Simultaneous Equations GCSE Questions

Answer $x = 7,\ y = 3$

1. $\displaystyle 2x = 14$Add: $+y$ and $-y$ cancel
2. $\displaystyle x = 7$
3. $\displaystyle 7 + y = 10 \;\Rightarrow\; y = 3$Substitute into the first equation

• M1 Adds the equations: $2x = 14$
• A1 $x = 7$
• A1 $y = 3$

Answer $x = 4,\ y = 3$

1. $\displaystyle x = 4$Subtract: both have $+y$, so the $y$ terms cancel
2. $\displaystyle 4 + y = 7 \;\Rightarrow\; y = 3$

• M1 Subtracts the equations: $x = 4$
• A1 $x = 4$
• A1 $y = 3$

Answer $x = 4,\ y = 2$

1. $\displaystyle 4x = 16$Add: $+2y$ and $-2y$ cancel
2. $\displaystyle x = 4$
3. $\displaystyle 4 - 2y = 0 \;\Rightarrow\; y = 2$

• M1 Adds the equations: $4x = 16$
• A1 $x = 4$
• A1 $y = 2$

Answer $x = 2,\ y = 3$

1. $\displaystyle 2y = 6$Subtract: both have $2x$
2. $\displaystyle y = 3$
3. $\displaystyle 2x + 3 = 7 \;\Rightarrow\; x = 2$

• M1 Subtracts the equations: $2y = 6$
• A1 $y = 3$
• A1 $x = 2$

Answer $x = 5,\ y = 10$

1. $\displaystyle x + 2x = 15$Replace $y$ with $2x$
2. $\displaystyle 3x = 15 \;\Rightarrow\; x = 5$
3. $\displaystyle y = 2 \times 5 = 10$

• M1 Substitutes: $x + 2x = 15$
• A1 $x = 5$
• A1 $y = 10$

Answer $x = 5,\ y = 2$

1. $\displaystyle 6x + 4y = 38$Multiply the first equation by $2$ to match the $4y$
2. $\displaystyle 5x = 25$Subtract $x + 4y = 13$
3. $\displaystyle x = 5$
4. $\displaystyle 5 + 4y = 13 \;\Rightarrow\; y = 2$

• M1 Multiplies the first by $2$ and subtracts: $5x = 25$, or equivalent
• A1 $x = 5$
• A1 $y = 2$

Answer $x = 2,\ y = -\tfrac{1}{2}$

1. $\displaystyle 10x - 4y = 22$Multiply the first equation by $2$
2. $\displaystyle 13x = 26$Add: $-4y$ and $+4y$ cancel
3. $\displaystyle x = 2$
4. $\displaystyle 3(2) + 4y = 4 \;\Rightarrow\; 4y = -2 \;\Rightarrow\; y = -\tfrac{1}{2}$

• M1 Multiplies the first by $2$: $10x - 4y = 22$
• M1 Adds: $13x = 26$
• A1 $x = 2$
• A1 $y = -\tfrac{1}{2}$ oe

Answer $x = -1,\ y = 2$

1. $\displaystyle 8x + 6y = 4$Multiply the first equation by $2$
2. $\displaystyle 9x + 6y = 3$Multiply the second equation by $3$
3. $\displaystyle x = -1$Subtract the first from the second
4. $\displaystyle 4(-1) + 3y = 2 \;\Rightarrow\; 3y = 6 \;\Rightarrow\; y = 2$

• M1 Multiplies both to match a coefficient, e.g. $8x + 6y = 4$ and $9x + 6y = 3$
• M1 Subtracts correctly: $x = -1$
• A1 $x = -1$
• A1 $y = 2$

Answer $x = 6,\ y = 13$

1. $\displaystyle 3x - 5 = x + 7$Both equal $y$, so they equal each other
2. $\displaystyle 2x = 12 \;\Rightarrow\; x = 6$
3. $\displaystyle y = 6 + 7 = 13$

