Linear Inequalities GCSE Questions

Answer $x > 4$

1. $\displaystyle 3x > 12$Add $5$ to both sides
2. $\displaystyle x > 4$Divide both sides by $3$: positive, so the sign stays the same

• M1 $3x > 12$
• A1 $x > 4$

Answer $x \le 10$

1. $\displaystyle \frac{x}{2} \le 5$Subtract $3$ from both sides
2. $\displaystyle x \le 10$Multiply both sides by $2$

• M1 $\dfrac{x}{2} \le 5$
• A1 $x \le 10$

Answer $x < 4$

1. $\displaystyle x + 1 < 5$Divide both sides by $4$
2. $\displaystyle x < 4$

• M1 $4x + 4 < 20$ or $x + 1 < 5$
• A1 $x < 4$

Answer $-2,\ -1,\ 0,\ 1,\ 2$

1. $\displaystyle -2 \le n \;\Rightarrow\; -2 \text{ is included}$$\le$ includes the end
2. $\displaystyle n < 3 \;\Rightarrow\; 3 \text{ is not included}$$<$ does not include the end
3. $\displaystyle -2,\ -1,\ 0,\ 1,\ 2$

• B2 All five correct and no extras

Answer $x \ge 5$

1. $\displaystyle 3x + 2 \ge 17$Subtract $2x$ from both sides
2. $\displaystyle 3x \ge 15 \;\Rightarrow\; x \ge 5$

• M1 $3x \ge 15$
• A1 $x \ge 5$

Answer $x < 3$

1. $\displaystyle -2x > -6$Subtract $7$ from both sides
2. $\displaystyle x < 3$Divide by $-2$: a negative, so flip the sign

• M1 $-2x > -6$ or $6 > 2x$
• A1 $x < 3$

Answer $-2 < x \le 3$

1. $\displaystyle -4 < 2x \le 6$Subtract $3$ from all three parts
2. $\displaystyle -2 < x \le 3$Divide all three parts by $2$

• M1 $-4 < 2x \le 6$
• A1 $-2 < x$
• A1 $x \le 3$, with both parts combined: $-2 < x \le 3$

Answer $2,\ 3,\ 4$

1. $\displaystyle 6 \le 3x < 15$Add $1$ to all three parts
2. $\displaystyle 2 \le x < 5$Divide by $3$
3. $\displaystyle 2,\ 3,\ 4$$2$ is included; $5$ is not

• M1 $6 \le 3x < 15$
• A1 $2 \le x < 5$
• A1 $2,\ 3,\ 4$

Answer $4$

1. $\displaystyle 4x > 14 \;\Rightarrow\; x > 3.5$
2. $\displaystyle \text{smallest integer} = 4$$3$ is too small, because $3 < 3.5$

• M1 $x > 3.5$
• A1 $4$

Answer $x \ge 2$

1. $\displaystyle 4(x + 1) - 3(x - 2) \ge 12$Multiply every term by $12$ (positive, so the sign stays)
2. $\displaystyle 4x + 4 - 3x + 6 \ge 12$Careful: $-3(x - 2) = -3x + 6$
3. $\displaystyle x + 10 \ge 12 \;\Rightarrow\; x \ge 2$

• M1 Multiplies every term by $12$: $4(x + 1) - 3(x - 2) \ge 12$
• M1 $x + 10 \ge 12$
• A1 $x \ge 2$

Answer $\{x : x > 2\}$

1. $\displaystyle 3x > 6 \;\Rightarrow\; x > 2$
2. $\displaystyle \{x : x > 2\}$Read as "the set of all $x$ such that $x > 2$"

• M1 $x > 2$
• A1 $\{x : x > 2\}$

Answer $3,\ 4,\ 5$

1. $\displaystyle 2n > 4 \;\Rightarrow\; n > 2$
2. $\displaystyle 3n < 18 \;\Rightarrow\; n < 6$
3. $\displaystyle 2 < n < 6 \;\Rightarrow\; n = 3,\ 4,\ 5$Neither end is included

• M1 $n > 2$
• M1 $n < 6$
• A1 $3,\ 4,\ 5$

Answer $10$ miles

1. $\displaystyle 4 + 2m \le 25$The cost must be at most £25
2. $\displaystyle 2m \le 21 \;\Rightarrow\; m \le 10.5$
3. $\displaystyle \text{greatest whole number} = 10$$11$ miles would cost £$26$

• M1 $4 + 2m \le 25$
• A1 $m \le 10.5$
• A1 $10$

Answer Dividing by $-3$ means the sign must be turned round. $x > -2$.

