Professor Dave again, let’s solve some more equations.
We now know the very basics about how algebra
works.
We have equations which contain one or more variables, and we can solve for these variables
by manipulating the equation in various ways.
But we will see that the strategies we need to utilize to solve these equations will get
more complex as the equations get more complex.
Just as with everything else in math, this gets easier with practice, so let’s look
at situations that are slightly trickier than the simplest ones we have learned.
If we have three X plus seven equals sixteen, we see that there are two things we have to
do, since X is involved in both a product and a sum.
But what do we do first?
If we divide by three first, things get tricky, because we have to divide both sides by three,
and on the left, that means the whole side, not just three X, so we would get three X
plus seven over three equals sixteen thirds.
That just made things worse, so instead, let’s subtract seven first.
That gives us three X equals nine, and then we can divide to get X equals three.
So we can see that it will typically be the case that we should do addition and subtraction
before we do multiplication and division when solving algebraic equations.
Now let’s try another one.
Here we have four X plus three equals X minus six.
We can see that since there is now more than one X term, we will need to perform operations
to get a single X term by itself in order to solve the equation.
When we have several terms like this, our strategy should be to do whatever we need
to do to get all of the X terms together first.
In this case, the easiest way to do that is to subtract X from both sides.
If we do that, we get four X minus X plus three on the left, and when we see something
like this, we can combine like terms.
We have one X term and another X term, so we can combine them to get three X, since
four minus one equals three.
The X doesn’t go away, it’s the common term that allows us to subtract, just the
way four apples minus three apples equals one apple, not just the number one.
So this leaves us with three X plus three.
On the right, the X will go away, because X minus X equals zero.
Now we have three X plus three equals negative six.
We will want to get all the numbers together, and we should do it in such a way that gets
the X term by itself.
That means our next move should be to subtract three from both sides.
That will get rid of this three on the left, and after we combine like terms, we get negative
nine on the right.
Now for the last step.
Since we have three X, we will need to divide by three.
That leaves us with X equals negative three.
So that problem took three steps, which is certainly more than one, but we will later
see plenty of examples that require much more manipulation than this, so let’s try some
even harder examples.
How about this one.
The quantity of three minus three X over four equals three times the quantity of two X plus
one.
We want to combine the X terms, but we can’t do this the way things look right now.
First let’s distribute the three on the right side and get six X plus three.
Now we have this fraction on the left that we would like to get rid of.
To do that, we can multiply both sides by four.
At this point, let’s acknowledge that an easier way to conceptualize a step like this
is to acknowledge that whatever is in the denominator of one side can be pulled up to
the numerator of the other side, and anything in the numerator of one side can be pulled
down to the denominator of the other side.
Pulling this four up over here is the same thing as multiplying both sides by four, because
it will cancel out on the left, and then appear on the right, but instead of having to write
fours and cancel fours and perform all of these extra steps each time, it’s good to
get used this simpler way of thinking, and realize that we can just grab the four, and
put it over here, making sure to put it outside of some parentheses, since it must multiply
the entire expression.
We multiply through and get twenty-four X plus twelve.
Now we are ready to combine X terms.
Adding three X to both sides, we get twenty-seven X over here, and then we can subtract twelve
from both sides, and we get negative nine equals twenty-seven X.
At this point, let’s just flip the equation around so that the X term is on the left side,
as we just prefer to read equations that way.
Lastly we can divide by twenty-seven, and negative nine over twenty-seven can be reduced
to negative one third.
So that took a couple more steps than the last example, but each individual step was
pretty easy.
We will find that a lot of algebra works this way.
We may have to develop a strategy to solve an equation, and that strategy could have
lots of steps, which means lots of opportunities to make a mistake, but as long as we are meticulous
as we move through each step and perform careful arithmetic, it is typically not too difficult
to find the answer. Let's check comprehension.