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ACT Math 3.1 Elementary Algebra 268 Views
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Description:
ACT Math: Elementary Algebra Drill 3, Problem 1. Solve for a, given the two equations in the video.
Transcript
- 00:02
Here's your shmoop du jour:
- 00:04
Solve for a. 3a + 16b = 42 6a + 20b = 12
- 00:11
And here are the potential answers...
- 00:16
Alright. it's really checking whether we realize that we can ADD equations.
- 00:21
But if we just add the two equations above, it doesn't really do anything for us -- so
Full Transcript
- 00:25
what we really want to do is add them in such a way so that we GET RID OF a variable.
- 00:30
If we just have ONE variable, the problem is a whole lot easier to solve.
- 00:34
Okay, let's set our sights on crushing a.
- 00:37
We can start by multiplying both sides of the first equation by... negative 2.
- 00:41
Then, when we add the two equations, we'll have no a and a whole lotta b.
- 00:46
So here we go.
- 00:47
Negative 2 times 3a plus 16b equals 42... which gets us negative 6a minus 32b equals negative 84.
- 00:56
Now we just add the two equations; our a's cancel out... negative 32b plus 20b is negative
- 01:02
12b, and negative 84 plus 12 is negative 72.
- 01:06
Multiply both sides by negative 1 and we have 12b equals 72.
- 01:11
Divide both sides by 12 and we have b equals 6.
- 01:15
We can now just plug in 6 for b, and 6a plus 120 equals 12.
- 01:19
Subtract 120 from both sides and we get 6a equals negative 108; divide both sides by
- 01:25
6 and we get a equals negative 18.
- 01:28
The answer is A.
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