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# Module 8

### Compound Inequalities

Question | Answer |
---|---|

Solve x-5<7 and 3x+4<8 | Solve each side first in which you add 5 on left side and get x<12. Subtract 4 on left side which gives you 4 and then divide by 3. x<3/4 or (12,3/4) |

Solve x>5 or x>3 | You would put them with negative infinities. Therefore (-infinity,5)U(-infinity,3) |

Solve -12>4(x-3)<2 | Firts multiply x-3 by 4. -12>4x-12<2 Then you add 12 to both sides of the equation which = 0>4x<14. Divide the 4 from both sides. Answer: (0,7/2) |

Solve -6x<-30 and x-10>5 | First get both x's alone. Divide by -6 on left side and add 10 on right side. Answer: x<10 and x>15 |

Solve x+ 3>11 and 4x+5>3 | Subtract 3 from 11 on left side, which gives you x>8. On right side subtract 5 from 3 and divide by 4. x> -1/2. Answer: (8,-1/2) |

Solve x<2 or x<4 | Negative infinity first for both due to infinite many solutions. (-infinity,2)U(-infinity,4) |

Solve -2x>20 and x+6>18 | On left side you would divide both sides by -2 which gives you x>-10. On right side subtract 6 from both sides and you get x>12. Answer: (-10,12) |

Solve 3x+2<11 and x-3<4 | Subtract 2 on both sides for the left equation and you get 3x<9 in which you divide both sides by 3 to get x alone, x<3. On right side add 3 to both sides, x<7. Answer: (3,7) |

Solve -5x<-15 or 4x-21>-5 | On left side divide both sides by -5 in which you get x<3. The answer would then be (3,infinity) |

Solve x+4<2 and 2x+2<4 | Subtract 4 on left side, x<-2. Subtract the 2 on right side then divide by 2, x<1. Answer: (2,1) |

Created by:
bryankoz19