Solved on Sep 27, 2023

Solve the rational equation $\frac{16}{8x-8}+\frac{1}{8}=\frac{2}{x-1}$ and find the valid solutions.

STEP 1

Assumptions1. The equation is $\frac{16}{8x-8}+\frac{1}{8}=\frac{}{x-1}$ . We need to find the restrictions on the variable $x$ and solve the equation.

STEP 2

First, we need to find the restrictions on the variable $x$ . The restrictions are the values of $x$ that would make the denominator of any fraction in the equation equal to zero, because division by zero is undefined.

STEP 3

The denominators in the equation are $8x-8$ and $x-1$ . So, we need to find the values of $x$ that would make these equal to zero.

STEP 4

Set the first denominator $8x-8$ equal to zero and solve for $x$ .
$8x-8=0$

STEP 5

implify the equation to find the restriction on $x$ .
$x=1$

STEP 6

Set the second denominator $x-1$ equal to zero and solve for $x$ .
$x-1=0$

STEP 7

implify the equation to find the restriction on $x$ .
$x=1$

STEP 8

So, the restriction on the variable $x$ is $x \neq1$ .

STEP 9

Now, we need to solve the equation. To do this, we first need to get rid of the fractions by multiplying each term by the least common denominator (LCD) of all the denominators. The LCD of $8x-8$ and $x-$ is $8x-8$ .

STEP 10

Multiply each term by the LCD.
$(8x-8)\left(\frac{16}{8x-8}\right)+(8x-8)\left(\frac{}{8}\right)=(8x-8)\left(\frac{2}{x-}\right)$

STEP 11

implify the equation.
$16+(x-)=16$

STEP 12

olve the equation for $x$ .
$x=$

STEP 13

However, $x=$ is a restriction on the variable $x$ , so the solution set is empty, denoted by $\varnothing$ .
So, the answer is C. $x \neq ; \varnothing$ .

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Solve the rational equation $\frac{16}{8x-8}+\frac{1}{8}=\frac{2}{x-1}$ and find the valid solutions.

STEP 1

Assumptions1. The equation is $\frac{16}{8x-8}+\frac{1}{8}=\frac{}{x-1}$ . We need to find the restrictions on the variable $x$ and solve the equation.

STEP 2

First, we need to find the restrictions on the variable $x$ . The restrictions are the values of $x$ that would make the denominator of any fraction in the equation equal to zero, because division by zero is undefined.

STEP 3

The denominators in the equation are $8x-8$ and $x-1$ . So, we need to find the values of $x$ that would make these equal to zero.

STEP 4

Set the first denominator $8x-8$ equal to zero and solve for $x$ .
$8x-8=0$

STEP 5

implify the equation to find the restriction on $x$ .
$x=1$

STEP 6

Set the second denominator $x-1$ equal to zero and solve for $x$ .
$x-1=0$

STEP 7

implify the equation to find the restriction on $x$ .
$x=1$

STEP 8

So, the restriction on the variable $x$ is $x \neq1$ .

STEP 9

Now, we need to solve the equation. To do this, we first need to get rid of the fractions by multiplying each term by the least common denominator (LCD) of all the denominators. The LCD of $8x-8$ and $x-$ is $8x-8$ .

STEP 10

Multiply each term by the LCD.
$(8x-8)\left(\frac{16}{8x-8}\right)+(8x-8)\left(\frac{}{8}\right)=(8x-8)\left(\frac{2}{x-}\right)$

STEP 11

implify the equation.
$16+(x-)=16$

STEP 12

olve the equation for $x$ .
$x=$

STEP 13

However, $x=$ is a restriction on the variable $x$ , so the solution set is empty, denoted by $\varnothing$ .
So, the answer is C. $x \neq ; \varnothing$ .

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Solve the rational equation 168x−8+18=2x−1\frac{16}{8x-8}+\frac{1}{8}=\frac{2}{x-1}8x−816​+81​=x−12​ and find the valid solutions.

STEP 1

Assumptions1. The equation is 168x−8+18=x−1\frac{16}{8x-8}+\frac{1}{8}=\frac{}{x-1}8x−816​+81​=x−1​ . We need to find the restrictions on the variable xxx and solve the equation.

STEP 2

First, we need to find the restrictions on the variable xxx. The restrictions are the values of xxx that would make the denominator of any fraction in the equation equal to zero, because division by zero is undefined.

STEP 3

The denominators in the equation are 8x−88x-88x−8 and x−1x-1x−1. So, we need to find the values of xxx that would make these equal to zero.

STEP 4

Set the first denominator 8x−88x-88x−8 equal to zero and solve for xxx.8x−8=08x-8=08x−8=0

STEP 5

implify the equation to find the restriction on xxx.x=1x=1x=1

STEP 6

Set the second denominator x−1x-1x−1 equal to zero and solve for xxx.x−1=0x-1=0x−1=0

STEP 7

implify the equation to find the restriction on xxx.x=1x=1x=1

STEP 8

So, the restriction on the variable xxx is x≠1x \neq1x=1.

STEP 9

Now, we need to solve the equation. To do this, we first need to get rid of the fractions by multiplying each term by the least common denominator (LCD) of all the denominators. The LCD of 8x−88x-88x−8 and x−x-x− is 8x−88x-88x−8.

STEP 10

Multiply each term by the LCD.(8x−8)(168x−8)+(8x−8)(8)=(8x−8)(2x−)(8x-8)\left(\frac{16}{8x-8}\right)+(8x-8)\left(\frac{}{8}\right)=(8x-8)\left(\frac{2}{x-}\right)(8x−8)(8x−816​)+(8x−8)(8​)=(8x−8)(x−2​)

STEP 11

implify the equation.16+(x−)=1616+(x-)=1616+(x−)=16

STEP 12

olve the equation for xxx.x=x=x=

STEP 13

However, x=x=x= is a restriction on the variable xxx, so the solution set is empty, denoted by ∅\varnothing∅.So, the answer is C. x≠;∅x \neq ; \varnothingx=;∅.

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Solve the rational equation $\frac{16}{8x-8}+\frac{1}{8}=\frac{2}{x-1}$ and find the valid solutions.

Assumptions1. The equation is $\frac{16}{8x-8}+\frac{1}{8}=\frac{}{x-1}$ . We need to find the restrictions on the variable $x$ and solve the equation.

First, we need to find the restrictions on the variable $x$ . The restrictions are the values of $x$ that would make the denominator of any fraction in the equation equal to zero, because division by zero is undefined.

The denominators in the equation are $8x-8$ and $x-1$ . So, we need to find the values of $x$ that would make these equal to zero.

Set the first denominator $8x-8$ equal to zero and solve for $x$ .
$8x-8=0$

implify the equation to find the restriction on $x$ .
$x=1$

Set the second denominator $x-1$ equal to zero and solve for $x$ .
$x-1=0$

implify the equation to find the restriction on $x$ .
$x=1$

So, the restriction on the variable $x$ is $x \neq1$ .

Now, we need to solve the equation. To do this, we first need to get rid of the fractions by multiplying each term by the least common denominator (LCD) of all the denominators. The LCD of $8x-8$ and $x-$ is $8x-8$ .

Multiply each term by the LCD.
$(8x-8)\left(\frac{16}{8x-8}\right)+(8x-8)\left(\frac{}{8}\right)=(8x-8)\left(\frac{2}{x-}\right)$

implify the equation.
$16+(x-)=16$

olve the equation for $x$ .
$x=$

However, $x=$ is a restriction on the variable $x$ , so the solution set is empty, denoted by $\varnothing$ .
So, the answer is C. $x \neq ; \varnothing$ .