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Mathematics 5% exam weight

Algebraic Processes

Part of the JAMB UTME study roadmap. Mathematics topic math-2 of Mathematics.

Algebraic Processes

🟢 Lite — Quick Review (1h–1d)

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Algebraic Processes — Quick Facts for JAMB

Indices and Logarithms: Laws of indices: $a^m \times a^n = a^{m+n}$, $a^m ÷ a^n = a^{m-n}$, $(a^m)^n = a^{mn}$, $a^0 = 1$, $a^{-n} = 1/a^n$.

Logarithm laws: $\log_a(xy) = \log_a x + \log_a y$, $\log_a(x/y) = \log_a x - \log_a y$, $\log_a(x^n) = n\log_a x$, $\log_a 1 = 0$, $\log_a a = 1$.

Change of base: $\log_a x = \frac{\log_b x}{\log_b a}$. This is essential for solving problems where you need to convert between bases.

Quadratic Equations: $ax^2 + bx + c = 0$. Solution: $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$. Discriminant $D = b^2 - 4ac$:

  • $D > 0$: two distinct real roots
  • $D = 0$: one repeated real root
  • $D < 0$: no real roots (two complex conjugates)

Sum of roots = $-b/a$, Product of roots = $c/a$.

Exam tip: For quadratic word problems, always check if the solutions make sense in context (e.g., a negative length or time is invalid).


🟡 Standard — Regular Study (2d–2mo)

Standard content for students with a few days to months.

Algebraic Processes — JAMB UTME Study Guide

Surds: Simplify $\sqrt{50} = \sqrt{25 \times 2} = 5\sqrt{2}$. Rationalise denominators: $\frac{1}{\sqrt{3}} = \frac{\sqrt{3}}{3}$. For expressions like $\frac{1}{1+\sqrt{2}}$: multiply numerator and denominator by $1-\sqrt{2}$: $\frac{1-\sqrt{2}}{(1)^2 - (\sqrt{2})^2} = \frac{1-\sqrt{2}}{-1} = \sqrt{2} - 1$.

Linear Equations: Solve simultaneously: substitution or elimination. Example: $3x + 2y = 7$ and $2x - y = 3$. Multiply second equation by 2: $4x - 2y = 6$. Add to first: $7x = 13$, so $x = 13/7$. Then $y = 2x - 3 = 26/7 - 3 = 5/7$.

Quadratic Equations — Factorisation: $2x^2 - 5x + 2 = 0$. Find two numbers that multiply to $2 \times 2 = 4$ and add to $-5$: $-4$ and $-1$. Rewrite: $2x^2 - 4x - x + 2 = 0$. Factor: $2x(x-2) - 1(x-2) = 0$. So $(2x-1)(x-2) = 0$. $x = 1/2$ or $x = 2$.

Word Problems: Example: “A number exceeds three times its reciprocal by 2. Find the number.” Let $x$ be the number. $x = 3(1/x) + 2$. $x^2 = 3 + 2x$. $x^2 - 2x - 3 = 0$. $(x-3)(x+1) = 0$. $x = 3$ or $x = -1$. Check: 3 exceeds $3 \times 1/3 = 1$ by 2 ✓. $(-1)$ exceeds $3 \times (-1) = -3$ by 2 ✓.

Arithmetic and Geometric Sequences:

  • Arithmetic: $T_n = a + (n-1)d$. Sum $S_n = n/2(2a + (n-1)d) = n(a+l)$ where $l$ = last term.
  • Geometric: $T_n = ar^{n-1}$. Sum $S_n = a(r^n - 1)/(r-1)$ for $r \neq 1$. Sum to infinity = $a/(1-r)$ if $|r| < 1$.

Binomial Theorem: $(a+b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k$. Pascal’s triangle or $\binom{n}{k} = \frac{n!}{k!(n-k)!}$. $(x+1)^4 = x^4 + 4x^3 + 6x^2 + 4x + 1$.


🔴 Extended — Deep Study (3mo+)

Comprehensive coverage for students on a longer study timeline.

