Math / Algebra

QuestionAdd. $\frac{7}{4x} + \frac{x-5}{6x^2}$
Select the correct choice below and fill in any answer boxes within your choice. (Simplify your answer. Use integers or fractions for any numbers in the expression. Use a comma to separate answers as needed.)
A. $\frac{7}{4x} + \frac{x-5}{6x^2} =$ $\quad$ , $x \ne$ $\quad$ B. $\frac{7}{4x} + \frac{x-5}{6x^2} =$ $\quad$ , no numbers must be excluded.

Studdy Solution

STEP 1

What is this asking? We need to add two fractions with variables and figure out which values of $x$ would make the denominator zero. Watch out! Don't forget to find the least common denominator and watch out for any values of $x$ that would make the denominator zero, because that's a no-no in math!

STEP 2

1. Find the Least Common Denominator
2. Add the Fractions
3. Simplify and Identify Excluded Values

STEP 3

Alright, let's find the least common denominator (LCD)!
We have $4x$ and $6x^2$ as our denominators.
The LCD is the smallest expression that both denominators divide into evenly.

STEP 4

Think of it like this: $4x$ is the same as $2 \cdot 2 \cdot x$ , and $6x^2$ is the same as $2 \cdot 3 \cdot x \cdot x$ .
We need to take the highest power of each factor present in either denominator.
So, we need two 2's, one 3, and two $x$ 's.
That gives us $2 \cdot 2 \cdot 3 \cdot x \cdot x = 12x^2$ .
Our LCD is $12x^2$ !

STEP 5

Now, we need to rewrite each fraction so it has the LCD of $12x^2$ as its denominator.
For the first fraction, $\frac{7}{4x}$ , we multiply the top and bottom by $3x$ because $4x \cdot 3x = 12x^2$ .
This gives us $\frac{7 \cdot 3x}{4x \cdot 3x} = \frac{21x}{12x^2}$ .

STEP 6

For the second fraction, $\frac{x-5}{6x^2}$ , we multiply the top and bottom by $2$ because $6x^2 \cdot 2 = 12x^2$ .
This gives us $\frac{(x-5) \cdot 2}{6x^2 \cdot 2} = \frac{2x-10}{12x^2}$ .

STEP 7

Now, we can add the fractions!
Since they have the same denominator, we can just add the numerators: $\frac{21x}{12x^2} + \frac{2x-10}{12x^2} = \frac{21x + 2x - 10}{12x^2}$

STEP 8

Let's simplify the numerator: $21x + 2x - 10 = 23x - 10$ .
So, our fraction becomes $\frac{23x - 10}{12x^2}$ .

STEP 9

Now, let's think about excluded values.
We can't have a zero in the denominator, so we need to find any values of $x$ that would make $12x^2$ equal to zero.
If $12x^2 = 0$ , then $x^2 = 0$ , which means $x = 0$ .
So, $x$ cannot be equal to zero.

STEP 10

So, our final answer is $\frac{23x - 10}{12x^2}$ , and $x \ne 0$ .
This matches choice A.

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Studdy Solution

STEP 1

What is this asking? We need to add two fractions with variables and figure out which values of $x$ would make the denominator zero. Watch out! Don't forget to find the least common denominator and watch out for any values of $x$ that would make the denominator zero, because that's a no-no in math!

STEP 2

1. Find the Least Common Denominator
2. Add the Fractions
3. Simplify and Identify Excluded Values

STEP 3

Alright, let's find the least common denominator (LCD)!
We have $4x$ and $6x^2$ as our denominators.
The LCD is the smallest expression that both denominators divide into evenly.

STEP 4

STEP 5

Now, we need to rewrite each fraction so it has the LCD of $12x^2$ as its denominator.
For the first fraction, $\frac{7}{4x}$ , we multiply the top and bottom by $3x$ because $4x \cdot 3x = 12x^2$ .
This gives us $\frac{7 \cdot 3x}{4x \cdot 3x} = \frac{21x}{12x^2}$ .

STEP 6

For the second fraction, $\frac{x-5}{6x^2}$ , we multiply the top and bottom by $2$ because $6x^2 \cdot 2 = 12x^2$ .
This gives us $\frac{(x-5) \cdot 2}{6x^2 \cdot 2} = \frac{2x-10}{12x^2}$ .

