Elementary algebra — simple equations

Every formula you touch as a power engineer — heat balances, pump head, steam tables, code wall-thickness — is an equation with one unknown you must isolate and solve. Get the algebra wrong and the boiler relief setting, the makeup-water rate, or the safe shell thickness comes out wrong. A misplaced negative sign in a heat-balance calculation can make a shortfall look like a surplus; a bracket removed carelessly in a code wall-thickness formula can under-state the minimum safe thickness. The chief's rule: rearrange the formula symbolically before you plug in numbers. It makes errors easier to catch and keeps your units honest.

Variable cheat-sheet — keep this open while you compute.

This objective builds in the order the exam builds it: signed numbers first (every later rule depends on getting a sign right), then brackets and order of operations, then indices and roots, then the transposition rules that let you rearrange any formula, then the harder equation types (roots, powers, fractions), and finally an organized method for presenting any multi-part problem.

Rules for signed (positive and negative) quantities. A negative quantity is always preceded by a minus sign; integers are all the positive and negative whole numbers plus zero. The four operations each have their own rule:

• Addition — like signs: add the two quantities and keep the common sign. Unlike signs: subtract the smaller quantity from the larger and keep the sign of the larger. Example: $(+5) + (+3) = +8$; $(-5) + (-3) = -8$; $(+5) + (-3) = +2$; $(-5) + (+3) = -2$.
• Subtraction — subtracting a quantity gives the same result as adding its opposite. Change the sign of the quantity being subtracted, then add. Example: $(+5) - (-3) = (+5) + (+3) = +8$; $(-5) - (+3) = (-5) + (-3) = -8$.
• Multiplication and division — like signs give a positive answer; unlike signs give a negative answer. Example: $(-3) \times (-8) = +24$; $(-3) \times (+8) = -24$; $(-24) \div (-6) = +4$; $(-24) \div (+6) = -4$.

Worked example — combining signed corrections. A drum-level transmitter reading of 620 mm must be adjusted by three signed corrections logged over a shift: $-12$ mm (temperature), $+5$ mm (density), and $-9$ mm (instrument bias). Add them in order, applying the addition rule at each step: $620 + (-12) = 608$; $608 + (+5) = 613$; $613 + (-9) = 604$. The corrected reading is 604 mm. Notice each step used the unlike-signs addition rule (subtract the smaller from the larger, keep the sign of the larger) except the second, which added two positives outright.

These four rules are the foundation for every later step in this lesson — bracket removal, transposition, and equation-solving are really just the addition/subtraction/multiplication/division rules applied inside a more complex expression.

Brackets and the order of operations (BEDMAS). An algebraic expression (such as $3a + 7 - (3 \times 2^2)$) can be simplified but not solved — it has no equals sign. The order in which you carry out operations inside and around brackets is fixed by the mnemonic BEDMAS: Brackets, Exponents, Division and Multiplication (co-equal — whichever comes first, left to right), Addition and Subtraction (also co-equal, left to right). BEDMAS is not "always divide before you multiply" — division and multiplication share one tier, and addition and subtraction share the next tier.

Rules for removing brackets:

• A positive sign (or no sign) directly in front of a bracket: remove the bracket, no sign changes. $a + (b - c) = a + b - c$.
• A negative sign in front of a bracket: remove the bracket and flip every sign inside. $a - (b - c) = a - b + c$.
• A multiplier in front of a bracket: distribute it across every term inside, applying the multiplication sign rule to each. $a - 3(b - c) = a - 3b + 3c$.

Worked example — removing brackets. Simplify $5 + 3(2x - 4) - (x - 6)$. Distribute the $+3$: $3(2x-4) = 6x - 12$. Distribute the implied $-1$ in front of the second bracket (flip both signs): $-(x - 6) = -x + 6$. Now the expression reads $5 + 6x - 12 - x + 6$. Gather like terms: $(6x - x) + (5 - 12 + 6) = 5x - 1$.

Worked example — BEDMAS on a numeric expression. Evaluate $4 + 2(9 - 3 \times 2) \div 4$. Brackets first, and inside the bracket BEDMAS still applies (multiply before subtract): $9 - (3 \times 2) = 9 - 6 = 3$. The expression is now $4 + 2(3) \div 4$. Multiplication and division are co-equal, left to right: $2 \times 3 = 6$, then $6 \div 4 = 1.5$. Finally addition: $4 + 1.5 = 5.5$.

Laws of indices and roots. When a quantity is multiplied by itself repeatedly, write it with an index (exponent): $a \times a \times a = a^3$. Three laws govern combining powers of the same base:

• Multiplying: add the indices — $a^m \times a^n = a^{m+n}$.
• Dividing: subtract the indices — $a^m \div a^n = a^{m-n}$.
• A power of a power: multiply the indices — $(a^m)^n = a^{mn}$.

