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Systems of Equations Videos 73 videos
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ACT Math 2.5 Elementary Algebra 283 Views
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Description:
ACT Math Section: Elementary Algebra Drill 2, Problem 5. Use these two equations to solve for both x and y.
- Elementary Algebra / Linear equations
- Product Type / ACT Math
- Foreign Language / Arabic Subtitled
- Foreign Language / Korean Subtitled
- Foreign Language / Spanish Subtitled
- Foreign Language / Chinese Subtitled
- Algebra / Solve systems of equations
- Algebra / Solve systems of equations
- Algebra / Solve systems of equations
- Algebra / Solve systems of equations
- Elementary Algebra / Evaluation of algebraic expressions through substitution
Transcript
- 00:02
Here's your shmoop du jour:
- 00:05
Solve for x and y: 3x + 2y = 13 6x + y = 20
- 00:11
And here are the potential answers...
- 00:20
OK. Not much of a brain-squeezer.
- 00:23
Since y, stands alone in the bottom equation, let's solve for y.
Full Transcript
- 00:27
We get y equals 20 minus 6x.
- 00:30
Now we just substitute 20 minus 6x anywhere we see a y.
- 00:34
Ok so rewrite the above equation with that substitution and we have
- 00:37
3x plus 40 minus 12x equals 13
- 00:41
Simplify to get negative 9x equals 13 minus 40 or negative 9x equals negative 27; divide
- 00:48
both sides by negative 9 and we get x equals 3.
- 00:52
Phew. Great. Only A and B are possible answers.
- 00:55
Now just plug in 3 for x in the second equation
- 00:58
and we get 6 times 3 is 18... plus y... equals 20.
- 01:02
Now subtract 18 from both sides and we get y equals 2.
- 01:06
Here we go. So boom! The answer is B.
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You don't want to leave a bunch of unsolved rational equations sitting around.