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Teachers & SchoolsACT Math Elementary Algebra Drill 4, Problem 2. Solve for x.

ACT Math | Elementary Algebra |

ACT Mathematics | Elementary Algebra |

Basic Operations | Exponents |

Elementary Algebra | Solving linear inequalities with one variable |

Foreign Language | Arabic Subtitled Chinese Subtitled Korean Subtitled Spanish Subtitled |

Intermediate Algebra | Absolute value equations and inequalities |

Language | English Language |

Product Type | ACT Math |

We multiply both sides by 2 to get x plus 1 is less than 6.

Subtract 1 from both sides and we have x is less than 5, so our potential answers are

only B and C.

Now let's figure out what this problem would be if x were the most negative it could be.

What would make things under the absolute value lines to equal NEGATIVE 3?

We'd have x plus 1 over 2 is less than negative 3, but with no absolute value lines now.

We multiply both sides by 2 and we have x plus 1 is less than negative 6.

Then subtract 1 from both sides and we have x is less than negative 7.

So we have a positive 5 and a negative 7 as the range here.

Which is... this range... and our answer is C.