Certainly! Let's solve the equation:
[tex]\[
\frac{c-1}{x} + d = c d
\][/tex]
Step-by-step:
1. Isolate the fraction term:
Move [tex]\( d \)[/tex] from the left side of the equation to the right side by subtracting [tex]\( d \)[/tex] from both sides:
[tex]\[
\frac{c-1}{x} = c d - d
\][/tex]
2. Factor out the common factor [tex]\( d \)[/tex] on the right side:
[tex]\[
c d - d = d (c - 1)
\][/tex]
So the equation now is:
[tex]\[
\frac{c-1}{x} = d (c - 1)
\][/tex]
3. Assuming [tex]\( c \neq 1 \)[/tex]:
If [tex]\( c = 1 \)[/tex], then the numerator of the fraction is 0, making the left side zero. But assuming [tex]\( c \neq 1 \)[/tex], we can proceed to isolate [tex]\( x \)[/tex].
4. Divide both sides of the equation by [tex]\( c-1 \)[/tex]:
Divide both sides by [tex]\( c-1 \)[/tex] to isolate [tex]\( \frac{1}{x} \)[/tex]:
[tex]\[
\frac{c-1}{x} \div (c-1) = [d (c - 1)] \div (c - 1)
\][/tex]
Simplifying, we get:
[tex]\[
\frac{1}{x} = d
\][/tex]
5. Solve for [tex]\( x \)[/tex]:
Invert both sides to solve for [tex]\( x \)[/tex]:
[tex]\[
x = \frac{1}{d}
\][/tex]
So, the solution to the equation
[tex]\[
\frac{c-1}{x} + d = c d
\][/tex]
is
[tex]\[
x = \frac{1}{d}
\][/tex]
