There are two ways to handle simultaneous equations.

This is the addition (or subtraction) method. Simultaneous Equations involve two equations with two unknowns. The substitution method involves:

• Rearrange one equation so that one unknown is on one side

• In the other equation, replace the unknown from the first equation

• Solve this second equation (with one unknown)

• Replace the solved value in the first equation

• Solve the first equation

• Check the values are correct

Note that it is not important which equation is used to start.

If one equation is a multiple of another, then there will be no solution (i.e. they are parallel lines).

Solve the simultaneous equations `4x-y=6` and `5x+2y=1`

Rearrange the first equation to get it in terms of `y`

`4x` | `-` | `y` | `=` | `6` | |||

Add `y` to both sides: | `4x` | `=` | `6` | `+` | `y` | ||

Subtract 6 from both sides: | `4x` | `-` | `6` | `=` | `y` |

Replace `y` in the second equation `4x-6`

Second equation: | `5x` | `+` | `2y` | `=` | `1` | ||

substitute for `y`: | `5x` | `+` | `2(4x-6)` | `=` | `1` | ||

expand brackets: | `5x` | `+` | `8x` | `-` | `12` | `=` | `1` |

add x terms: | `13x` | `-` | `12` | `=` | `1` | ||

add 12 to both sides: | `13x` | `=` | `13` | ||||

divide both sides by 13 | `x` | `=` | `1` |

Replace `x` in the first equation with the found value of `x`

First equation: | `4x` | `-` | `y` | `=` | `6` | ||

substitute for `x`: | `4(1)` | `-` | `y` | `=` | `6` | ||

`4` | `-` | `y` | `=` | `6` | |||

add `y` to both sides: | `4` | `=` | `6` | `+` | `y` | ||

subtract 6 from both sides: | `-2` | `=` | `y` |

Finally, check by putting both values in the other equation

Check using 2nd equation: | `5(1)+2(-2)=1`✔ |

Answer: `x=1, y=-2`

Solve the simultaneous equations `4x+3y=14` and `6x+2y=11`.

Rearrange the first equation to be in terms of `y`

2nd equation to get `y` | `6x` | + | `2y` | = | 11 | ||

subtract 6x from both sides: | `2y` | = | 11 | - | `6x` | ||

divide all terms by 2 | `y` | = | `frac(11)(2)` | - | `3x` |

Solve the equation for `x`

First equation: | `4x` | + | `3y` | = | 14 | ||

Replace the value `y`: | `4x` | + | `3(frac(11)(2)-3x)` | = | 14 | ||

Multiply out the brackets: | `4x` | `+` | `frac(33)(2)` | `-` | `9x` | `=` | `14` |

subtract the fraction from both sides: | `4x` | `` | `` | `-` | `9x` | `=` | `-frac(5)(2)` |

add the `x`-values: | `-5x` | `` | `` | `` | `` | `=` | `-frac(5)(2)` |

divide both sides by 5: | `-x` | `` | `` | `` | `` | `=` | `-frac(1)(2)` |

multiply both sides by -1: | `x` | `` | `` | `` | `` | `=` | `frac(1)(2)` |

Substitute the found value of `x`

Second equation | `6x` | `+` | `2y` | `=` | `11` | ||

Substitute known value of `x` | `6(frac(1)(2))` | `+` | `2y` | `=` | `11` | ||

`3` | `+` | `2y` | `=` | `11` | |||

Subtract 3 from both sides | `` | `` | `2y` | `=` | `8` | ||

Divide both sides by 2 | `` | `` | `y` | `=` | `4` |

Check the answer

Check using 1st equation | 4(`frac(1)(2)`)+3(4)=14` ✔ |

Answer: `x=frac(1)(2)` and `y=4`

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