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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.

Language:
English Language

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

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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