Math / Algebra

Question(10) $\frac{2}{x^{2}-x}=\frac{1}{x-1}$

Studdy Solution

STEP 1

What is this asking? We need to find the value of $x$ that makes this equation true! Watch out! Be careful not to divide by zero.
We'll need to check if any solutions make the denominators zero.

STEP 2

1. Simplify the equation
2. Solve for $x$
3. Check for extraneous solutions

STEP 3

Let's factor the denominator of the first term: $x^2 - x = x \cdot (x-1)$ .
So, our equation becomes $\frac{2}{x \cdot (x-1)} = \frac{1}{x-1}.$

STEP 4

To get rid of those pesky fractions, we'll multiply both sides of the equation by $x \cdot (x-1)$ .
This gives us $x \cdot (x-1) \cdot \frac{2}{x \cdot (x-1)} = x \cdot (x-1) \cdot \frac{1}{x-1}.$

STEP 5

Now we can simplify by dividing to one.
On the left side, both $x$ and $(x-1)$ divide to one, leaving us with just $2$ .
On the right side, $(x-1)$ divides to one, leaving us with $x$ .
So our simplified equation is $2 = x.$

STEP 6

Let's plug our potential solution $x = \mathbf{2}$ back into the original equation.
The denominators are $x^2 - x$ and $x - 1$ . If $x = 2$ , then $x^2 - x = 2^2 - 2 = 4 - 2 = 2$ , and $x - 1 = 2 - 1 = 1$ .
Neither of these is zero, so $x = \mathbf{2}$ is a valid solution!

STEP 7

Our solution is $x = \mathbf{2}$ !

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Question(10) $\frac{2}{x^{2}-x}=\frac{1}{x-1}$

Studdy Solution

STEP 1

What is this asking? We need to find the value of $x$ that makes this equation true! Watch out! Be careful not to divide by zero.
We'll need to check if any solutions make the denominators zero.

STEP 2

1. Simplify the equation
2. Solve for $x$
3. Check for extraneous solutions

STEP 3

Let's factor the denominator of the first term: $x^2 - x = x \cdot (x-1)$ .
So, our equation becomes $\frac{2}{x \cdot (x-1)} = \frac{1}{x-1}.$

STEP 4

To get rid of those pesky fractions, we'll multiply both sides of the equation by $x \cdot (x-1)$ .
This gives us $x \cdot (x-1) \cdot \frac{2}{x \cdot (x-1)} = x \cdot (x-1) \cdot \frac{1}{x-1}.$

STEP 5

Now we can simplify by dividing to one.
On the left side, both $x$ and $(x-1)$ divide to one, leaving us with just $2$ .
On the right side, $(x-1)$ divides to one, leaving us with $x$ .
So our simplified equation is $2 = x.$

STEP 6

Let's plug our potential solution $x = \mathbf{2}$ back into the original equation.
The denominators are $x^2 - x$ and $x - 1$ . If $x = 2$ , then $x^2 - x = 2^2 - 2 = 4 - 2 = 2$ , and $x - 1 = 2 - 1 = 1$ .
Neither of these is zero, so $x = \mathbf{2}$ is a valid solution!

STEP 7

Our solution is $x = \mathbf{2}$ !

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Question(10) 2x2−x=1x−1\frac{2}{x^{2}-x}=\frac{1}{x-1}x2−x2​=x−11​

Studdy Solution

STEP 1

What is this asking? We need to find the value of xxx that makes this equation true! Watch out! Be careful not to divide by zero.We'll need to check if any solutions make the denominators zero.

STEP 2

1. Simplify the equation2. Solve for xxx3. Check for extraneous solutions

STEP 3

Let's **factor** the denominator of the first term: x2−x=x⋅(x−1)x^2 - x = x \cdot (x-1)x2−x=x⋅(x−1).So, our equation becomes 2x⋅(x−1)=1x−1.\frac{2}{x \cdot (x-1)} = \frac{1}{x-1}.x⋅(x−1)2​=x−11​.

STEP 4

STEP 5

Now we can **simplify** by dividing to one.On the left side, both xxx and (x−1)(x-1)(x−1) divide to one, leaving us with just 222.On the right side, (x−1)(x-1)(x−1) divides to one, leaving us with xxx.So our simplified equation is 2=x.2 = x.2=x.

STEP 6

STEP 7

Our solution is x=2x = \mathbf{2}x=2!

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Question(10) $\frac{2}{x^{2}-x}=\frac{1}{x-1}$

What is this asking? We need to find the value of $x$ that makes this equation true! Watch out! Be careful not to divide by zero.
We'll need to check if any solutions make the denominators zero.

1. Simplify the equation
2. Solve for $x$
3. Check for extraneous solutions

Let's factor the denominator of the first term: $x^2 - x = x \cdot (x-1)$ .
So, our equation becomes $\frac{2}{x \cdot (x-1)} = \frac{1}{x-1}.$

Now we can simplify by dividing to one.
On the left side, both $x$ and $(x-1)$ divide to one, leaving us with just $2$ .
On the right side, $(x-1)$ divides to one, leaving us with $x$ .
So our simplified equation is $2 = x.$

Our solution is $x = \mathbf{2}$ !