$(x-2)^2 = x^2 - 4x + 4$ or $(y+2)^2 = y^2 + 4y + 4$ | M: 3 or 4 terms | M1 |

$(x-2)^2 + x^2 = 10$ or $y^2 + (y+2)^2 = 10$ | M: Substitute | M1 |

$2x^2 - 4x - 6 = 0$ or $2y^2 + 4y - 6 = 0$ | Correct 3 terms | A1 |

$(x-3)(x+1) = 0$, $x = \ldots$ or $(y+3)(y-1) = 0$, $y = \ldots$ | | M1 |

$x = 3, x = -1$ or $y = -3, y = 1$ | | A1 |

$y = 1, y = -3$ or $x = -1, x = 3$ | | M1, A1 | (7 marks)

(Allow equivalent fractions such as: $x = \frac{6}{2}$ for $x = 3$)

**Total: 7 marks**

**Guidance:**
- 1st M: 'Squaring a bracket', needs 3 or 4 terms, one of which must be an $x^2$ or $y^2$ term.
- 2nd M: Substituting to get an equation in one variable (awarded generously).
- 1st A: Accept equivalent forms, e.g. $2x^2 - 4x = 6$.
- 3rd M: Attempting to solve a 3-term quadratic, to get 2 solutions.
- 4th M: Attempting at least one y value (or x value).
- If y solutions are given as x values, or vice-versa, penalise at the end, so that it is possible to score M1 M1 A1 M1 A1 M0 A0.

**"Non-algebraic" solutions:**
- No working, and only one correct solution pair found: M0 M0 A0 M0 A0 M1 A0
- No working, and both correct solution pairs found, but not demonstrated: M0 M0 A0 M1 A1 M1 A1
- Both correct solution pairs found, and demonstrated, perhaps in a table of values: Full marks

**Squaring individual terms, e.g.:**
- $y^2 = x^2 + 4$ | M0 |
- $x^2 + 4 + x^2 = 10$ | M1 A0 | (Eqn. in one variable)
- $x = \sqrt{3}$ | M0 A0 | (Not solving 3-term quad.)
- $y^2 = x^2 + 4 = 7$, $y = \sqrt{7}$ | M1 A0 | (Attempting one y value)

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