TCE Mathematics Methods – Foundation Mastery Pack

1. Algebraic language and the order of operations

Algebra uses symbols to describe quantities and relationships. In 4x²−3x+7, x is the variable, 4 and −3 are coefficients, and 7 is a constant term. An expression has no equality sign; an equation states that two expressions are equal. This distinction controls what can be done: an expression may be simplified, whereas an equation may be solved for values that make the statement true.

Substitution means replacing a symbol by a value while preserving the structure of the expression. Use brackets around a negative replacement. For x=−2, 3x²−5x+1 becomes 3(−2)²−5(−2)+1=12+10+1=23. Writing 3×−2² risks applying the square only to 2 or losing the sign. Powers are evaluated before multiplication, then addition and subtraction, unless brackets set another order.

Like terms have the same variable part, including the same powers. Thus 5x²−2x+3x²+7x simplifies to 8x²+5x; x² and x are not like terms. Distribution removes brackets: a(b+c)=ab+ac, so −3(2x−5)=−6x+15. A minus before a bracket changes every term. A useful check is to substitute a simple value, such as x=1, into the original and simplified forms. Equal outputs do not prove an identity, but unequal outputs expose an error.

Units remain part of meaning. If C=12+4n gives a cost in dollars, then C(5)=32 dollars, not merely “32”. Algebraic steps should retain exact fractions until a decimal is requested. The official information sheet supplies many formulas, but accurate use still requires identifying each symbol, substituting consistently, and reporting an answer in the context of the question.

2. Substituting into expressions and formulae

When several variables occur, write the formula first and substitute all given values in one clear line. For A=(1/2)(a+b)h with a=5, b=9 and h=4, A=(1/2)(5+9)(4)=28. The brackets show that the sum 5+9 is multiplied by both one half and 4. Substituting piecemeal can hide an omitted factor or attach the height to only one parallel side.

Signed values need particular care. For v²=u²+2as with u=7, a=−2 and s=6, v²=7²+2(−2)(6)=49−24=25. Therefore v=±5 algebraically. If v represents speed, the model or question may select v=5; if v is signed velocity, more context is needed. Algebra produces candidate values, while the quantity's meaning determines which are admissible.

A formula can be evaluated repeatedly without being rearranged. If f(t)=18t−4.9t² describes a height, then f(2)=36−19.6=16.4. The input replaces every occurrence of t. For a two-variable rule g(x,y)=x²−3y, g(−2,5)=4−15=−11. The comma separates two inputs; it is not multiplication. State calculator rounding only at the end so that intermediate rounding does not accumulate.

Dimensional checking is a strong error test. In d=vt, metres per second multiplied by seconds gives metres. In A=πr², substituting r=3 cm gives 9π cm² because an area unit is squared. Dimensional agreement cannot guarantee that the numerical work is right, but incompatible units show that the formula or substitution has been mishandled.

For a composite measurement, substitute before simplifying units. If E=(1/2)mv² with m=1.6 kg and v=3.5 m/s, then E=(1/2)(1.6)(3.5)²=9.8 kg·m²/s², or 9.8 J. Squaring 3.5 is essential; squaring the product mv instead would change the formula. Estimating first, (1/2)(1.6)(about 12)≈9.6, supports the calculated magnitude.

3. Rearranging formulae by inverse operations

To make a variable the subject, treat a formula as an equation and perform the same valid operation on both sides. Work outward from the required variable. From y=mx+c, subtract c to get y−c=mx, then divide by m: x=(y−c)/m, provided m≠0. The whole numerator must be divided by m. The incorrect x=y−c/m divides only c.

Where the required variable occurs in a denominator, first clear the denominator. From P=k/t, multiply both sides by t to obtain Pt=k, then divide by P: t=k/P, with the original restrictions t≠0 and P≠0 for this rearrangement. From 1/f=1/u+1/v, subtract 1/u, combine fractions and invert: 1/v=(u−f)/(fu), so v=fu/(u−f). The value u=f must be excluded because it would make the new denominator zero.

For a squared variable, isolate the square before taking roots. From A=πr², r²=A/π and r=√(A/π) when r is a physical radius. The algebraic equation r²=A/π has two roots, but the model restricts radius to r≥0. From s=ut+(1/2)at², making a the subject gives 2(s−ut)=at² and a=2(s−ut)/t² for t≠0.

Check a rearrangement by reversing the operations or substituting test values. If y=3x+8 and y=20, the rearranged expression gives x=(20−8)/3=4; the original then gives 3(4)+8=20. A symbolic check is stronger: substitute x=(y−c)/m back into mx+c to obtain y. Always state restrictions introduced by division or square roots rather than silently accepting invalid values.

Literal equations with several occurrences of the subject may require collecting terms. From A=xp+xq, factor x to obtain A=x(p+q), then x=A/(p+q), provided p+q≠0. Dividing the original equation term-by-term too early is less clear. The factorisation shows both the rearrangement and the restriction in one step.

