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|
|Foreign Language||Arabic Subtitled|
|Polynomials||Adding and Subtracting Polynomials|
|Product Type||ACT Math|
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.'