negative rates) and workers who start late. The verification step at the end ensures the answer is physically reasonable.
Choosing the Right Progression Formula
A decision tree that guides the student from a raw sequence to the correct progression type and the applicable formulas. The critical S∞ branch confirms that infinite geometric series only converges when |r| < 1.
Approximate Distribution of Algebra Topics on MSTE Board Exam
Approximate percentage distribution of algebra question types based on analysis of PRC Civil Engineer board exams from 2015 to 2023. Quadratic equations and word problems are the highest-yield topics (25% each), followed by progressions (20%). Allocate review time proportionally to these weights.
Binomial Theorem — Finding a Specific Term
A procedural flowchart for the three variants of binomial theorem problems: finding the rth term directly, finding the term with a specific power of x, and finding the constant term. The r-validity check prevents errors from non-integer solutions.
In summary
Algebra and Fundamentals is the foundation upon which all subsequent MSTE topics — trigonometry, analytic geometry, calculus, differential equations, and engineering economics — are built. The five most exam-critical skills from this chapter are: (1) applying the quadratic formula and discriminant correctly, (2) adding rates (not times) in work problems, (3) using log laws to decompose composite arguments, (4) computing the nth term and sum of AP and GP with |r| < 1 for infinite GP, and (5) extracting specific terms from binomial expansions using the general term formula. Mastery means being able to execute any of these in under 2 minutes — the target pace for MSTE board exam problems. Use the Vieta's check after every quadratic, verify domain restrictions after every logarithmic equation, and always confirm |r| < 1 before writing S∞. These habits eliminate the most common avoidable errors. Work through each practice problem under timed conditions, review the flowcharts until the decision process is automatic, and cross-reference every formula against the master mind map to ensure zero gaps before exam day. Consistent, formula-focused practice on these fundamentals will give you the confidence and speed needed to dominate the algebra section of the PRC Civil Engineer Licensure Examination.