• M1 $3x - 5 = x + 7$
• A1 $x = 6$
• A1 $y = 13$

Answer $x = 3,\ y = 9$ and $x = -2,\ y = 4$

1. $\displaystyle x^2 = x + 6$Substitute $y = x + 6$ into $y = x^2$
2. $\displaystyle x^2 - x - 6 = 0$
3. $\displaystyle (x - 3)(x + 2) = 0 \;\Rightarrow\; x = 3 \text{ or } x = -2$
4. $\displaystyle x = 3: y = 9;\quad x = -2: y = 4$Use the linear equation to find each $y$

• M1 $x^2 = x + 6$
• A1 $x^2 - x - 6 = 0$
• M1 $(x - 3)(x + 2) = 0$
• A1 $x = 3$ and $x = -2$
• A1 $y = 9$ and $y = 4$, correctly paired

Answer $x = 1,\ y = -2$ and $x = 4,\ y = 4$

1. $\displaystyle x^2 - 3x = 2x - 4$
2. $\displaystyle x^2 - 5x + 4 = 0$
3. $\displaystyle (x - 1)(x - 4) = 0 \;\Rightarrow\; x = 1 \text{ or } x = 4$
4. $\displaystyle x = 1: y = -2;\quad x = 4: y = 4$

• M1 $x^2 - 3x = 2x - 4$
• A1 $x^2 - 5x + 4 = 0$
• M1 $(x - 1)(x - 4) = 0$
• A1 $x = 1$ and $x = 4$
• A1 $y = -2$ and $y = 4$, correctly paired

Answer Coffee £$2.50$, cake £$1.50$

1. $\displaystyle 4c + 3k = 1450,\quad 2c + 5k = 1250$Work in pence to avoid decimals
2. $\displaystyle 4c + 10k = 2500$Multiply the second equation by $2$
3. $\displaystyle 7k = 1050 \;\Rightarrow\; k = 150$Subtract the first equation
4. $\displaystyle 4c + 450 = 1450 \;\Rightarrow\; c = 250$
5. $\displaystyle \text{coffee } £2.50,\ \text{cake } £1.50$

• M1 $4c + 3k = 1450$ and $2c + 5k = 1250$ (in pence) oe
• M1 Correct method to eliminate one letter, e.g. $7k = 1050$
• A1 Cake £$1.50$
• A1 Coffee £$2.50$

Answer $25$ and $17$

1. $\displaystyle x + y = 42,\quad x - y = 8$
2. $\displaystyle 2x = 50 \;\Rightarrow\; x = 25$Add the equations
3. $\displaystyle y = 42 - 25 = 17$

• M1 $x + y = 42$ and $x - y = 8$
• M1 $2x = 50$
• A1 $25$ and $17$

Answer She added the right-hand sides ($11 + 7$) instead of subtracting them. $y = 4$, $x = 3$.

1. $\displaystyle (x + 2y) - (x + y) = 11 - 7$Subtract both sides: the numbers on the right are subtracted too
2. $\displaystyle y = 4$
3. $\displaystyle x + 4 = 7 \;\Rightarrow\; x = 3$

• B1 Explains that $11 - 7 = 4$ should be used, not $11 + 7$
• A1 $y = 4$
• A1 $x = 3$

Answer $20$ chickens and $10$ sheep

1. $\displaystyle c + s = 30$Each animal has one head
2. $\displaystyle 2c + 4s = 80$Chickens have $2$ legs, sheep have $4$
3. $\displaystyle 2c + 2s = 60$Multiply the first equation by $2$
4. $\displaystyle 2s = 20 \;\Rightarrow\; s = 10,\quad c = 20$Subtract

