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# | English & Math | Ref |
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1 | Substitute x = b - a into a + x gives us | substitution, addition is commutative, definition of negative element, addition is associative, definition of zero element |

a + b - a = b + a - a = b + 0 = b | ||

2 | Therefore from (1), x = b - a is a solution to the question's equation | |

3 | Suppose there is an arbitrary solution x1 | by assumption |

a + x1 = b | ||

4 | Adding the negative of a to both sides | you can add the same term to both sides of an equation as the equality of lhs = rhs means that the resultant sum of the lhs and the term added and rhs and the term added are equal to each other |

a + x1 - a = b - a | ||

5 | Re-arranging: | addition is commutative, definition of negative element, definition of zero element |

x1 + a - a = b - a = x1 + 0 = b - a x1 = b - a | ||

6 | Therefore if a solution exists, it will be = b - a and accordingly is unique as there is only one solution | |

7 | Alternatively we can do steps (3..5) for a second arbitrary x2 and then show | proving uniqueness by showing two arbitrary items are equal to each other |

x1 = x2 | ||

8 | However, we proved in (2) that b - a is a solution and therefore it is the unique solution to the equation. |