Algebraic Processes — Comprehensive Mathematics Notes

Simultaneous Equations — Elimination and Substitution:

Example with three variables: $x + y + z = 6$ $2x - y + 3z = 13$ $x - 2y + z = -2$

Step 1: From equation 1: $z = 6 - x - y$. Step 2: Substitute into equation 2: $2x - y + 3(6-x-y) = 13$. $2x - y + 18 - 3x - 3y = 13$. $-x - 4y = -5$. $x + 4y = 5$…(4) Step 3: Substitute into equation 3: $x - 2y + (6-x-y) = -2$. $x - 2y + 6 - x - y = -2$. $-3y = -8$. $y = 8/3$. Step 4: From (4): $x = 5 - 4(8/3) = 5 - 32/3 = (15-32)/3 = -17/3$. Step 5: $z = 6 - (-17/3) - 8/3 = 6 + 17/3 - 8/3 = 6 + 9/3 = 6 + 3 = 9$.

Partial Fractions: Decompose rational functions into simpler fractions.

Case 1 — Linear factors: $\frac{5x+1}{(x-1)(x+2)} = \frac{A}{x-1} + \frac{B}{x+2}$. $5x+1 = A(x+2) + B(x-1)$. At $x=1$: $6 = 3A$, $A=2$. At $x=-2$: $-9 = -3B$, $B=3$. So $\frac{2}{x-1} + \frac{3}{x+2}$.

Case 2 — Repeated linear factors: $\frac{x^2+1}{(x-1)^3} = \frac{A}{x-1} + \frac{B}{(x-1)^2} + \frac{C}{(x-1)^3}$.

Case 3 — Irreducible quadratic: $\frac{5x^2+3x+6}{(x-1)(x^2+2)} = \frac{A}{x-1} + \frac{Bx+C}{x^2+2}$.

Matrices and Determinants: For a $2 \times 2$ matrix $A = \begin{pmatrix} a & b \ c & d \end{pmatrix}$: $\det(A) = ad - bc$. If $\det(A) \neq 0$, the inverse exists: $A^{-1} = \frac{1}{ad-bc}\begin{pmatrix} d & -b \ -c & a \end{pmatrix}$.

For $3 \times 3$ determinant (Sarrus’ rule): $\det = a(ei-fh) - b(di-fg) + c(dh-eg)$.

Complex Numbers: $z = a + bi$, where $i = \sqrt{-1}$. $|z| = \sqrt{a^2 + b^2}$ (modulus). Arg $z = \tan^{-1}(b/a)$ (argument). $z^* = a - bi$ (complex conjugate). $z \times z^* = |z|^2 = a^2 + b^2$.

Multiplication: $(a+bi)(c+di) = ac + adi + bci + bd(i^2) = (ac-bd) + (ad+bc)i$.

Division: $\frac{a+bi}{c+di} = \frac{(a+bi)(c-di)}{c^2+d^2}$.

Quadratic Equations — Applications:

Example: “Two taps fill a tank in 20 and 30 minutes respectively. A drainage tap empties it in 40 minutes. If all three are open, how long to fill the tank?”

  • Rates: Tap A fills $1/20$ per minute. Tap B fills $1/30$ per minute. Drain empties $1/40$ per minute.
  • Net rate = $1/20 + 1/30 - 1/40 = (6+4-3)/120 = 7/120$ per minute.
  • Time = $1/(7/120) = 120/7 = 17\frac{1}{7}$ minutes = 17 minutes 8.6 seconds.

AP and GP — Advanced:

Sum of AP: $S_n = \frac{n}{2}(2a + (n-1)d)$. Also $S_n = n(a+l)/2$ where $l$ is last term.

For GP: $T_n = ar^{n-1}$. If $|r| < 1$, $S_\infty = a/(1-r)$.

Example: $0.\overline{3} = 0.3 + 0.03 + 0.003 + … = 3/10 + 3/100 + 3/1000 + … = \frac{3/10}{1-1/10} = \frac{3/10}{9/10} = 3/9 = 1/3$.

JAMB Pattern Analysis: JAMB questions frequently test: (1) Solving quadratic equations (factorisation or formula), (2) Logarithm evaluation and laws, (3) Simultaneous equations (2 variables), (4) AP/Gp problems, (5) Binomial expansion. Common error: forgetting to rationalise denominators with surds. JAMB 2022: “Simplify $\frac{\sqrt{3}}{1+\sqrt{3}}$.” Answer: $\frac{\sqrt{3}(1-\sqrt{3})}{1-3} = \frac{\sqrt{3}-3}{-2} = \frac{3-\sqrt{3}}{2}$.



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Subjects4 subjects (language + 3 for course)
Time2 hours
Marking+1 per correct answer
Score400 max (used for university admission)
RegistrationJanuary – February each year

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