# ACT Math 1.4 Elementary Algebra

Math Elementary Algebra: Drill 1, Problem 4. Solve for *y* using substitution.

ACT Math | Elementary Algebra |

ACT Mathematics | Elementary Algebra |

Elementary Algebra | Evaluation of algebraic expressions through substitution Linear equations Substitution |

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

Language | English Language |

Polynomials | Adding and Subtracting Polynomials |

Product Type | ACT Math |

### Transcript

We have to solve for y by getting rid of x.

to draw any undue attention.

Then we just solve for y as in any normal, one variable problem.

So let's take the first equation and rewrite to get 3x equals 15 minus 4y.

Now divide both sides by 3 and we have x equals 5 minus 4/3 y

Then we just substitute 5 minus four-thirds y inside of x in the next equation

and we get negative 2 times the quantity 5 minus four-thirds y...plus 6y equals 12.

We distribute the negative 2 to get negative 10 PLUS eight-thirds y plus 6y equals 12.

Simplify a bit more by rewriting as negative 10 plus 8y over 3 plus 18y over 3... because

6 can also be written as 18 over 3, which gives us common denominators... equals 12.

Dealing with just the fractions we get 18 plus 8 which is 26y... over 3.

Add 10 to both sides of the equation and we can simplify everything to 26 y over 3 equals 22.

Multiply both sides by 3 over 26, and we get y equals 66 over 26.

The greatest common factor of 66 and 26 is 2 so we can simplify this to 33 over 13.

So our correct answer for y is B.

As in, 'Burlap sack.'