# JOINT VARIATION — MEANING, EXPLANATION AND EXAMPLES

Joint variation is a type of variation where one variable depends on two or more other variables. The relationship between the variables can come in many forms.

Let's illustrate some joint variation

1. If $x$ varies directly as $y$ and $z$, we write $x\alpha yz$

2. If $z$ varies directly as $a$ and inversely as the square of $b$, we write $z\alpha\frac{a}{b^2}$

3. If $z$ varies directly as the square of $y$ and inversely as the cube of $x$, we write $z\alpha\frac{y^2}{x^3}$

👉This post extends our series on variation.

#### Example 1

$x$ varies directly as $y$ and inversely as $z$ and $x=42$ when $y=7$, $z=3$.

Determine the equation connecting $x$, $y$ and $z$. Hence, find $x$ when $y=5$ and $z=9$

Solution:

$x\alpha\frac{y}{z}$

$x=\frac{ky}{z}$

$x=42$, $y=7$, $z=3$.

$42=\frac{k(7)}{3}$

$126=7k$

Divide both sides by $7$

$\frac{126}{7}=\frac{7k}{7}$

$18=k$

$k=18$

The equation connecting $x$, $y$ and $z$ is:

$x=\frac{18y}{z}$

Let's find $x$ when $y=5$ and $z=9$

$x=\frac{18(5)}{9}$

$x=10$

#### Example 2

If $a$ varies directly as $b$ and $c$, when $b=2$ and $c=3$, $a=30$. Find $a$ when $b=4$ and $c=6$

Solution:

$a\alpha bc$

$a=kbc$

$30=k(2)(3)$

$30=k6$

Divide both side by $6$

$\frac{30}{6}=\frac{k6}{6}$

$k=5$

$a=5bc$

Now, let's find $a$ when $b=4$ and $c=6$

$a=5(4)(6)$

$a=120$

#### Example 3

$x$ varies directly as $y$ and inversely as the square root of $z$, and $x=15$ when $y=20$ and $z=16$

1. Find the relation between $x$, $y$ and $z$

2. Find $z$ when $y=35$ and $x=21$

Solution:

$x\alpha\frac{y}{\sqrt{z}}$

$x=\frac{ky}{\sqrt{z}}$

$x=15$, $y=20$ and $z=16$

$15=\frac{k(20)}{\sqrt{16}}$

$15=\frac{k(20)}{4}$

$15\times4=k(20)$

$60=20k$

Divide both aide by $20$

$\frac{60}{20}=\frac{20k}{20}$f

$3=k$

$k=3$

The relation connecting $x$, $y$ and $z$.

$x=\frac{3y}{\sqrt{z}}$

Let's find $z$ when $y=35$ and $x=21$

$21=\frac{3(35)}{\sqrt{z}}$

$21=\frac{105}{\sqrt{z}}$

$21\sqrt{z}=105$

Divide both side by $21$

$\frac{21\sqrt{z}}{21}=\frac{105}{21}$

$\sqrt{z}=5$

To get rid of square root, we square both sides

$(\sqrt{z})^2=5^2$

$z=25$

#### Example 4

$a$ varies directly as $b$ and inversely as $c$, when $b=9$ and $c=2$, $a=27$

1. Find the formula connecting $a$, $b$ and $c$.

2. Find $a$ when $b=14$ and $c=12$

Solution:

$a\alpha\frac{b}{c}$

$a=\frac{kb}{c}$

$27=\frac{k(9)}{2}$

Cross multiply

$27\times2=9k$

$54=9k$

Divide both side by $9$

$\frac{54}{9}=\frac{9k}{9}$

$k=6$

Therefore, the formula connecting $a$, $b$ and $c$ is

$a=\frac{6b}{c}$

Let's solve for $a$ when $b=14$ and $c=12$

$a=\frac{6(14)}{12}$

$a=\frac{84}{12}$

$a=7$

Related post

#### Example 5

A variable($x$) varies directly as the cube root of $y$ and inversely as the square of $z$. Given that $x=\frac{1}{4}$ when $y=64$ and $z=2$. Find $z$ when $x=\frac{1}{12}$ and $y=27$

Solution

$x=\frac{k(\sqrt[3]{y}}{z^2}$

$x=\frac{1}{4}$, $y=64$, $z=2$.

$\frac{1}{4}=\frac{k\sqrt[3]{64}}{4}$

$\frac{1}{4}=\frac{k(4)}{4}$

$\frac{1}{4}=k$

$k=\frac{1}{4}$

$x=\frac{(\sqrt[3]{y}}{4z^2}$

Let's find $z$ when $x=\frac{1}{12}$ and $y=27$

$\frac{1}{12}=\frac{\sqrt[3]{27}}{4z^2}$

$\frac{1}{12}=\frac{3}{4z^2}$

Cross multiply

$1\times4z^2=12\times3$

$4z^2=36$

Divide both side by $4$

$\frac{4z^2}{4}=\frac{36}{4}$

$z^2=9$

To get rid of the square, we square root both sides

$\sqrt{z^2}=\sqrt{9}$

$z=3$.

That is all for joint variation. We are going to continue this series on variation when we will discuss partial variation.

"Connecting with like-minded people is one of the fastest ways to excel". Want to succeed in maths or any other subject, join our Telegram community

Help us grow our readership by sharing this post