### Video Transcript

In this video, we will learn how to
use properties of proportions to find an unknown value in a proportional
relationship and prove algebraic statements.

Two or more numbers are said to be
in proportion if the ratios of the pairs of numbers are equal. For example, the numerators and
denominators of equivalent fractions are in proportion. Letβs imagine we are told that the
ratio seven to 14 is the same as the ratio 21 to π₯. One way to find the value of π₯ is
to calculate the proportionality coefficient of the first ratio. This value π is equal to 14
divided by seven, which is equal to two. The proportionality coefficient
must be the same for the second ratio. So π₯ over 21 is equal to two. Multiplying through by 21, we see
that π₯ is equal to 42. The ratio seven to 14 is equal to
the ratio 21 to 42. Letβs now consider a more formal
definition of this.

If the ratio π to π is the same
as the ratio π to π, then we say that π, π, π, and π are proportional. In particular, since π over π is
equal to π over π, we have ππ is equal to ππ. The values π, π, π, and π are
called the terms of the proportion. And we label them as the first
proportional, the second proportional, third proportional, and fourth
proportional. The outer terms π and π are
called the extremes, and the inner terms π and π are the means. This means that the equation ππ
is equal to ππ can be thought of as the product of the extremes is equal to the
product of the means. We will now look at an example
where we need to use this property.

If eight and three are in the same
ratio as 96 and π₯, then find the value of π₯.

We begin by recalling that if the
ratios of two pairs of numbers are the same, then their proportions are the
same. This means that the quotient of
each pair of numbers are equal. Eight over three is equal to 96
over π₯. We can multiply both sides of this
equation by three π₯. The left-hand side becomes eight
π₯. And the right-hand side is 96
multiplied by three, which is 288. Our final step to calculate the
value of π₯ is to divide through by eight. And this is equal to 36. If eight and three are in the same
ratio as 96 and π₯, the value of π₯ is 36.

It is worth noting that in
questions like this, we can quote the following result. If π over π is equal to π over
π, then ππ is equal to ππ. The product of eight and π₯ must be
equal to the product of three and 96.

We will now move on to consider a
list of three or four terms in continued proportion. A list of terms is said to be in
continued proportion if the ratio between successive terms is constant. Any number of quantities can be in
continued proportion. However, for the purposes of this
video, we will deal with three or four terms.

If we consider three terms π, π,
and π that are in continued proportion, then π over π is equal to π over π. This also means that ππ is equal
to π squared, where the middle term π is called the middle proportion or mean and
π and π are known as the extremes. If four terms π, π, π, and π
are in continued proportion, then π over π is equal to π over π, which is equal
to π over π, where π and π are known as the extremes and π and π are the
means. We can also label these as the
first, second, third, and fourth proportionals.

Letβs now consider one example of
each type.

If π is the middle proportion
between π and π, then which of the following is equal to π squared plus π
squared over π squared plus π squared? Is it (A) π over π, option (B) π
over π, option (C) two π over π, or option (D) two π over π?

We begin by recalling that if three
numbers are in continued proportion, where π is the middle proportion between π
and π, then π over π is equal to π over π. Cross multiplying, this means that
ππ is equal to π squared. The expression that we need to
simplify in this question is π squared plus π squared over π squared plus π
squared. We will begin by replacing π
squared with ππ. On the numerator, we have π
squared plus ππ, and on the denominator, ππ plus π squared.

Our next step is to factor out a
common factor of π on the numerator and a common factor of π on the
denominator. This leaves us with π multiplied
by π plus π over π multiplied by π plus π. As addition is commutative, we can
cancel a common factor of π plus π. And this leaves us with a
simplified expression of π over π. The correct answer is therefore
option (A). If π is the middle proportion
between π and π, then π squared plus π squared over π squared plus π squared
is equal to π over π.

In our next example, we will use
the properties of four numbers in proportion.

If π, π, π, and π are
proportional, which of the following equals the square root of six π squared minus
nine π squared over six π squared minus nine π squared? Is it option (A) π over π, option
(B) π over π, option (C) π over π, or option (D) π over π?

