Flashcards · 15 cards

Expressions, formulae & change of subject

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Know it
Substituting $x = 4$ and $y = -2$ into $x^2 - y^3$ : what do you get?
Answer
Put negative numbers in brackets: $y^3 = (-2)^3 = -8$ , so $x^2 - y^3 = 16 - (-8) = 24$ .
From the lesson: Substituting and expanding
Rule
Given $m : n = 3 : 2$ , how do you evaluate $\dfrac{m^2 + n^2}{m^2 - n^2}$ ?
Answer
Write $m = 3k$ and $n = 2k$ , substitute, and the $k$ 's cancel: $\dfrac{9k^2 + 4k^2}{9k^2 - 4k^2} = \dfrac{13}{5}$ .
One block sizem : n = 3 : 2 means m = 3k and n = 2k
From the lesson: Substituting and expanding
Rule
$(a + b)(c + d) = \,?$
Answer
$ac + ad + bc + bd$ : every term of one bracket times every term of the other.
Four productsEvery term of one bracket times every term of the other
From the lesson: Substituting and expanding
Know it
$(p + q)^2 = \,?$ and $(p - q)^2 = \,?$
Answer
$p^2 + 2pq + q^2$ and $p^2 - 2pq + q^2$ . Not $p^2 + q^2$ : the middle term matters.
From the lesson: Substituting and expanding
Rule
$f(x) = 2x - 5$ . What is $f(3a)$ ?
Answer
Put $3a$ wherever $x$ is: $f(3a) = 2(3a) - 5 = 6a - 5$ .
Whatever is in the bracketreplaces x everywhere in the rule
From the lesson: Substituting and expanding
Rule
How do you find the inverse of $f(x) = 2x - 5$ ?
Answer
Undo each step, last one first: add 5, then divide by 2. So $f^{-1}(x) = \dfrac{x + 5}{2}$ .
The inverse machineUndo each step of f, last one first
From the lesson: Substituting and expanding
Know it
Name three ways to factorise.
Answer
A common factor: $6x^2 - 9x = 3x(2x - 3)$ . Grouping: $ax + ay + bx + by = (a + b)(x + y)$ . The difference of two squares.
From the lesson: Factorising and algebraic fractions
Rule
Factorise $a^2 - b^2$ .
Answer
$(a - b)(a + b)$ .
Difference of two squaresCut b² from a², and the rest rearranges into (a + b)(a − b)
From the lesson: Factorising and algebraic fractions
Know it
When is $\dfrac{x + 1}{x^2 - 9}$ undefined?
Answer
When the denominator is 0: $x^2 - 9 = 0$ , so $x = 3$ or $x = -3$ .
From the lesson: Factorising and algebraic fractions
Know it
$\dfrac{1}{x - 3} - \dfrac{1}{x + 2} = \,?$
Answer
Use a common denominator and put the tops in brackets: $\dfrac{(x + 2) - (x - 3)}{(x - 3)(x + 2)} = \dfrac{5}{(x - 3)(x + 2)}$ .
From the lesson: Factorising and algebraic fractions
Rule
How do you split $\dfrac{5x - 1}{(x - 1)(x + 1)}$ into partial fractions?
Answer
Write it as $\dfrac{A}{x - 1} + \dfrac{B}{x + 1}$ and make each bracket zero in turn: $x = 1$ gives $A = 2$ , and $x = -1$ gives $B = 3$ .
Partial fractionsMake each bracket zero in turn to find A and B
From the lesson: Factorising and algebraic fractions
Rule
How do you change the subject of a formula?
Answer
Undo what was done to the letter, in reverse order, doing the same to both sides each time.
Build it, then undo itOpposite operations, last one first
From the lesson: Changing the subject of a formula
Know it
Make $w$ the subject of $y = \dfrac{3w + 1}{2w - 5}$ .
Answer
Multiply out, collect the $w$ terms on one side, then factorise out $w$ : $w(2y - 3) = 5y + 1$ , so $w = \dfrac{5y + 1}{2y - 3}$ .
From the lesson: Changing the subject of a formula
Which method?
WAEC 2024 · Paper 1 · Q5
Simplify: $(2p - q)^2 - (p + q)^2$ .
Expanding both brackets is slow. What is quicker?
Answer
It is a difference of two squares, $A^2 - B^2 = (A - B)(A + B)$ , with $A = 2p - q$ and $B = p + q$ .
From the lesson: Substituting and expanding Try the question
Which method?
WAEC 2022 · Paper 1 · Q18
Make $t$ the subject of $k = m\sqrt{\dfrac{t - p}{r}}$ .
What comes first?
Answer
Undo the outside first: divide by $m$ , then square both sides to remove the root. Then multiply by $r$ and add $p$ .
From the lesson: Changing the subject of a formula Try the question