Algebraic language and notation

(a) $4a$

1. Four lots of $a$ is $4a$.

• B1 The correct answer, $4a$.

(b) $3mn$

1. Write the number first and leave out the multiplication signs: $3mn$.

• B1 The correct answer, $3mn$.

(c) $45x + 30y$

1. $x$ pens cost $45x$ pence and $y$ rubbers cost $30y$ pence, so the total is $45x + 30y$.

• B1 One correct term, $45x$ or $30y$.
• B1 The full expression $45x + 30y$.

(a) $4p^{2}$

1. Repeated multiplication p $\times$ p is $p^{2}.$
2. Therefore 4 $\times$ p $\times$ p = $4p^{2}.$

• B1 Correct answer: $4p^{2}$

(a) (iv)

1. An identity is true for every value of $x$. $2(x + 1)$ expands to $2x + 2$ whatever $x$ is, so (iv) is an identity.

• B1 The correct answer, Statement (iv).

(b) (iii)

1. A formula gives one quantity in terms of others: $v$ in terms of $u$, $a$ and $t$.

• B1 Statement (iii).

(c) Subtracting $5x$ leaves $2 = 7$, which is never true.

1. Subtracting $5x$ from both sides gives $2 = 7$, which is false for every value of $x$. So no value of $x$ works.

• C1 Showing that the $x$ terms cancel to leave $2 = 7$ (or that $5x + 2$ is always 5 less than $5x + 7$), so it is never true.

(a) $12p+6e$

1. Form the cost of one pack, then multiply every term by three.
   $$3(4p+2e)$$
2. Therefore $12p+6e$.

• P1 Form the cost of one pack, then multiply every term by three.
• A1 Correct answer: $12p+6e$

(b) $138$ p

1. Substitute each price into its own term.
   $$12\times 6+6\times 11$$
2. Therefore $138$ p.

• M1 Substitute each price into its own term.
• A1 Correct answer: $138$ p

(a) $4x+6$

1. Add two lengths and two widths.
   $$2(x+3)+2x$$
2. Therefore $4x+6$.

• M1 Add two lengths and two widths.
• A1 Correct answer: $4x+6$

(b) $6$ cm

1. Set the perimeter expression equal to the measured perimeter.
   $$4x+6=30$$
2. Therefore $6$ cm.

• P1 Set the perimeter expression equal to the measured perimeter.
• A1 Correct answer: $6$ cm

(a) $C=12+4n$

1. Add the fixed charge once to the variable charge.
   $$12+4n$$
2. Therefore $C=12+4n$.

• M1 Add the fixed charge once to the variable charge.
• A1 Correct answer: $C=12+4n$

(b) £$5.71$

1. Divide the total charge by the number sharing it.
   $$\frac{12+4\times 7}{7}$$
2. Therefore £$5.71$.

• P1 Divide the total charge by the number sharing it.
• A1 Correct answer: £$5.71$

(a) $3(n+5)$

1. Therefore $3(n+5)$.

• B1 Correct answer: $3(n+5)$

(b) $9$

1. Form an equation from the expression.
   $$3(n+5)=42$$
2. Therefore $9$.

• M1 Form an equation from the expression.
• A1 Correct answer: $9$

(c) 3(n + 5) is 3 times the whole number n + 5, so it is a multiple of 3.

1. Therefore 3(n + 5) is 3 times the whole number n + 5, so it is a multiple of 3.

• C1 Correct conclusion with supporting reasoning: 3(n + 5) is 3 times the whole number n + 5, so it is a multiple of 3.

(a) $(6x + 6) - (4x + 8) = 2x - 2$

1. Perimeter of the rectangle:
   $$2(2x + 3) + 2x = 4x + 6 + 2x = 6x + 6$$
2. Perimeter of the square:
   $$4(x + 2) = 4x + 8$$
3. Difference:
   $$(6x + 6) - (4x + 8) = 6x + 6 - 4x - 8 = 2x - 2$$

• P1 Finding the rectangle's perimeter as $6x + 6$.
• P1 Finding the square's perimeter as $4x + 8$.
• A1 Subtracting with the bracket to reach $2x - 2$.

(b) $6x + 6 = 6(x + 1)$

1. $6x + 6 = 6(x + 1)$, and $x + 1$ is a whole number, so the perimeter is 6 times a whole number.

• C1 Factorising to $6(x + 1)$ and saying $x + 1$ is a whole number.