We begin by recalling that saying
that π, π, π, and π are proportional is equivalent to saying that the ratio of
π to π is equal to the ratio of π to π. In particular, their coefficients
of proportionality will be equivalent. So π is equal to π multiplied by
π, and π is equal to π multiplied by π for some constant π. We will begin by substituting these
expressions for π and π. This gives us the square root of
six multiplied by ππ all squared minus nine π squared over six multiplied by ππ
squared minus nine π squared.

The numerator of the fraction under
the square root can be rewritten as six π squared π squared minus nine π
squared. And the denominator is equal to six
π squared π squared minus nine π squared. Next, we can factor out π squared
from the numerator and π squared from the denominator. And after doing this, we can cancel
a shared factor of six π squared minus nine. Our expression simplifies to the
square root of π squared over π squared. Recalling that we can take the
square root of the numerator and denominator separately, this simplifies to π over
π.

At this stage, we note that this is
not one of the four options. We will therefore consider the two
proportionality equations we wrote earlier. Dividing these, we see that π over
π is equal to ππ over ππ. Since the shared factor π is a
nonzero constant, we can cancel this, leaving us with ππ is equal to ππ. We can therefore conclude that the
correct answer is option (B). If π, π, π, and π are
proportional, then the square root of six π squared minus nine π squared over six
π squared minus nine π squared is equal to π over π.

Before looking at one final
example, we will consider one further property of proportionality. The proportionality of the sum
states that if π, π, π, and π are proportional, then π over π is equal to π
over π, which is equal to π plus π over π plus π. We can simply add the numerators
and denominators of equivalent fractions separately without affecting their
value. Letβs now look at an example of
this in action.

If π over seven is equal to π
over four which is equal to π over 14 which is equal to six π minus seven π plus
two π over three π₯, find the value of π₯.

In order to answer this question,
we firstly recall that if four numbers π€, π₯, π¦, and π§ are proportional, then π€
over π₯ is equal to π¦ over π§ which is equal to π€ plus π¦ over π₯ plus π§. In this question, we are given
three equivalent fractions: π over seven, π over four, and π over 14. And we are asked to determine an
unknown in the fourth.

Unfortunately, we cannot apply the
property directly, as we get the equation shown, which we cannot solve for π₯. Instead, we will find equivalent
fractions for π over seven, π over four, and π over 14, so their numerators match
each term on the right-hand side. Multiplying the numerator and
denominator of the first fraction by six, we see that π over seven is equal to six
π over 42. In a similar way, π over four is
equivalent to negative seven π over negative 28. And π over 14 is equal to two π
over 28.

We now have six π over 42 is equal
to negative seven π over negative 28 which is equal to two π over 28 which is
equal to six π minus seven π plus two π over three π₯. Applying the property to the first
three fractions, we have six π minus seven π plus two π over 42 minus 28 plus
28. And we know this is equal to six π
minus seven π plus two π over three π₯. Since the numerators of both sides
of the equation are equal, their denominators must be equal. So we have three π₯ is equal to 42
minus 28 plus 28. This simplifies to three π₯ is
equal to 42. And dividing through by three, we
have π₯ is equal to 14.

If π over seven is equal to π
over four which is equal to π over 14 which is equal to six π minus seven π plus
two π over three π₯, then the value of π₯ is 14.

We will now summarize the key
points from this video. If π¦ is directly proportional to
π₯, then π¦ is equal to π multiplied by π₯, where the constant π is known as the
coefficient of proportionality. If two ratios π to π and π to π
are equal, then π over π is equal to π over π. And cross multiplying, π
multiplied by π is equal to π multiplied by π. Another way of writing this is as
follows. If π, π, π, and π are
proportional, then ππ is equal to ππ. And we also saw that π over π is
equal to π over π which is equal to π plus π over π plus π. We also saw that if three terms π,
π, and π are in continued proportion, then π over π is equal to π over π,
which means that ππ is equal to π squared, noting that terms are said to be in
continued proportion if the ratio between successive terms is constant.