Identities and algebraic reasoning

(a) No: $x + x + x = 3x$, but $x^3 = x \times x \times x$.

1. $x + x + x = 3x$, which is three lots of $x$ added. $x^3$ is three $x$'s multiplied.
2. For example, when $x = 2$: $3x = 6$ but $x^3 = 8$.

• C1 "No", with a reason such as $x + x + x = 3x$ or a value that gives different answers.

(b) $3x$

1. Three lots of $x$: $3x$.

• B1 The correct answer, $3x$.

(a) 3(x + 2) = 3x + 6

1. An identity is true for every allowed value of its variable.
2. Expanding 3(x + 2) gives 3x + 6 for every x.

• B1 Correct answer: 3(x + 2) = 3x + 6

(a) Yes. Both expressions simplify to 3x + 12.

1. $$4x+12-x$$
2. $$3x+12$$
3. Expand and simplify the first expression: 4x + 12 $-$ x = 3x + 12.
4. Expand the second expression: 3(x + 4) = 3x + 12.
5. Both give the same expression for every x, so Mina is correct.

• P1 Establishing $4x+12-x$ or an equivalent valid method.
• P1 Establishing $3x+12$ or an equivalent valid method.
• C1 Correct conclusion with the complete supporting argument: Yes. Both expressions simplify to 3x + 12.

(a) $2x+6$

1. Expand the bracket then combine the x terms.
   $$3x+6-x$$
2. Therefore $2x+6$.

• M1 Expand the bracket then combine the x terms.
• A1 Correct answer: $2x+6$

(b) Yes. Algebraic simplification makes the two sides identical, so they agree for every x.

1. Yes. Algebraic simplification makes the two sides identical, so they agree for every x.

• C1 Correct conclusion with supporting reasoning: Yes. Algebraic simplification makes the two sides identical, so they agree for every x.

(c) $5$

1. Use the simplified form and subtract its constant term.
   $$2x=10$$
2. Therefore $5$.

• M1 Use the simplified form and subtract its constant term.
• A1 Correct answer: $5$

(a) An identity: expanding gives x² + 2x + 1 - x² = 2x + 1 for every value of x.

1. Expand the square and simplify the left side.
   $$(x+1)^{2}-x^{2}=2x+1$$
2. An identity: expanding gives x² + 2x + 1 - x² = 2x + 1 for every value of x.

• M1 Expand the square and simplify the left side.
• C1 Correct conclusion with supporting reasoning: An identity: expanding gives x² + 2x + 1 - x² = 2x + 1 for every value of x.

(b) An equation: it is true only when x = 7.

1. Expand both sides and solve.
   $$4x-8=2x+6$$
2. An equation: it is true only when x = 7.

• M1 Expand both sides and solve.
• C1 Correct conclusion with supporting reasoning: An equation: it is true only when x = 7.

(a) (2n - 1) + (2n + 1) = 4n, which is 4 times a whole number.

1. Add the two expressions and simplify.
   $$(2n-1)+(2n+1)=4n$$
2. Therefore (2n - 1) + (2n + 1) = 4n, which is 4 times a whole number.

• M1 Add the two expressions and simplify.
• C1 Correct conclusion with supporting reasoning: (2n - 1) + (2n + 1) = 4n, which is 4 times a whole number.

(b) Yes. (2n - 1)(2n + 1) = 4n² − 1, which is one less than an even number, so it is odd.

1. Yes. (2n - 1)(2n + 1) = 4n² − 1, which is one less than an even number, so it is odd.

• C1 Correct conclusion with supporting reasoning: Yes. (2n - 1)(2n + 1) = 4n² − 1, which is one less than an even number, so it is odd.

(a) $n + (n + 1) + (n + 2) = 3n + 3 = 3(n + 1)$

1. Write three consecutive whole numbers algebraically.
   $$n, \quad n + 1, \quad n + 2$$
2. Add them.
   $$n + (n + 1) + (n + 2) = 3n + 3 = 3(n + 1)$$
3. $n + 1$ is a whole number, so the sum is 3 times a whole number: a multiple of 3.

• M1 Writing the numbers as $n$, $n + 1$, $n + 2$ (or $n - 1$, $n$, $n + 1$).
• M1 Simplifying the sum to $3n + 3$ (or $3n$).
• C1 Writing $3(n + 1)$ (or $3n$) and saying it is 3 times a whole number.

(b) $4n + 6 = 4(n + 1) + 2$, which always leaves remainder 2 when divided by 4.

1. Add four consecutive whole numbers.
   $$n + (n + 1) + (n + 2) + (n + 3) = 4n + 6$$
2. $4n + 6 = 4(n + 1) + 2$: it is always 2 more than a multiple of 4, so it is never a multiple of 4.

• M1 Finding the sum $4n + 6$.
• C1 Explaining that $4n + 6$ is 2 more than the multiple of 4, $4n + 4$ (or $4(n + 1) + 2$), so it is never a multiple of 4.

(a) $35$

1. $$2x^{2}+(a-6)x+b-3a$$
2. $$a-6=1$$
3. $$b-3a=7$$
4. Expand the left side: $2x^{2}$ + (a $-$ 6)x + b $-$ 3a.
5. Equal expressions have equal coefficients, so a $-$ 6 = 1 and b $-$ 3a = 7.
6. Thus a = 7 and b = 28, giving a + b = 35.

• P1 Establishing $2x^{2}+(a-6)x+b-3a$ or an equivalent valid method.
• P1 Establishing $a-6=1$ or an equivalent valid method.
• P1 Establishing $b-3a=7$ or an equivalent valid method.
• A1 Correct answer: $35$