Two special cases complete the set: any nonzero quantity to the power zero equals 1 ($a^0 = 1$), and a negative index means "take the reciprocal" ($a^{-n} = \tfrac{1}{a^n}$). A root is simply a fractional index: $\sqrt{a} = a^{1/2}$, and this is exactly why squaring undoes a square root — $(a^{1/2})^2 = a^{1} = a$, using the power-of-a-power rule.

Worked example — laws of indices. Simplify $\dfrac{P^3 \times P^2}{P^4}$. Multiply on top first (add indices): $P^3 \times P^2 = P^{3+2} = P^5$. Divide (subtract indices): $P^5 \div P^4 = P^{5-4} = P^1 = P$. Also, $(V^2)^3 = V^{2 \times 3} = V^6$ — multiply the indices for a power of a power, never add them.

An index applies only to the symbol directly in front of it: $ab^2$ squares only $b$; to square the whole product you must write $(ab)^2 = a^2b^2$.

What an equation is. An equation states that a quantity on the left equals a quantity on the right, joined by an equals sign (=). A simple equation has one unknown. The cardinal rule: whatever you do arithmetically to one side, you must do identically to the other side, or the equality breaks.

Transposing terms (addition and subtraction). Terms are quantities separated by + or - signs. To move a term across the equals sign, transfer it and flip its sign: + becomes -, and - becomes +.

Worked example — transposing a term. Solve $0.004 + D = 0.0042$ for $D$. Transpose the $0.004$ (it flips to $-0.004$): $D = 0.0042 - 0.004 = 0.0002$.

Transposing factors (multiplication and division) — cross-multiplication. When the unknown is tied up by multiplication or division, a factor multiplying on one side moves to divide on the other side, and vice versa. Applied to two fractions equal to each other, this is cross-multiplication: in $\dfrac{R_1}{R_2} = \dfrac{R_3}{R_4}$, each denominator moves diagonally to become a numerator on the other side, giving $R_1 R_4 = R_2 R_3$.

Worked example — cross-multiply (combined-gas form), with the intermediate step shown. Given $\dfrac{P_1 V_1}{T_1} = \dfrac{P_2 V_2}{T_2}$, find $T_2$ when $P_1 = 100$, $V_1 = 2$, $T_1 = 300$, $P_2 = 150$, $V_2 = 3$. First cross-multiply symbolically — bring $T_2$ up on the left and $T_1$ up on the right: $T_2 \times P_1 V_1 = T_1 \times P_2 V_2$ Now divide both sides by $P_1 V_1$ to finish isolating $T_2$: $T_2 = \frac{T_1 \, P_2 \, V_2}{P_1 \, V_1}$ Substitute: $T_2 = \dfrac{300 \times 150 \times 3}{100 \times 2} = \dfrac{135\,000}{200} = 675$. Notice the two separate moves — cross-multiply first, then divide out the coefficient — rather than jumping straight to the final rearranged formula.

When the unknown is a factor buried inside a term. If the unknown is a factor within a term (not standing alone), isolate the whole term first using the term-transposition rule, then isolate the factor. Do not split the factor out while the term is still buried with other terms.

Worked example — term with a buried factor. Solve $7.5 = 36 - 4.5p$ for $p$. You cannot move the $4.5$ alone, because $4.5p$ is one term being subtracted from $36$. Transpose the $36$ first: $4.5p = 36 - 7.5 = 28.5$. Now transpose the factor $4.5$ (divide): $p = 28.5 \div 4.5 = 6.33$.

Equations with roots and powers — a matched, contrastive pair. To undo a square (the unknown is squared), take the square root of both sides. To undo a square root (the unknown is under a root sign), square both sides. These are inverse moves for inverse situations — apply whichever one actually undoes what is being done to the unknown, never the other one by habit.

Worked example — undoing a square (root both sides). Solve $A^2 = 121$. Take the square root of both sides: $A = \sqrt{121} = 11$.

Worked example — undoing a square root (square both sides). Solve $\sqrt{D} = 6.4$ for $D$. The unknown is trapped under a root sign, so square both sides to remove the root: $(\sqrt{D})^2 = (6.4)^2$, giving $D = 40.96$. Compare this directly with the previous example: there the unknown was squared and you rooted; here the unknown is rooted and you square. Mixing these up — rooting when you should square, or vice versa — is one of the most common exam errors in this objective.

Worked example — chaining two techniques (clear a factor, then undo a power). In $A = \pi r^2$, solve for $r$ when $A = 50.27$. First transpose the factor $\pi$ to isolate the squared term: $r^2 = \frac{A}{\pi} = \frac{50.27}{3.1416} = 16.0$ Then take the square root of both sides to undo the power: $r = \sqrt{16.0} = 4.0$. The order matters — clear the factor first so the power stands alone, then undo the power. Reverse the order and you cannot isolate $r$ cleanly.

Fractional equations. Bring every fraction to the least common denominator (LCD), then drop the common denominator, leaving whole numbers to solve.

Try it yourself before reading on: solve $\dfrac{x}{3} + \dfrac{x}{6} = 5$ for $x$ — find the LCD of 3 and 6 first, then combine.

