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Inverse Proportion

An inversely proportional relationship is of the form $y=\frac{k}{x}$. It can be related to a rational function. For k=1, the function looks like this:

Let's discuss the case where x is positive. If x is positive, then:

As $x\rightarrow\infty$, $y\rightarrow0$ and vice versa. In words, this means as $x$ gets larger, $y$ gets smaller and vice versa.

Sometimes, they say $y$ is inversely proportional to $x^2$. Don't be scared of the $x^2$! This is simply $y=\frac{k}{x^2}$.

(Video) Inversely Proportional

by Meso Maths

Variables are inversely proportional when their product is always the same. That means that if the product is positive, $y$ will decrease as $x$ increases. If the product is negative, $y$ will increase as $x$ increases (notice that this is the opposite of direct proportionality).

Inversely proportional functions take the form:

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

In this equation, $k$ is the product of $x$ and $y$. In direct variation) equations, $k$ is the slope, but that is not the case with inversely proportional equations because they do not create a line. Instead they look like this:

Let't try an example problem. We are given the following information. $y$ is inversely proportional to $x$. When $y=2$, $x=3$. What is the value of $x$ when $y=5$? First, we set up the problem using the definition of inversely proportional. We know $y$ is inversely proportional to $x$. This means $$y=\frac{k}{x}$$

Now, we know when $y=2$, $x=3$. So, we can plug those values in to the equation and simplify. $$\begin{align} 2&=\frac{k}{3}\\ 2\color{blue}{\times{3}}&=\frac{k}{3}\color{blue}{\times{3}}\\ k&=6 \end{align}$$

Now, we want to find $x$ when $y=5$. So, we set up the the equation using the definition of inversely proportional again, but this time we know $k=6$. $$y=\frac{6}{x}$$ Now, we can plug in $5$ for $y$ and solve for $x$. $$\begin{align} y&=\frac{6}{x}\\ 5&=\frac{6}{x}\\ 5x&=6\\ x&=\frac{6}{5} \end{align}$$

Therefore, $x=\frac{6}{5}$. Now, you can try one on your own!

So by our proportionality equation, we know that:

$$t=k\left(\frac{1}{u}\right)$$

We also know that when $t=10$, $u=15$. In other words:

$$10 = k\left(\frac{1}{15}\right) 10 = \frac{k}{15}$$

We can use this to find $k$:

$$\begin{align} 10 &= \frac{k}{15}\\ 10 (15) & = k\\ 150 & = k \end{align}$$

Then, we can use $k$ and the value we know for $t$ to figure out $u$'s corresponding value:

$$\begin{align} 20 &= (150)\frac{1}{x}\\ 20 &= \frac{150}{x}\\ 20(x) &= 150\\ x &= \frac{150}{20}\\ x &= \frac{15}{2}\\ x &= 7\frac{1}{2}\\ \end{align}$$

Done!

(Video) Direct and Inverse Proportion

by HEGARTYMATHS

In this video, HEGARTYMATHS goes over the basics of solving problems involving direct proportionality.

If $y$ is directly proportional to $x$ we write

$$y=kx$$

In this equation, $k$ is some constant that links $y$ and $x$. When solving proportion problems, the first step is to usually solve for $k$. Once you have that, you can solve for any $x$ or $y$ value if you are given the other.

First, find $k$:

$$\begin{align} y &= kx\\\ \\ (320) &= k(80)\\ \frac{320}{80} &= k\\ \frac{4}{1} &= k\\ 4 &= k \end{align}$$

Next, plug $k$ and $y$ into the equation.

$$\begin{align} y &= kx\\\ \\ (100) &= (4)x\\ \frac{100}{4} &= x\\ 25 &= x \end{align}$$

Done!