• M1 $c + s = 30$
• M1 $2c + 4s = 80$
• A1 $s = 10$
• A1 $c = 20$

Answer $x = 3,\ y = 4$ and $x = -4,\ y = -3$

1. $\displaystyle x^2 + (x + 1)^2 = 25$Substitute the linear equation into the circle
2. $\displaystyle x^2 + x^2 + 2x + 1 = 25$
3. $\displaystyle 2x^2 + 2x - 24 = 0 \;\Rightarrow\; x^2 + x - 12 = 0$
4. $\displaystyle (x + 4)(x - 3) = 0 \;\Rightarrow\; x = 3 \text{ or } x = -4$
5. $\displaystyle x = 3: y = 4;\quad x = -4: y = -3$

• M1 $x^2 + (x + 1)^2 = 25$
• A1 $2x^2 + 2x - 24 = 0$ or $x^2 + x - 12 = 0$
• M1 $(x + 4)(x - 3) = 0$
• A1 $x = 3$ and $x = -4$
• A1 $y = 4$ and $y = -3$, correctly paired

Answer $x = 1,\ y = -2$ and $x = 2,\ y = -1$

1. $\displaystyle 2x^2 - 5x + 1 = x - 3$
2. $\displaystyle 2x^2 - 6x + 4 = 0 \;\Rightarrow\; x^2 - 3x + 2 = 0$Divide by $2$
3. $\displaystyle (x - 1)(x - 2) = 0 \;\Rightarrow\; x = 1 \text{ or } x = 2$
4. $\displaystyle x = 1: y = -2;\quad x = 2: y = -1$

• M1 $2x^2 - 5x + 1 = x - 3$
• A1 $2x^2 - 6x + 4 = 0$ or $x^2 - 3x + 2 = 0$
• M1 $(x - 1)(x - 2) = 0$
• A1 $x = 1$ and $x = 2$
• A1 $y = -2$ and $y = -1$, correctly paired

Answer $20$ adult tickets

1. $\displaystyle a + c = 40,\quad 8a + 5c = 260$
2. $\displaystyle 5a + 5c = 200$Multiply the first equation by $5$
3. $\displaystyle 3a = 60 \;\Rightarrow\; a = 20$Subtract

• M1 $a + c = 40$
• M1 $8a + 5c = 260$
• M1 Correct elimination, e.g. $3a = 60$
• A1 $20$

Answer $m = 3,\ c = 2$

1. $\displaystyle 5 = m + c,\quad 11 = 3m + c$Substitute each point into $y = mx + c$
2. $\displaystyle 6 = 2m \;\Rightarrow\; m = 3$Subtract the first from the second
3. $\displaystyle c = 5 - 3 = 2$

• M1 $5 = m + c$ and $11 = 3m + c$
• A1 $m = 3$
• A1 $c = 2$

Answer Jo is $14$, Sam is $6$

1. $\displaystyle j + 2 = 2(s + 2) \;\Rightarrow\; j = 2s + 2$
2. $\displaystyle j - 2 = 3(s - 2) \;\Rightarrow\; j = 3s - 4$
3. $\displaystyle 2s + 2 = 3s - 4 \;\Rightarrow\; s = 6$
4. $\displaystyle j = 2(6) + 2 = 14$Check: in $2$ years, $16 = 2 \times 8$; two years ago, $12 = 3 \times 4$

• M1 $j + 2 = 2(s + 2)$
• M1 $j - 2 = 3(s - 2)$
• M1 Simplifies to $j = 2s + 2$ and $j = 3s - 4$, and solves
• A1 Jo $14$ and Sam $6$

Answer $x = 1.59,\ y = 4.18$ and $x = -2.39,\ y = -3.78$

1. $\displaystyle x^2 + (2x + 1)^2 = 20$
2. $\displaystyle x^2 + 4x^2 + 4x + 1 = 20 \;\Rightarrow\; 5x^2 + 4x - 19 = 0$
3. $\displaystyle x = \frac{-4 \pm \sqrt{16 + 380}}{10} = \frac{-4 \pm \sqrt{396}}{10}$It doesn't factorise, so use the formula
4. $\displaystyle x = 1.589\ldots \quad\text{or}\quad x = -2.389\ldots$
5. $\displaystyle y = 2(1.589\ldots) + 1 = 4.179\ldots$Use the unrounded $x$ values
6. $\displaystyle y = 2(-2.389\ldots) + 1 = -3.779\ldots$
7. $\displaystyle (1.59,\ 4.18) \text{ and } (-2.39,\ -3.78)$

