Algebra

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Inverse Proportion

Inversely proportion is where one value increases, and the other decreases, at the same rate.

If two quantities are inversely proportional, then the rate at which one increases is the same as the rate at which the other decreases. For example, we would expect that as the temperature increases, sales of winter coats decrease.

We call a relationship between two quantities inversely proportional if the rate at which they change is the same. For example, if the temperature doubles, to be inversely proportional, the sales of coats has to halve.

Likewise, if the sales of coats double, the temperature has to:

If we call the number of coats $x$ and the temperature $y$ , we can describe the relationship using algebra.

Algebraically, if $y$ is inversely proportional to $x$ :

$y = \dfrac{k}{x}$

where $k$ is the constant of proportionality, representing the ratio between the proportional values.

Let's let $y=10$ and $x =100$

$y=\dfrac{k}{x}$

What would be the value of $k$ ?

As $k=1000$ , what is the value of $y$ when we double $x$ to be $200$ ?

Graphically, when y is proportional to $x$ , it is a straight line relationship, but inverse proportion is a non-linear relationship.

If two quantities are inversely proportional, then...

We can also use these ideas to solve algebraic problems.

For example, if $y$ is inversely proportional to $x$ and $y = 1$ when $x = 5$ , what is $k$ ? $y=\dfrac{k}{x}$

Rearrange the equation to make $k$ the subject

Since $y=\dfrac{k}{x}$ , we can multiply both sides by $x$ to find that $k=xy$ .

Put the values we know into $k=xy$

$5 \times 1 = 5$ . Therefore, we know that $k=5$ .

Form an equation

Use the value we just calculated in place of $k$ --> $y = \dfrac{5}{x}$

We can apply this to all values of $x$ and $y$

$y$ is always equal to $5 \div x$

If $y=\dfrac{5}{x}$ what is $x$ when $y= 5$ ?

$y=\dfrac{k}{x}$

If $y=20$ when $x = 4$ , what is $k$ ?

$y$ is inversely proportional to $x^3$ . When $x = 3$ , $y = 10$ . What is $y$ when $x = 10$ ?

We know they are inversely proportional

Therefore, we know that $y=\dfrac{k}{x^3}$

Input the values we know into the equation

In the question, we can see that when $x=3$ , $y=10$ . We can put these values into the equation: $10=\dfrac{k}{3^3}$ , so $10=\dfrac{k}{27}$ .

To find $k$ , rearrange the equation

To find $y$ when $x=10$ , we need the constant, $k$ . To find this, we can rearrange the equation we derived by multiplying both sides by $27$ . Therefore, since $10=\dfrac{k}{27}$ , $k=10 \times 27$ , so $k=270$ .

We can apply this to find $y$ when $x=10$

Since they are inversely proportional, $y=\dfrac{k}{x}$ . Now we know $k$ , we can fill in the equation as $y=\dfrac{270}{10}$ , so $y=27$

$y=\dfrac{k}{x}$

If $y=16$ when $x = 10$ , what is $k$ ?