Ask questions, share knowledge, and connect with a vibrant community on IDNLearn.com. Find accurate and detailed answers to your questions from our experienced and dedicated community members.
Sagot :
To determine the least common denominator (LCD) and write the sum of the numerators for the given problem:
[tex]\[ \frac{2}{x^2 - 3x - 4} + \frac{3}{x^2 - 6x + 8} \][/tex]
we need to follow these steps:
### Step 1: Factor Each Denominator
#### Factor [tex]\( x^2 - 3x - 4 \)[/tex]
Let's factorize [tex]\(x^2 - 3x - 4\)[/tex]:
We need to find two numbers that multiply to [tex]\(-4\)[/tex] and add up to [tex]\(-3\)[/tex]. These numbers are [tex]\(-4\)[/tex] and [tex]\(1\)[/tex].
So we can write:
[tex]\[ x^2 - 3x - 4 = (x - 4)(x + 1) \][/tex]
#### Factor [tex]\( x^2 - 6x + 8 \)[/tex]
Now, let's factorize [tex]\(x^2 - 6x + 8\)[/tex]:
We need to find two numbers that multiply to [tex]\(8\)[/tex] and add up to [tex]\(-6\)[/tex]. These numbers are [tex]\(-4\)[/tex] and [tex]\(-2\)[/tex].
So we can write:
[tex]\[ x^2 - 6x + 8 = (x - 4)(x - 2) \][/tex]
### Step 2: Find the Least Common Denominator (LCD)
To find the LCD, we need to collect all unique factors from both denominators:
From the denominator [tex]\(x^2 - 3x - 4 = (x - 4)(x + 1)\)[/tex], the factors are [tex]\(x - 4\)[/tex] and [tex]\(x + 1\)[/tex].
From the denominator [tex]\(x^2 - 6x + 8 = (x - 4)(x - 2)\)[/tex], the factors are [tex]\(x - 4\)[/tex] and [tex]\(x - 2\)[/tex].
Combining all unique factors, we get:
[tex]\[ (x - 4)(x + 1)(x - 2) \][/tex]
### Step 3: Write the LCD
Thus, the least common denominator (LCD) is:
[tex]\[ \boxed{(x - 4)(x + 1)(x - 2)} \][/tex]
### Rewrite Each Fraction with the LCD as the Denominator
Let's rewrite each fraction so they have the LCD as their denominator.
#### For [tex]\( \frac{2}{x^2 - 3x - 4} \)[/tex]:
We have:
[tex]\[ \frac{2}{x^2 - 3x - 4} = \frac{2}{(x - 4)(x + 1)} \][/tex]
To make the denominator [tex]\((x-4)(x+1)(x-2)\)[/tex], we multiply the numerator and the denominator by [tex]\((x - 2)\)[/tex]:
[tex]\[ \frac{2}{(x - 4)(x + 1)} \cdot \frac{(x - 2)}{(x - 2)} = \frac{2(x - 2)}{(x - 4)(x + 1)(x - 2)} \][/tex]
#### For [tex]\( \frac{3}{x^2 - 6x + 8} \)[/tex]:
We have:
[tex]\[ \frac{3}{x^2 - 6x + 8} = \frac{3}{(x - 4)(x - 2)} \][/tex]
To make the denominator [tex]\((x-4)(x+1)(x-2)\)[/tex], we multiply the numerator and the denominator by [tex]\((x + 1)\)[/tex]:
[tex]\[ \frac{3}{(x - 4)(x - 2)} \cdot \frac{(x + 1)}{(x + 1)} = \frac{3(x + 1)}{(x - 4)(x + 1)(x - 2)} \][/tex]
### Step 4: Write the Sum of the Numerators
Now, we can add the fractions:
[tex]\[ \frac{2(x - 2)}{(x - 4)(x + 1)(x - 2)} + \frac{3(x + 1)}{(x - 4)(x + 1)(x - 2)} = \frac{2(x - 2) + 3(x + 1)}{(x - 4)(x + 1)(x - 2)} \][/tex]
Simplify the numerator:
[tex]\[ 2(x - 2) + 3(x + 1) = 2x - 4 + 3x + 3 = 5x - 1 \][/tex]
So the sum of the fractions is:
[tex]\[ \frac{5x - 1}{(x - 4)(x + 1)(x - 2)} \][/tex]
### Final Answer
The least common denominator is:
[tex]\[ (x - 4)(x + 1)(x - 2) \][/tex]
And the sum of the numerators is:
[tex]\[ 5x - 1 \][/tex]
Therefore, the combined fraction is:
[tex]\[ \frac{5x - 1}{(x - 4)(x + 1)(x - 2)} \][/tex]
[tex]\[ \frac{2}{x^2 - 3x - 4} + \frac{3}{x^2 - 6x + 8} \][/tex]
we need to follow these steps:
### Step 1: Factor Each Denominator
#### Factor [tex]\( x^2 - 3x - 4 \)[/tex]
Let's factorize [tex]\(x^2 - 3x - 4\)[/tex]:
We need to find two numbers that multiply to [tex]\(-4\)[/tex] and add up to [tex]\(-3\)[/tex]. These numbers are [tex]\(-4\)[/tex] and [tex]\(1\)[/tex].