• M1 $x^2 + (2x + 1)^2 = 20$
• A1 $5x^2 + 4x - 19 = 0$
• M1 Correct substitution into the quadratic formula
• A1 $x = 1.59$ and $x = -2.39$
• M1 Substitutes their $x$ values into $y = 2x + 1$
• A1 $y = 4.18$ and $y = -3.78$, correctly paired

Answer $x^2 + 5 = x + 3$ gives $x^2 - x + 2 = 0$, and $x^2 - x + 2 = \left(x - \tfrac{1}{2}\right)^2 + \tfrac{7}{4}$, which is always at least $\tfrac{7}{4}$, so it is never $0$.

1. $\displaystyle x^2 + 5 = x + 3 \;\Rightarrow\; x^2 - x + 2 = 0$Where they meet, the $y$ values are equal
2. $\displaystyle x^2 - x + 2 = \left(x - \tfrac{1}{2}\right)^2 - \tfrac{1}{4} + 2 = \left(x - \tfrac{1}{2}\right)^2 + \tfrac{7}{4}$Complete the square
3. $\displaystyle \left(x - \tfrac{1}{2}\right)^2 + \tfrac{7}{4} \ge \tfrac{7}{4} > 0$A square is never negative
4. $\displaystyle \text{No solutions, so they do not meet}$

• M1 $x^2 - x + 2 = 0$
• M1 $\left(x - \tfrac{1}{2}\right)^2 + \tfrac{7}{4}$, or the formula with a negative under the square root
• A1 Conclusion: it can never equal $0$, so the line and curve do not meet

Answer $x = 2,\ y = 1$ and $x = \tfrac{3}{2},\ y = \tfrac{4}{3}$

1. $\displaystyle y = \frac{7 - 2x}{3}$Rearrange the linear equation
2. $\displaystyle x \cdot \frac{7 - 2x}{3} = 2 \;\Rightarrow\; 7x - 2x^2 = 6$
3. $\displaystyle 2x^2 - 7x + 6 = 0$
4. $\displaystyle (2x - 3)(x - 2) = 0 \;\Rightarrow\; x = \tfrac{3}{2} \text{ or } x = 2$
5. $\displaystyle x = 2: y = 1;\quad x = \tfrac{3}{2}: y = \tfrac{4}{3}$Use $xy = 2$, so $y = \tfrac{2}{x}$

• M1 Rearranges the linear equation, e.g. $y = \dfrac{7 - 2x}{3}$
• M1 Substitutes: $x \cdot \dfrac{7 - 2x}{3} = 2$
• A1 $2x^2 - 7x + 6 = 0$
• M1 $(2x - 3)(x - 2) = 0$
• A1 $x = 2$ and $x = \tfrac{3}{2}$
• A1 $y = 1$ and $y = \tfrac{4}{3}$, correctly paired

Answer Length $8$ cm, width $5$ cm

1. $\displaystyle l + w = 13,\quad lw = 40$Half the perimeter is $13$
2. $\displaystyle w(13 - w) = 40$Substitute $l = 13 - w$
3. $\displaystyle 13w - w^2 = 40 \;\Rightarrow\; w^2 - 13w + 40 = 0$
4. $\displaystyle (w - 5)(w - 8) = 0 \;\Rightarrow\; w = 5 \text{ or } w = 8$
5. $\displaystyle \text{The sides are } 8 \text{ cm and } 5 \text{ cm}$Both solutions give the same rectangle

• M1 $2l + 2w = 26$ (or $l + w = 13$) and $lw = 40$
• M1 Substitutes: $w(13 - w) = 40$
• A1 $w^2 - 13w + 40 = 0$
• M1 $(w - 5)(w - 8) = 0$
• A1 $8$ cm by $5$ cm