### Linear formulas and equations: Linear equations and inequalities

### General solution of a linear inequality

General solution to a linear inequalityLet #a# and #b# be numbers. We will take a look at the four possible inequalities with #a \cdot x +b# on the left hand side and #0# to the right hand side. All four inequalities have four possible forms of solutions, dependent on the value of #a# and #b#.

**The inequality #a \cdot x+b \gt 0#**

Case |
Solutions |

#a \gt 0# | #x \gt -\frac{b}{a}# |

#a \lt 0# | #x \lt -\frac{b}{a}# |

#a=0# and #b \gt 0# | every #x# |

#a=0# and #b \leq 0# | none |

**The inequality #a \cdot x+b \geq 0#**

Case |
Solutions |

#a \gt 0# | #x \geq -\frac{b}{a}# |

#a \lt 0# | #x \leq -\frac{b}{a}# |

#a=0# and #b\geq 0# | every #x# |

#a=0# and #b\lt 0# | none |

**The inequality #a \cdot x+b \lt 0#**

Case |
Solutions |

#a \gt 0# | #x \lt -\frac{b}{a}# |

#a \lt 0# | #x \gt -\frac{b}{a}# |

#a=0# and #b \lt 0# | every #x# |

#a=0# and #b \geq 0# | none |

**The inequality #a \cdot x+b \leq 0#**

Case |
Solutions |

#a \gt 0# | #x \leq -\frac{b}{a}# |

#a \lt 0# | #x \geq -\frac{b}{a}# |

#a=0# and #b\leq 0# | every #x# |

#a=0# and #b\gt 0# | none |

In practice, we do not have to remember these tables. We can solve the linear equations by means of reduction. If the #x# is removed from the inequality and the inequality of the numbers is true, then it will hold for every #x#. If it is not true, then there is no value of #x# for which the inequality holds.

Determine all #x# for which #6\cdot x+1\leq 6\cdot x+4#.

Give your answer in one of the following forms with a suitable number for #a#. Simplify #a# as much as possible.

Give your answer in one of the following forms with a suitable number for #a#. Simplify #a# as much as possible.

- #x \lt a#
- #x\gt a#
- #x\le a#
- #x\ge a#

#all#

This can be reduced as follows:

\[\begin{array}{rcl} 6x+1 &\leq & 6x+4\\&&\phantom{xxx}\blue{\text{given}} \\

1 &\leq& 4\\ &&\phantom{xxx}\blue{6x \text{ subtracted on both sides }}\\

\end{array}\]

#x# was eliminated from the inequality. Because the remaining inequality is correct, all values of #x# are solutions.

This can be reduced as follows:

\[\begin{array}{rcl} 6x+1 &\leq & 6x+4\\&&\phantom{xxx}\blue{\text{given}} \\

1 &\leq& 4\\ &&\phantom{xxx}\blue{6x \text{ subtracted on both sides }}\\

\end{array}\]

#x# was eliminated from the inequality. Because the remaining inequality is correct, all values of #x# are solutions.

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