inverse of y=(3x-1)/(2x+8) — step-by-step solution

1) For a function

\frac{a x + b}{c x + d}

, two inputs u and v with the same output satisfy

(a d - b \cdot c) \cdot (u - v) = 0

after multiplying by the denominators. Here,

3 \cdot 8 - (-1 \cdot 2)

26

≠ 0, so u = v. The function therefore has an inverse on its range

Express input

x

in terms of output

y

y = \frac{3 x - 1}{2 x + 8}

⇥1.1) Let's solve in

ℝ

\frac{3 x - 1}{2 x + 8} = y

⇥1.2) Let's add fractions:

\frac{3 x - 1}{2 x + 8} - y = 0

⇥ⓘ To add fractions, put them over the same denominator, then add the numerators:

\frac{3 x - 1}{2 x + 8} - y

\frac{(3 x - 1) - y \cdot (2 x + 8)}{2 x + 8}

⇥

\frac{(3 x - 1) - y \cdot (2 x + 8)}{2 x + 8} = 0

⇥

\frac{3 x - 1 - y \cdot (2 x + 8)}{2 x + 8} = 0

⇥A quotient is zero if and only if its numerator is zero and its denominator is not: we solve the numerator equal to 0

⇥⇥1.2.1) Let's solve in

ℝ

3 x - 1 - y \cdot (2 x + 8) = 0

⇥⇥ⓘ Expand by multiplying each term inside the parentheses:

- y \cdot (2 x + 8)

- (y \cdot 2 x + y \cdot 8)

⇥⇥

3 x - 1 - (y \cdot 2 x + y \cdot 8) = 0

⇥⇥ⓘ A minus sign before a parenthesis changes the sign of each of its terms:

- (y \cdot 2 x + y \cdot 8)

- y \cdot 2 x - y \cdot 8

⇥⇥

3 x

- 1

- y \cdot 2 x

- y \cdot 8 = 0

⇥⇥

3 x - y \cdot 2 x = 1 + y \cdot 8

⇥⇥

(3 - y \cdot 2) \cdot x = 1 + y \cdot 8

⇥⇥ⓘ Both sides are divided by

(3 - y \cdot 2)

, which must be nonzero:

y \neq \frac{3}{2}

⇥⇥Solution obtained:

x = \frac{8 y + 1}{3 - 2 y}

⇥⇥1.2.2) To prevent the denominator from being zero, let's solve:

2 x + 8 \neq 0

⇥⇥

2 x \neq -8

⇥⇥ⓘ Both sides are divided by

2

⇥⇥

x \neq - \frac{8}{2}

⇥⇥ⓘ We simplify the fraction by dividing the numerator and the denominator by

2

, their greatest common divisor:

- \frac{8}{2}

-4

⇥⇥Domain condition:

x \neq -4

Renaming the output as

x

, the inverse is

\frac{8 x + 1}{3 - 2 x}

, with domain

x \neq \frac{3}{2}

, the condition found when isolating the input

▶Inverse function:

\frac{8 x + 1}{3 - 2 x}