4. Function notation and evaluating outputs

A function assigns exactly one output to each permitted input. Writing y=f(x) names the rule f and identifies x as the input. If f(x)=2x²−3x+1, then f(4)=2(4)²−3(4)+1=21. The notation f(4) does not mean f multiplied by 4. It means the output produced by the rule when its input is 4.

An algebraic input replaces x everywhere. For f(x)=2x²−3x+1, f(a)=2a²−3a+1, while f(a+1)=2(a+1)²−3(a+1)+1=2a²+a. Expanding too early without brackets is a common source of error. The square applies to all of a+1, so (a+1)²=a²+2a+1, not a²+1.

Function values can be combined after each input is evaluated. If f(x)=x²−4, then f(3)+f(−1)=5+(−3)=2. An expression used as input needs brackets: f(2x)=(2x)²−4=4x²−4, while 2f(x)=2(x²−4)=2x²−8. These are different instructions and should not be interchanged.

A table or graph can also define a function. If the point (2,7) lies on y=f(x), then f(2)=7. To solve f(x)=7, find every permitted input whose output is 7; there may be more than one. A function can assign one y-value to each x while different x-values share the same y-value. Keep the direction of the mapping clear when reading coordinates.

Difference notation is also useful. For f(x)=x²−3x, f(x+2)=(x+2)²−3(x+2)=x²+x−2, so f(x+2)−f(x)=4x−2. This is not f(x)+2. The input has changed before the rule is applied, and brackets preserve that change through every occurrence of x.

If f(x)=3x−2 and f(a)=13, then 3a−2=13 and a=5. This reverses evaluation: the output is known and the input is sought. For a nonlinear function the same output may correspond to several inputs, so solving f(x)=k must find every permitted solution rather than assume the function can be read backwards uniquely.

5. Equations, identities and checking solutions

An equation is true for particular values, while an identity is true for every value in its common domain. The equation 3x+5=20 has the solution x=5. The statement 3(x+2)=3x+6 is an identity because distribution makes both sides equal for all real x. An apparent solution must satisfy the original equation, especially when denominators or powers are involved.

Solving a linear equation preserves equality. For 5−2(3x−1)=17, expand to 7−6x=17, subtract 7, and divide by −6 to obtain x=−5/3. Substitution checks it: 5−2(−5+1)=13? That reveals a careless mental simplification. Correctly, 3x−1=−5−1=−6, so 5−2(−6)=17. A check must reproduce the original structure exactly.

Equations with fractions can be cleared by multiplying every term by a common denominator. For (x−1)/3+(x+2)/2=5, multiply the entire equation by 6: 2(x−1)+3(x+2)=30. This gives 5x+4=30 and x=26/5. Multiplying only the fractions on one side changes the equation. Restrictions such as a denominator being nonzero must be recorded before clearing it.

When an equation models a situation, checking includes interpretation. A negative time, length or item count may be algebraically correct but inadmissible. State exact solutions unless the context requires approximation, retain enough working to show the method, and include both coordinates when solving a system. A numerical substitution check confirms a candidate; a derivation explains why all candidates have been found.

A solution set should distinguish an exact result from a check. For (x−1)/3+(x+2)/2=5, substituting x=26/5 produces 7/5+18/5=5, confirming the candidate. The equality works because both fractional terms were preserved. Writing the set as {26/5} records the complete real solution.

6. A reliable workflow for modelling and communication

Translate a sentence by defining symbols before forming an expression. If a taxi cost is a fixed $6 plus $2.40 per kilometre d, write C=6+2.40d. “Three less than twice n” is 2n−3, whereas “three times the difference between n and 2” is 3(n−2). Word order alone is unreliable; identify operations and grouping from the meaning.

Suppose the taxi fare is $34.80. Substitute the output and solve: 34.80=6+2.40d, so 28.80=2.40d and d=12 km. Check both arithmetic and reasonableness: 12 kilometres contributes $28.80, which becomes $34.80 after the fixed charge. If the calculated distance were negative, the model or algebra would need review.

Choose a form suited to the task. A formula with the output isolated is convenient for evaluation; a rearranged form is convenient when an input is unknown. Exact fractions make equivalence visible, while a final decimal may suit measurement. Label equations, use equality signs only between equal expressions, and avoid chains such as “x=4+3=7x” in which adjacent statements are not equivalent.

A complete response usually contains a definition or formula, substitution, algebra, the requested value and a contextual statement. The information sheet reduces memory load but not the need to choose and apply a relationship. Before finishing, ask: Did every input replace the correct symbol? Were negative values bracketed? Were restrictions and units retained? Does the answer satisfy the original relationship?

In a second model, temperature F and C are related by F=(9/5)C+32. For F=68, C=(5/9)(68−32)=20. The subtraction must occur before multiplication by 5/9. Checking gives (9/5)(20)+32=68, and the units identify which scale belongs to each value.