So we can write:
[tex]\[ x^2 - 3x - 4 = (x - 4)(x + 1) \][/tex]
#### Factor [tex]\( x^2 - 6x + 8 \)[/tex]
Now, let's factorize [tex]\(x^2 - 6x + 8\)[/tex]:
We need to find two numbers that multiply to [tex]\(8\)[/tex] and add up to [tex]\(-6\)[/tex]. These numbers are [tex]\(-4\)[/tex] and [tex]\(-2\)[/tex].
So we can write:
[tex]\[ x^2 - 6x + 8 = (x - 4)(x - 2) \][/tex]
### Step 2: Find the Least Common Denominator (LCD)
To find the LCD, we need to collect all unique factors from both denominators:
From the denominator [tex]\(x^2 - 3x - 4 = (x - 4)(x + 1)\)[/tex], the factors are [tex]\(x - 4\)[/tex] and [tex]\(x + 1\)[/tex].
From the denominator [tex]\(x^2 - 6x + 8 = (x - 4)(x - 2)\)[/tex], the factors are [tex]\(x - 4\)[/tex] and [tex]\(x - 2\)[/tex].
Combining all unique factors, we get:
[tex]\[ (x - 4)(x + 1)(x - 2) \][/tex]
### Step 3: Write the LCD
Thus, the least common denominator (LCD) is:
[tex]\[ \boxed{(x - 4)(x + 1)(x - 2)} \][/tex]
### Rewrite Each Fraction with the LCD as the Denominator
Let's rewrite each fraction so they have the LCD as their denominator.
#### For [tex]\( \frac{2}{x^2 - 3x - 4} \)[/tex]:
We have:
[tex]\[ \frac{2}{x^2 - 3x - 4} = \frac{2}{(x - 4)(x + 1)} \][/tex]
To make the denominator [tex]\((x-4)(x+1)(x-2)\)[/tex], we multiply the numerator and the denominator by [tex]\((x - 2)\)[/tex]:
[tex]\[ \frac{2}{(x - 4)(x + 1)} \cdot \frac{(x - 2)}{(x - 2)} = \frac{2(x - 2)}{(x - 4)(x + 1)(x - 2)} \][/tex]
#### For [tex]\( \frac{3}{x^2 - 6x + 8} \)[/tex]:
We have:
[tex]\[ \frac{3}{x^2 - 6x + 8} = \frac{3}{(x - 4)(x - 2)} \][/tex]
To make the denominator [tex]\((x-4)(x+1)(x-2)\)[/tex], we multiply the numerator and the denominator by [tex]\((x + 1)\)[/tex]:
[tex]\[ \frac{3}{(x - 4)(x - 2)} \cdot \frac{(x + 1)}{(x + 1)} = \frac{3(x + 1)}{(x - 4)(x + 1)(x - 2)} \][/tex]
### Step 4: Write the Sum of the Numerators
Now, we can add the fractions:
[tex]\[ \frac{2(x - 2)}{(x - 4)(x + 1)(x - 2)} + \frac{3(x + 1)}{(x - 4)(x + 1)(x - 2)} = \frac{2(x - 2) + 3(x + 1)}{(x - 4)(x + 1)(x - 2)} \][/tex]
Simplify the numerator:
[tex]\[ 2(x - 2) + 3(x + 1) = 2x - 4 + 3x + 3 = 5x - 1 \][/tex]
So the sum of the fractions is:
[tex]\[ \frac{5x - 1}{(x - 4)(x + 1)(x - 2)} \][/tex]
### Final Answer
The least common denominator is:
[tex]\[ (x - 4)(x + 1)(x - 2) \][/tex]
And the sum of the numerators is:
[tex]\[ 5x - 1 \][/tex]
Therefore, the combined fraction is:
[tex]\[ \frac{5x - 1}{(x - 4)(x + 1)(x - 2)} \][/tex]
Your presence in our community is highly appreciated. Keep sharing your insights and solutions. Together, we can build a rich and valuable knowledge resource for everyone. Find clear answers at IDNLearn.com. Thanks for stopping by, and come back for more reliable solutions.
