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Description:
Math Elementary Algebra: Drill 1, Problem 4. Solve for y using substitution.
- Elementary Algebra / Substitution
- Elementary Algebra / Linear equations
- Product Type / ACT Math
- Foreign Language / Korean Subtitled
- Foreign Language / Chinese Subtitled
- Foreign Language / Spanish Subtitled
- Foreign Language / Arabic Subtitled
- Elementary Algebra / Evaluation of algebraic expressions through substitution
Transcript
- 00:03
Shmoopy-Doo, where are you?
- 00:07
Solve for y using substitution: 3x + 4y = 15 -2x + 6y = 12
- 00:15
And here are the potential answers...
- 00:23
OK so what is this question asking?
- 00:24
It's just another isolation and plugging problem.
Full Transcript
- 00:27
We have to solve for y by getting rid of x.
- 00:30
to draw any undue attention.
- 00:30
Then we just solve for y as in any normal, one variable problem.
- 00:34
So let's take the first equation and rewrite to get 3x equals 15 minus 4y.
- 00:41
Now divide both sides by 3 and we have x equals 5 minus 4/3 y
- 00:48
Then we just substitute 5 minus four-thirds y inside of x in the next equation
- 00:57
and we get negative 2 times the quantity 5 minus four-thirds y...plus 6y equals 12.
- 01:07
We distribute the negative 2 to get negative 10 PLUS eight-thirds y plus 6y equals 12.
- 01:16
Simplify a bit more by rewriting as negative 10 plus 8y over 3 plus 18y over 3... because
- 01:23
6 can also be written as 18 over 3, which gives us common denominators... equals 12.
- 01:31
Dealing with just the fractions we get 18 plus 8 which is 26y... over 3.
- 01:36
Add 10 to both sides of the equation and we can simplify everything to 26 y over 3 equals 22.
- 01:43
Multiply both sides by 3 over 26, and we get y equals 66 over 26.
- 01:56
The greatest common factor of 66 and 26 is 2 so we can simplify this to 33 over 13.
- 02:03
So our correct answer for y is B.
- 02:05
As in, 'Burlap sack.'
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