Linear equations, including both sides

(a) $7$

1. $$6x=42$$
2. Subtract 5 from each side: 6x = 42.
3. Divide each side by 6: x = 7.

• M1 Establishing $6x=42$ or an equivalent valid method.
• A1 Correct answer: $7$

(a) $x = 5$

1. Subtract 7: $4x = 20$. Divide by 4: $x = 5$.

• M1 Writing $4x = 20$.
• A1 The correct answer, $x = 5$.

(b) $y = 9$

1. Divide by 5: $y - 2 = 7$. Add 2: $y = 9$.

• M1 Writing $y - 2 = 7$ or $5y - 10 = 35$.
• A1 The correct answer, $y = 9$.

(c) $a = 4$

1. Collect the $a$ terms on one side and the numbers on the other.
   $$7a - 3a = 12 + 4 \Rightarrow 4a = 16$$
2. So $a = 4$.

• M1 Collecting to get $4a = 16$ (or $4a - 4 = 12$).
• A1 The correct answer, $a = 4$.

(a) $6$

1. $$10x-15=4x+21$$
2. $$6x=36$$
3. Expand: 10x $-$ 15 = 4x + 21.
4. Subtract 4x and add 15 to obtain 6x = 36.
5. Divide by 6: x = 6.

• M1 Establishing $10x-15=4x+21$ or an equivalent valid method.
• M1 Establishing $6x=36$ or an equivalent valid method.
• A1 Correct answer: $6$

(a) $10$

1. $$3x-2=28$$
2. Multiply both sides by 4: 3x $-$ 2 = 28.
3. Add 2 then divide by 3: x = 30/3 = 10.

• M1 Establishing $3x-2=28$ or an equivalent valid method.
• A1 Correct answer: $10$

(a) $5$

1. Expand both brackets and preserve every sign.
   $$12x-12=8x+8$$
2. Collect x terms on one side and constants on the other.
   $$4x=20$$
3. Therefore $5$.

• M1 Expand both brackets and preserve every sign.
• M1 Collect x terms on one side and constants on the other.
• A1 Correct answer: $5$

(b) Substitute the value into both original bracketed sides and confirm that the two totals are equal.

1. Substitute the value into both original bracketed sides and confirm that the two totals are equal.

• C1 Correct conclusion with supporting reasoning: Substitute the value into both original bracketed sides and confirm that the two totals are equal.

(a) $6$

1. Multiply every term by 6.
   $$2(x-6)+3(x+3)=27$$
2. Collect terms and isolate x.
   $$5x=30$$
3. Therefore $6$.

• M1 Multiply every term by 6.
• M1 Collect terms and isolate x.
• A1 Correct answer: $6$

(b) $11$

1. Use the solved value in the requested expression.
   $$2\times 6-1$$
2. Therefore $11$.

• M1 Use the solved value in the requested expression.
• A1 Correct answer: $11$

(a) $x = 16$

1. Multiply both sides by 5.
   $$2x + 3 = 35$$
2. $$2x = 32 \Rightarrow x = 16$$

• M1 Multiplying the whole of both sides by 5: $2x + 3 = 35$.
• A1 The correct answer, $x = 16$.

(b) $x = 6$

1. Expand both sides.
   $$6x - 3 = 4x + 8 + 1 = 4x + 9$$
2. Collect.
   $$2x = 12 \Rightarrow x = 6$$

• M1 Expanding both brackets correctly: $6x - 3$ and $4x + 8$.
• M1 Collecting terms to reach $2x = 12$.
• A1 The correct answer, $x = 6$.

(a) $x = 2$

1. Multiply every term by 12, the lowest common multiple of 3 and 4.
   $$4(x + 4) - 3(x - 2) = 24$$
2. Expand, taking care with the minus sign.
   $$4x + 16 - 3x + 6 = 24$$
3. $$x + 22 = 24 \Rightarrow x = 2$$

• M1 Multiplying through by 12 (or a common multiple) to clear both fractions, including the 2 on the right.
• M1 Expanding to $4x + 16 - 3x + 6$.
• A1 The correct answer, $x = 2$.

(b) $x = \dfrac{10}{3}$

1. Add the four expressions.
   $$(2x + 1) + (3x - 2) + (x + 7) + 10 = 6x + 16$$
2. Form an equation for the mean.
   $$\frac{6x + 16}{4} = 3x - 1 \Rightarrow 6x + 16 = 12x - 4$$
3. $$20 = 6x \Rightarrow x = \frac{10}{3}$$

• P1 Finding the total $6x + 16$.
• P1 Forming $\frac{6x + 16}{4} = 3x - 1$ (or $6x + 16 = 4(3x - 1)$).
• A1 $x = \frac{10}{3}$ (or $3\frac{1}{3}$).