# 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.

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