Tip:
Highlight text to annotate it
X
(male narrator) In this video,
we will look at solving formulas for a variable,
which involve fractions.
Remember from solving linear equations,
we could clear fractions by multiplying...
by the least common denominator.
We will follow this same strategy with the formulas.
As we do, remember, we may have to distribute first...
before we can multiply by the LCD
to be sure we arrive at the correct solution.
Let's try some examples
where we clear fractions by multiplying by the LCD.
In this problem, you might make the 4a into a fraction
by putting it over 1.
Now, as I look at my fractions,
I might notice that the least common denominator is the x.
So we multiply each term by this x.
Notice in the first and last fractions,
the x's will divide out completely.
This will leave 5 plus 4ax, equals b.
Now that we've cleared the fractions,
we can begin solving for the variable,
just as we have with other formulas.
We balance through the equal sign
and get rid of the term without the x.
Here, it's a +5.
The opposite of a +5 is a -5.
We will do the same thing to both sides,
leaving 4ax on the left side.
On the right side, you notice we do not have like terms.
Thus, we will leave the subtraction problem: b minus 5.
Remember, we want the x alone.
That means we need to get rid of the factors 4 and a.
Right now, the 4 and a are multiplied by x,
so we get rid of them by dividing both sides by 4a.
The 4s and a's divide out, and we're left with:
x equals the expression, b minus 5, over 4a.
With the variable alone, this becomes our final solution.
Let's try another problem-- clearing fractions--
that may be a little more involved.
In this problem, we're solving for b.
Notice we have fractions
in front of parentheses in this problem.
Recall that we must always distribute
before we clear the fractions.
Let's start by doing that.
We'll distribute the 1/2h onto both terms.
We now have: a equals 1/2 of hb, plus 1/2 of hc.
Now that we have cleared the parentheses,
we're ready to clear the fractions
by multiplying by the least common denominator.
Don't forget the A on the left side is also a term,
maybe putting it over 1.
The least common denominator is 2.
We will multiply both sides, every term, by 2.
As we do, on the left side, we have 2A.
On the right side, the 2s divide out,
and we're left with hb plus hc.
Remember, we're solving for the b
on the right side of the equation.
First, we will add and subtract
the term that does not have any b's on it.
Right now, we have plus hc,
so we will subtract hc from both sides of the equation.
On the left, we do not have like terms,
so we have 2A minus hc, equals hb.
Finally, to get the b alone,
we will divide both sides by the h.
Now, the b is alone on the right side,
and on the left, we have the fraction:
2A minus hc, over h.
Again, it is important to remember,
we cannot divide out the h's
in the numerator and denominator
because of the subtraction
in the problem.
We are not allowed to reduce
if there is any adding or subtracting in the fraction.
Multiplying by the least common denominator
can still clear fractions.