Linear/quadratic simultaneous equations

(a) {(−1, −1), (3, 7)}

1. $$x^{2}-2=2x+1$$
2. $$(x-3)(x+1)=0$$
3. Equate the expressions for y: 2x + 1 = $x^{2}$ $-$ 2.
4. Rearrange and factorise: $x^{2}$ $-$ 2x $-$ 3 = (x $-$ 3)(x + 1) = 0.
5. The x-values are 3 and $-1$; substituting into y = 2x + 1 gives y = 7 and $-1$ respectively.

• M1 Establishing $x^{2}-2=2x+1$ or an equivalent valid method.
• M1 Establishing $(x-3)(x+1)=0$ or an equivalent valid method.
• A1 Correct answer: {(−1, −1), (3, 7)}

(a) $x + 1 = x^2 - 5 \Rightarrow x^2 - x - 6 = 0$

1. Set the two expressions for $y$ equal.
   $$x + 1 = x^2 - 5$$
2. Rearrange.
   $$x^2 - x - 6 = 0$$

• M1 Equating: $x + 1 = x^2 - 5$.
• A1 Rearranging to $x^2 - x - 6 = 0$.

(b) $(3, 4)$

1. $(x - 3)(x + 2) = 0$, so $x = 3$ or $x = -2$.
2. When $x = 3$: $y = 3 + 1 = 4$.

• M1 Factorising to $(x - 3)(x + 2)$.
• A1 The correct answer, $(3, 4)$.

(c) $(-2, -1)$

1. When $x = -2$: $y = -2 + 1 = -1$. Check: $(-2)^2 - 5 = -1$.

• B1 The correct answer, $(-2, -1)$.

(a) {(−7, −4), (4, 7)}

1. $$x^{2}+(x+3)^{2}=65$$
2. $$(x+7)(x-4)=0$$
3. Substitute y = x + 3 into the circle: $x^{2}$ + (x + 3)$^{2}$ = 65.
4. Simplify: $2x^{2}$ + 6x $-$ 56 = 0, so $x^{2}$ + 3x $-$ 28 = (x + 7)(x $-$ 4) = 0.
5. x = $-7$ gives y = $-4$; x = 4 gives y = 7.
6. The points are ($-7$, $-4$) and (4, 7).

• P1 Establishing $x^{2}+(x+3)^{2}=65$ or an equivalent valid method.
• P1 Establishing $(x+7)(x-4)=0$ or an equivalent valid method.
• A1 Correct answer: {(−7, −4), (4, 7)}

(a) $(2,4)$ and $(4,16)$

1. Equate the two expressions for y.
   $$x^{2}=6x-8$$
2. Factorise to obtain both x values.
   $$(x-2)(x-4)=0$$
3. Use the original curve to recover the matching y values.
   $$y=2^{2}$$
4. Therefore $(2,4)$ and $(4,16)$.

• M1 Equate the two expressions for y.
• M1 Factorise to obtain both x values.
• M1 Use the original curve to recover the matching y values.
• A1 Correct answer: $(2,4)$ and $(4,16)$

(b) $6$

1. Use rise over run from the two points.
   $$\frac{16-4}{4-2}$$
2. Therefore $6$.

• M1 Use rise over run from the two points.
• A1 Correct answer: $6$

(a) (2, 4) and (−2, −4)

1. Substitute y = 2x into the first equation.
   $$x^{2}+(2x)^{2}=20$$
2. Solve for x.
   $$x^{2}=4$$
3. Therefore (2, 4) and (−2, −4).

• M1 Substitute y = 2x into the first equation.
• M1 Solve for x.
• A1 Correct answer: (2, 4) and (−2, −4)

(b) They are the two points where the line y = 2x crosses the circle of radius √20 centred at the origin.

1. They are the two points where the line y = 2x crosses the circle of radius √20 centred at the origin.

• C1 Correct conclusion with supporting reasoning: They are the two points where the line y = 2x crosses the circle of radius √20 centred at the origin.

(a) Substituting y = 3 - x gives x(3 - x) = −10, so 3x - x² = −10 and x² − 3x - 10 = 0.

1. Substitute the line into the curve.
   $$x(3-x)=-10$$
2. Substituting y = 3 - x gives x(3 - x) = −10, so 3x - x² = −10 and x² − 3x - 10 = 0.

• M1 Substitute the line into the curve.
• C1 Correct conclusion with supporting reasoning: Substituting y = 3 - x gives x(3 - x) = −10, so 3x - x² = −10 and x² − 3x - 10 = 0.

(b) (5, −2) and (−2, 5)

1. Factorise and pair each x with its own y-value.
   $$(x-5)(x+2)=0$$
2. Therefore (5, −2) and (−2, 5).

• M1 Factorise and pair each x with its own y-value.
• A1 Correct answer: (5, −2) and (−2, 5)

(a) $k = 2$

1. Where they meet:
   $$x^2 + 3 = 2x + k \Rightarrow x^2 - 2x + (3 - k) = 0$$
2. Complete the square.
   $$(x - 1)^2 + 2 - k = 0$$
3. For exactly one solution, the only solution must be $x = 1$, so $2 - k = 0$ and $k = 2$.

• P1 Forming $x^2 - 2x + 3 - k = 0$.
• P1 Using the condition for one solution: $(x - 1)^2 = k - 2$ has one solution only when $k - 2 = 0$ (or $b^2 - 4ac = 0$).
• A1 The correct answer, $k = 2$.

(b) $(1, 4)$

1. $x = 1$ and $y = 1^2 + 3 = 4$ (also $2 \times 1 + 2 = 4$).

• B1 The correct answer, $(1, 4)$.

(a) $-10, 2$

1. $$x^{2}-(k+4)x+9=0$$
2. $$(k+4)^{2}-36=0$$
3. At an intersection, $x^{2}$ $-$ 4x + 7 = kx $-$ 2, so $x^{2}$ $-$ (k + 4)x + 9 = 0.
4. Exactly one intersection means this quadratic has a repeated root, so its discriminant is zero.
5. (k + 4)$^{2}$ $-$ 36 = 0, giving k + 4 = 6 or k + 4 = $-6.$
6. Therefore k = 2 or k = $-10.$

• P1 Establishing $x^{2}-(k+4)x+9=0$ or an equivalent valid method.
• P1 Establishing $(k+4)^{2}-36=0$ or an equivalent valid method.
• A1 Correct answer: $-10, 2$