STEP 7

Now, we can add the fractions!
Since they have the same denominator, we can just add the numerators: $\frac{21x}{12x^2} + \frac{2x-10}{12x^2} = \frac{21x + 2x - 10}{12x^2}$

STEP 8

Let's simplify the numerator: $21x + 2x - 10 = 23x - 10$ .
So, our fraction becomes $\frac{23x - 10}{12x^2}$ .

STEP 9

Now, let's think about excluded values.
We can't have a zero in the denominator, so we need to find any values of $x$ that would make $12x^2$ equal to zero.
If $12x^2 = 0$ , then $x^2 = 0$ , which means $x = 0$ .
So, $x$ cannot be equal to zero.

STEP 10

So, our final answer is $\frac{23x - 10}{12x^2}$ , and $x \ne 0$ .
This matches choice A.

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Studdy Solution

STEP 1

STEP 2

1. Find the Least Common Denominator2. Add the Fractions3. Simplify and Identify Excluded Values

STEP 3

Alright, let's **find the least common denominator (LCD)**!We have 4x4x4x and 6x26x^26x2 as our denominators.The LCD is the smallest expression that both denominators divide into evenly.

STEP 4

STEP 5

STEP 6

For the second fraction, x−56x2\frac{x-5}{6x^2}6x2x−5​, we multiply the top and bottom by 222 because 6x2⋅2=12x26x^2 \cdot 2 = 12x^26x2⋅2=12x2.This gives us (x−5)⋅26x2⋅2=2x−1012x2\frac{(x-5) \cdot 2}{6x^2 \cdot 2} = \frac{2x-10}{12x^2}6x2⋅2(x−5)⋅2​=12x22x−10​.

STEP 7

Now, we can **add the fractions**!Since they have the same denominator, we can just add the numerators: 21x12x2+2x−1012x2=21x+2x−1012x2 \frac{21x}{12x^2} + \frac{2x-10}{12x^2} = \frac{21x + 2x - 10}{12x^2} 12x221x​+12x22x−10​=12x221x+2x−10​

STEP 8

Let's **simplify** the numerator: 21x+2x−10=23x−1021x + 2x - 10 = 23x - 1021x+2x−10=23x−10.So, our fraction becomes 23x−1012x2\frac{23x - 10}{12x^2}12x223x−10​.

STEP 9

Now, let's think about **excluded values**.We can't have a zero in the denominator, so we need to find any values of xxx that would make 12x212x^212x2 equal to zero.If 12x2=012x^2 = 012x2=0, then x2=0x^2 = 0x2=0, which means x=0x = 0x=0.So, xxx cannot be equal to **zero**.

STEP 10

So, our final answer is 23x−1012x2\frac{23x - 10}{12x^2}12x223x−10​, and x≠0x \ne 0x=0.This matches choice A.

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1. Find the Least Common Denominator
2. Add the Fractions
3. Simplify and Identify Excluded Values

Alright, let's find the least common denominator (LCD)!
We have $4x$ and $6x^2$ as our denominators.
The LCD is the smallest expression that both denominators divide into evenly.

For the second fraction, $\frac{x-5}{6x^2}$ , we multiply the top and bottom by $2$ because $6x^2 \cdot 2 = 12x^2$ .
This gives us $\frac{(x-5) \cdot 2}{6x^2 \cdot 2} = \frac{2x-10}{12x^2}$ .

Now, we can add the fractions!
Since they have the same denominator, we can just add the numerators: $\frac{21x}{12x^2} + \frac{2x-10}{12x^2} = \frac{21x + 2x - 10}{12x^2}$

Let's simplify the numerator: $21x + 2x - 10 = 23x - 10$ .
So, our fraction becomes $\frac{23x - 10}{12x^2}$ .

Now, let's think about excluded values.
We can't have a zero in the denominator, so we need to find any values of $x$ that would make $12x^2$ equal to zero.
If $12x^2 = 0$ , then $x^2 = 0$ , which means $x = 0$ .
So, $x$ cannot be equal to zero.

So, our final answer is $\frac{23x - 10}{12x^2}$ , and $x \ne 0$ .
This matches choice A.