Worked example — basic fractional equation. The LCD of 3 and 6 is 6: $\dfrac{2x}{6} + \dfrac{x}{6} = \dfrac{30}{6}$. Drop the common denominator: $2x + x = 30$, so $3x = 30$ and $x = 10$.

Worked example — the harder case: the unknown inside a denominator. Solve $\dfrac{V}{R} = C - \dfrac{V}{r}$ for $R$, given $V = 48$, $C = 9$, $r = 320$. Substitute: $\dfrac{48}{R} = 9 - \dfrac{48}{320}$. Now bring both fractions to a common denominator that includes the unknown, $320R$: $\frac{320 \times 48}{320R} = \frac{9 \times 320R}{320R} - \frac{48R}{320R}$ Drop the common denominator: $15\,360 = 2880R - 48R = 2832R$. Transpose: $R = 15\,360 \div 2832 = 5.42$. This case is harder than the first because the unknown's own denominator has to be folded into the LCD symbolically before you can drop it — do this even when it looks like you could shortcut the arithmetic, because in problems where the unknown appears in two denominators, the shortcut disappears and only this method works.

Apply an organized, systematic approach to solving a problem. Every later chapter — mechanics, thermodynamics, code calculations — asks you to solve a multi-part numeric problem and present the answer clearly. Follow this 8-step method every time, not just when a problem looks hard:

1. Analyze the problem — read carefully; if there are multiple parts, note how they relate.
2. Determine the required units for the answer(s).
3. Determine the formula(s) needed — one, or more than one?
4. Find the givens and convert their units if necessary.
5. Identify any other needed constants (density, a conversion factor) not given directly.
6. Sketch the problem, labelling every given measurement.
7. Write down any other known factors identified in step 5.
8. Write the base equation(s) with the unknown isolated on the left, then solve, showing every step and a clearly labelled final answer.

Worked example — the organized method, three parts. A vertical cylindrical day tank has a flat bottom, a diameter of 1.2 m, and a height of 2.5 m. (a) How many cubic metres of diesel will it hold when full? (b) How many kilograms of diesel does it hold, given a diesel density of 840 kg/m³? (c) A second, rectangular tank has a square base 1.8 m on a side — how tall must it be to hold the same volume?

Following steps 1–2: this is a three-part problem needing units of m³, kg, and m respectively. Step 3: (a) needs $V = \pi r^2 h$; (b) needs $m = \rho V$; (c) needs $V = L \times W \times h$. Step 4: radius $r = 0.6$ m, height $h = 2.5$ m; the rectangular base is $1.8 \text{ m} \times 1.8 \text{ m}$. Step 5: density $\rho = 840$ kg/m³ (given). Steps 6–8, solved in order:

(a) $V = \pi r^2 h = 3.1416 \times (0.6)^2 \times 2.5 = 3.1416 \times 0.36 \times 2.5 = 2.827$ m³ (Ans.)

(b) $m = \rho V = 840 \times 2.827 = 2375$ kg (Ans.)

(c) The rectangular tank must hold the same 2.827 m³. Its base area is $1.8 \times 1.8 = 3.24$ m². Transpose $V = L W h$ for $h$: $h = \dfrac{V}{LW} = \dfrac{2.827}{3.24} = 0.873$ m (Ans.)

Notice how the same organized method carries a single computed volume from part (a) forward into parts (b) and (c) without re-deriving it — this is exactly why step 1 (identify how the parts relate) matters before you touch a calculator.

Common misconceptions and exam traps. (1) Forgetting to flip the sign when transposing a term — moving $+8$ across always makes it $-8$. (2) Half-transposing a buried factor, such as splitting $4.5p$ apart before isolating the whole term. (3) Performing an operation on only one side of the equation, silently breaking the equality. (4) Substituting numbers before rearranging symbolically — transpose first; it is cleaner and catches unit errors sooner. (5) In a two-technique problem, undoing the power before clearing the factor — always clear the factor first. (6) Reading BEDMAS as "always divide before you multiply" — division and multiplication are one co-equal tier, worked strictly left to right; the same is true of addition and subtraction. (7) Distributing a negative multiplier into only the first term of a bracket instead of every term inside it. (8) Applying the wrong index law — adding indices when the operation is a power of a power (which needs multiplication), or multiplying indices when combining two separate multiplied powers (which needs addition). (9) Confusing "undo a square root by squaring" with "undo a square by rooting" — apply the operation that is the true inverse of what is actually being done to the unknown, not the one you did on the last question. (10) In the unknown-in-a-denominator case, doing the surrounding arithmetic first and skipping the symbolic common-denominator step — it happens to work when the unknown appears only once, but the shortcut fails as soon as the unknown appears in more than one denominator, so build the habit of always folding it into the LCD.

Source: PanGlobal Third Class, Part A1, Chapter 1 (Algebraic Operations, Logarithms, and Problem Solving), Objectives 1–5 and 7; SOPEEC 3rd Class Paper 3A1.