Answer :
Sure, let's perform the indicated operations step-by-step:
We are given the expression:
[tex]\[ \frac{x+1}{3y} + \frac{x-2}{4y} - \frac{x+3}{6y} \][/tex]
First, let's find a common denominator for these fractions. The denominators are [tex]\(3y\)[/tex], [tex]\(4y\)[/tex], and [tex]\(6y\)[/tex]. The least common multiple (LCM) of these denominators is [tex]\(12y\)[/tex].
Next, let's rewrite each fraction with this common denominator:
1. [tex]\[ \frac{x+1}{3y} \rightarrow \frac{(x+1) \times 4}{3y \times 4} = \frac{4(x+1)}{12y} \][/tex]
2. [tex]\[ \frac{x-2}{4y} \rightarrow \frac{(x-2) \times 3}{4y \times 3} = \frac{3(x-2)}{12y} \][/tex]
3. [tex]\[ \frac{x+3}{6y} \rightarrow \frac{(x+3) \times 2}{6y \times 2} = \frac{2(x+3)}{12y} \][/tex]
Now we can rewrite the original expression with these new fractions:
[tex]\[ \frac{4(x+1)}{12y} + \frac{3(x-2)}{12y} - \frac{2(x+3)}{12y} \][/tex]
Since all terms have the common denominator [tex]\(12y\)[/tex], we can combine the numerators over this common denominator:
[tex]\[ \frac{4(x+1) + 3(x-2) - 2(x+3)}{12y} \][/tex]
Let's expand and combine the terms in the numerator:
[tex]\[ 4(x+1) + 3(x-2) - 2(x+3) \][/tex]
Expanding each term:
[tex]\[ 4x + 4 + 3x - 6 - 2x - 6 \][/tex]
Combining like terms:
[tex]\[ (4x + 3x - 2x) + (4 - 6 - 6) = 5x - 8 \][/tex]
So the expression simplifies to:
[tex]\[ \frac{5x - 8}{12y} \][/tex]
Therefore, the result of the indicated operations is:
[tex]\[ \frac{x-2}{4y} + \frac{x+1}{3y} - \frac{x+3}{6y} = \frac{5x-8}{12y} \][/tex]
We are given the expression:
[tex]\[ \frac{x+1}{3y} + \frac{x-2}{4y} - \frac{x+3}{6y} \][/tex]
First, let's find a common denominator for these fractions. The denominators are [tex]\(3y\)[/tex], [tex]\(4y\)[/tex], and [tex]\(6y\)[/tex]. The least common multiple (LCM) of these denominators is [tex]\(12y\)[/tex].
Next, let's rewrite each fraction with this common denominator:
1. [tex]\[ \frac{x+1}{3y} \rightarrow \frac{(x+1) \times 4}{3y \times 4} = \frac{4(x+1)}{12y} \][/tex]
2. [tex]\[ \frac{x-2}{4y} \rightarrow \frac{(x-2) \times 3}{4y \times 3} = \frac{3(x-2)}{12y} \][/tex]
3. [tex]\[ \frac{x+3}{6y} \rightarrow \frac{(x+3) \times 2}{6y \times 2} = \frac{2(x+3)}{12y} \][/tex]
Now we can rewrite the original expression with these new fractions:
[tex]\[ \frac{4(x+1)}{12y} + \frac{3(x-2)}{12y} - \frac{2(x+3)}{12y} \][/tex]
Since all terms have the common denominator [tex]\(12y\)[/tex], we can combine the numerators over this common denominator:
[tex]\[ \frac{4(x+1) + 3(x-2) - 2(x+3)}{12y} \][/tex]
Let's expand and combine the terms in the numerator:
[tex]\[ 4(x+1) + 3(x-2) - 2(x+3) \][/tex]
Expanding each term:
[tex]\[ 4x + 4 + 3x - 6 - 2x - 6 \][/tex]
Combining like terms:
[tex]\[ (4x + 3x - 2x) + (4 - 6 - 6) = 5x - 8 \][/tex]
So the expression simplifies to:
[tex]\[ \frac{5x - 8}{12y} \][/tex]
Therefore, the result of the indicated operations is:
[tex]\[ \frac{x-2}{4y} + \frac{x+1}{3y} - \frac{x+3}{6y} = \frac{5x-8}{12y} \][/tex]
