rational equations |

__Rational Equations (fractions)__

There are two types of equations with rational expressions or fractions. The 1st type have ** constants or real numbers as denominators** and the 2nd type have

Equations which have **constants** or real numbers as **denominators** are solved easily if we

**multiply through by the lowest common multiple **(**lcm) **of the denominator constants.

Once you've done this, **the equation becomes a linear or quadratic equation** which can be solved using one of the techniques described in those lessons. (6.1 and 6.2)

Rational equations which have **variables** in their **denominators** are solved in a similar manner, however **we must check that none of our solutions make any denominator equal zero since division by zero is undefined**. If one or more of our answers make some denominator equal zero, we must eliminate those values from the solution set.

examples-1 |
examples-2 |
practice |
solutions |

__examples with constant denominators__

a) 4( 4 4 |
a) The Multiply through by 12. Remove brackets Transpose, collect like terms and solve. |

b) 2(2 4 5 |
b) Multiply through by 6 the Remove brackets. Transpose, collect like terms and solve. |

c)
4( 4 ( |
c) The Multiply through by 12, Remove brackets, collect like terms. Solve by factoring the trinomial. |

examples-1 |
examples-2 |
practice |
solutions |

__examples with variable denominators__

a) 4( 4 |
a) The Multiply, remove brackets, transpose, solve. Check that denominator = zero. |

b) (3 6 2 |
b) The Multiply through by the brackets, collect like terms and solve. |

c)
( 0 = – 2 no solutions |
c) We factor the denominators the Multiply by the which cannot be true. This equation has no solutions since 0 cannot equal –2. |

d)
If ( therefore there are no solutions. |
d) We factor the denominators the Multiply by the Check if m = 1 makes any denominators = 0. It makes both denominators = 0, therefore there are no solutions. |

examples-1 |
examples-2 |
practice |
solutions |

Solve the following equations. Check your answers.

1)

2)

3)

4)

5) What number must be added to both numerator and denominator of the fraction to change its value to ?

6) Find 3 **consecutive** **even integers** such that one-quarter of the 1st, plus one-third of the 2nd, plus one-fifth of the 3rd equals 14. (hint: to get from one **even** integer to the next, add 2)

examples-1 |
examples-2 |
practice |
solutions |

Solve the following equations. Check your answers.

1) 4(2 8 5 |
2) 8( 8 –4 |

3)
7 (7 |
4) 8 2 ( |

5) What number must be added to both numerator and denominator of the fraction to change its value to ?

- Let

- The new fraction with value will look like this:

- When we cross multiply we get: 3( 7 +

- This then becomes 21 + 3

- We must add – 5 to both numerator and denominator to make the fraction =

- 6) Find 3 consecutive

- Let

- therefore

- and

- 15

- 47

- 47

- The 3 consecutive

examples-1 |
examples-2 |
practice |
solutions |

(*all content **© MathRoom Learning Service; 2004 - *).