# ACT Math 3.1 Elementary Algebra

ACT Math: Elementary Algebra Drill 3, Problem 1. Solve for a, given the two equations in the video.

ACT Math | Elementary Algebra |

ACT Mathematics | Elementary Algebra |

Elementary Algebra | Linear equations Understanding algebraic operations |

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

Language | English Language |

One-Variable Equations and Inequalities | Solving Equations |

Product Type | ACT Math |

Systems of Equations | Linear and Nonlinear Equations |

### Transcript

what we really want to do is add them in such a way so that we GET RID OF a variable.

If we just have ONE variable, the problem is a whole lot easier to solve.

Okay, let's set our sights on crushing a.

We can start by multiplying both sides of the first equation by... negative 2.

Then, when we add the two equations, we'll have no a and a whole lotta b.

So here we go.

Negative 2 times 3a plus 16b equals 42... which gets us negative 6a minus 32b equals negative 84.

Now we just add the two equations; our a's cancel out... negative 32b plus 20b is negative

12b, and negative 84 plus 12 is negative 72.

Multiply both sides by negative 1 and we have 12b equals 72.

Divide both sides by 12 and we have b equals 6.

We can now just plug in 6 for b, and 6a plus 120 equals 12.

Subtract 120 from both sides and we get 6a equals negative 108; divide both sides by

6 and we get a equals negative 18.

The answer is A.