if y varies inversely as x and y=4 when x=12, find y when x=2

1) As

y

varies inversely as

x

, a constant

k

exists such that:

y = \frac{k}{x}

With

y = 4

when

x = 12

, we get:

4 = \frac{k}{12}

To find

y

when

x = 2

, let's solve the system:

{y = \frac{k}{x}, 4 = \frac{k}{12}, x = 2}

in terms of

y

⇥1.1) You can rearrange the equations to improve solving:

⇥ [1]:

4 = \frac{k}{12}

⇥ [2]:

x = 2

⇥ [3]:

y = \frac{k}{x}

⇥1.2) Let's solve in

ℝ

{4 = \frac{k}{12}, x = 2, y = \frac{k}{x}}

⇥1.3) Solving for k:

4 = \frac{k}{12}

[1]

⇥

\frac{k}{12} = 4

⇥

k = 4 \cdot 12

⇥ⅰ

4 \cdot 12

4 \cdot (10 + 2)

4 \cdot 10 + 4 \cdot 2

40 + 8

48

⇥Solution found [1]:

k = 48

⇥1.4) Solving for x:

x = 2

[2]

⇥Solution found [2]:

x = 2

⇥1.5) Solving for y:

y = \frac{k}{x}

[3] using

k = 48

and

x = 2

⇥ⓘ Simplify the fraction by dividing the numerator and the denominator by

2

, their greatest common divisor:

\frac{48}{2}

24

⇥Solution found [3]:

y = 24

▶So when

x = 2

y

is:

24

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