range of x/(6x-5) — step-by-step solution

1) Let's find the real values taken by

\frac{x}{6 x - 5}

, for real

x

Let's find the domain of

\frac{x}{6 x - 5}

For the expression

\frac{x}{6 x - 5}

to be defined, we must have:

6 x - 5 \neq 0

6 x \neq 5

Solution obtained:

x \neq \frac{5}{6}

The input domain is

x \in ℝ \ {\frac{5}{6}}

Set

y = \frac{x}{6 x - 5}

and solve for

x

. The determinant

-5

is nonzero: the inverse is unique

2) Let's solve in

ℝ

\frac{x}{6 x - 5} = y

Let's sum the fractions:

\frac{x}{6 x - 5} - y = 0

\frac{x - y \cdot (6 x - 5)}{6 x - 5} = 0

3) Let's solve in

ℝ

x - y \cdot (6 x - 5) = 0

x - (y \cdot 6 x + y \cdot (-5)) = 0

x - y \cdot 6 x - y \cdot (-5) = 0

x - y \cdot 6 x + y \cdot 5 = 0

x - y \cdot 6 x = - y \cdot 5

(1 - y \cdot 6) \cdot x = - y \cdot 5

x = \frac{- y \cdot 5}{1 - y \cdot 6}

x = - \frac{y \cdot 5}{1 - y \cdot 6}

x = \frac{y \cdot 5}{- 1 + y \cdot 6}

Solution found:

x = \frac{5 y}{6 y - 1}

4) To avoid a zero denominator, let's solve:

6 x - 5 \neq 0

6 x \neq 5

Solution excluded:

x \neq \frac{5}{6}

Every value in the inverse's domain is attained, so its domain gives the range

5) Let's evaluate the domain of

\frac{5 y}{6 y - 1}

For the expression

\frac{5 y}{6 y - 1}

to exist, you must check:

6 y - 1 \neq 0

6 y \neq 1

Solution found:

y \neq \frac{1}{6}

▶Range:

y \in ℝ \ {\frac{1}{6}}