1. $\displaystyle -3x < 6$Tom is right so far
2. $\displaystyle x > -2$Dividing by $-3$ flips $<$ to $>$

• B1 Explains that dividing by a negative number reverses the inequality
• B1 $x > -2$

Answer $0 < x < 8$

1. $\displaystyle 2x + 2(x + 4) < 40$Perimeter: two of each side
2. $\displaystyle 4x + 8 < 40 \;\Rightarrow\; 4x < 32 \;\Rightarrow\; x < 8$
3. $\displaystyle 0 < x < 8$A length must also be more than $0$

• M1 $2x + 2(x + 4) < 40$ oe
• A1 $x < 8$
• B1 $x > 0$, because a length must be positive

Answer $3,\ 4,\ 5,\ 6,\ 7$

1. $\displaystyle 3x - 2 > x + 3 \;\Rightarrow\; 2x > 5 \;\Rightarrow\; x > 2.5$
2. $\displaystyle 5x \le 2x + 21 \;\Rightarrow\; 3x \le 21 \;\Rightarrow\; x \le 7$
3. $\displaystyle 2.5 < x \le 7 \;\Rightarrow\; 3,\ 4,\ 5,\ 6,\ 7$$7$ is included; $2.5$ is not a whole number anyway

• M1 $2x > 5$
• A1 $x > 2.5$
• A1 $x \le 7$
• A1 $3,\ 4,\ 5,\ 6,\ 7$

Answer $\{x : 1 \le x < 4.5\}$

1. $\displaystyle 2 \le 2x < 9$Add $5$ to all three parts
2. $\displaystyle 1 \le x < 4.5$Divide all three parts by $2$
3. $\displaystyle \{x : 1 \le x < 4.5\}$

• M1 $2 \le 2x < 9$
• A1 $1 \le x < 4.5$
• A1 $\{x : 1 \le x < 4.5\}$

Answer $41$ students

1. $\displaystyle 300 + 12n \le 800$
2. $\displaystyle 12n \le 500 \;\Rightarrow\; n \le 41.66\ldots$
3. $\displaystyle \text{greatest number} = 41$Round down: $42$ students would cost £$804$

• M1 $300 + 12n \le 800$
• A1 $n \le 41.6\ldots$
• A1 $41$

Answer $6$

1. $\displaystyle 3n - 4 > 11 \;\Rightarrow\; 3n > 15 \;\Rightarrow\; n > 5$
2. $\displaystyle 2n + 1 < 15 \;\Rightarrow\; 2n < 14 \;\Rightarrow\; n < 7$
3. $\displaystyle 5 < n < 7 \;\Rightarrow\; n = 6$

• M1 $3n - 4 > 11$ and $2n + 1 < 15$
• A1 $n > 5$ and $n < 7$
• A1 $6$

Answer $4 \le x \le 7$

1. $\displaystyle \text{perimeter} = x + 2x + x + 2x = 6x$
2. $\displaystyle 24 \le 6x \le 42$"At least" means $\ge$; "at most" means $\le$
3. $\displaystyle 4 \le x \le 7$Divide all three parts by $6$

• M1 Perimeter $= 6x$
• M1 $24 \le 6x \le 42$
• A1 $4 \le x \le 7$

Answer $(2, 1),\ (3, 1),\ (4, 1),\ (5, 1),\ (3, 2),\ (4, 2)$

1. $\displaystyle y = 1: \quad x > 1 \text{ and } x \le 5 \;\Rightarrow\; x = 2, 3, 4, 5$Work through each possible value of $y$
2. $\displaystyle y = 2: \quad x > 2 \text{ and } x \le 4 \;\Rightarrow\; x = 3, 4$
3. $\displaystyle y = 3: \quad x > 3 \text{ and } x \le 3 \;\Rightarrow\; \text{none}$Larger $y$ values give no pairs either

• B3 All six pairs and no extras (B2 for at least four correct pairs with at most one wrong; B1 for at least two correct pairs)

Answer $\{x : x > 3\}$

1. $\displaystyle 3 - 2x \le x + 12 \;\Rightarrow\; -9 \le 3x \;\Rightarrow\; x \ge -3$The $x$ is in all three parts, so solve each half separately
2. $\displaystyle x + 12 < 4x + 3 \;\Rightarrow\; 9 < 3x \;\Rightarrow\; x > 3$
3. $\displaystyle \{x : x > 3\}$Every $x > 3$ also satisfies $x \ge -3$

• M1 Splits into $3 - 2x \le x + 12$ and $x + 12 < 4x + 3$
• A1 $x \ge -3$
• A1 $x > 3$
• A1 $\{x : x > 3\}$, the values that satisfy both

Answer $x > \tfrac{4}{3}$

1. $\displaystyle 5(x + 2) > 2(x + 2) + 2(5)$Area $>$ perimeter
2. $\displaystyle 5x + 10 > 2x + 14$
3. $\displaystyle 3x > 4 \;\Rightarrow\; x > \tfrac{4}{3}$This also makes the side $x + 2$ positive

• M1 $5(x + 2) > 2(x + 2) + 10$
• M1 $5x + 10 > 2x + 14$
• A1 $x > \tfrac{4}{3}$ oe

Answer $k = 4$

1. $\displaystyle 3x - 2x > 5 + k \;\Rightarrow\; x > k + 5$Solve with $k$ still in it
2. $\displaystyle k + 5 = 9$Compare with $x > 9$
3. $\displaystyle k = 4$

• M1 $x > k + 5$
• M1 $k + 5 = 9$
• A1 $k = 4$