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To solve the problem [tex]\(\frac{2p + 2}{4p + 10} \div \frac{4p + 4}{2p + 5}\)[/tex], we will follow these steps:
1. Rewrite the division as multiplication by the reciprocal:
[tex]\[ \frac{2p + 2}{4p + 10} \div \frac{4p + 4}{2p + 5} = \frac{2p + 2}{4p + 10} \times \frac{2p + 5}{4p + 4} \][/tex]
2. Simplify each of the fractions:
- For the numerator of the first fraction, factor out common factors:
[tex]\[ 2p + 2 = 2(p + 1) \][/tex]
- For the denominator of the first fraction:
[tex]\[ 4p + 10 = 2(2p + 5) \][/tex]
- For the numerator of the second fraction:
[tex]\[ 4p + 4 = 4(p + 1) \][/tex]
So, we rewrite each fraction with the factored forms:
[tex]\[ \frac{2(p + 1)}{2(2p + 5)} \times \frac{2p + 5}{4(p + 1)} \][/tex]
3. Cancel out common factors where possible:
- In the first fraction, [tex]\(2\)[/tex] in the numerator and [tex]\(\frac{1}{2}\)[/tex] in the denominator can be simplified:
[tex]\[ \frac{2(p + 1)}{2(2p + 5)} = \frac{p + 1}{2p + 5} \][/tex]
- In the second fraction, the factor [tex]\((p + 1)\)[/tex] appears in both numerator and denominator; so they can be canceled:
[tex]\[ \frac{(2p + 5)}{4(p + 1)} = \frac{(2p + 5)}{4(p + 1)} \][/tex]
Now we have:
[tex]\[ \frac{p + 1}{2p + 5} \times \frac{2p + 5}{4(p + 1)} \][/tex]
4. More cancellation:
- The terms [tex]\(2p + 5\)[/tex] in the numerator of the second fraction and the denominator of the first fraction cancel each other out:
[tex]\[ \frac{p + 1}{2p + 5} \times \frac{2p + 5}{4(p + 1)} = \frac{\cancel{(p + 1)}}{\cancel{(2p + 5)}} \times \frac{\cancel{(2p + 5)}}{4\cancel{(p + 1)}} \][/tex]
5. Rewriting the remaining terms:
- The simplified fraction becomes:
[tex]\[ \frac{1}{4} \][/tex]
Thus, the simplified result of [tex]\(\frac{2p+2}{4p+10} \div \frac{4p+4}{2p+5}\)[/tex] is:
[tex]\[ \boxed{\frac{1}{4}} \][/tex]
1. Rewrite the division as multiplication by the reciprocal:
[tex]\[ \frac{2p + 2}{4p + 10} \div \frac{4p + 4}{2p + 5} = \frac{2p + 2}{4p + 10} \times \frac{2p + 5}{4p + 4} \][/tex]
2. Simplify each of the fractions:
- For the numerator of the first fraction, factor out common factors:
[tex]\[ 2p + 2 = 2(p + 1) \][/tex]
- For the denominator of the first fraction:
[tex]\[ 4p + 10 = 2(2p + 5) \][/tex]
- For the numerator of the second fraction:
[tex]\[ 4p + 4 = 4(p + 1) \][/tex]
So, we rewrite each fraction with the factored forms:
[tex]\[ \frac{2(p + 1)}{2(2p + 5)} \times \frac{2p + 5}{4(p + 1)} \][/tex]
3. Cancel out common factors where possible:
- In the first fraction, [tex]\(2\)[/tex] in the numerator and [tex]\(\frac{1}{2}\)[/tex] in the denominator can be simplified:
[tex]\[ \frac{2(p + 1)}{2(2p + 5)} = \frac{p + 1}{2p + 5} \][/tex]
- In the second fraction, the factor [tex]\((p + 1)\)[/tex] appears in both numerator and denominator; so they can be canceled:
[tex]\[ \frac{(2p + 5)}{4(p + 1)} = \frac{(2p + 5)}{4(p + 1)} \][/tex]
Now we have:
[tex]\[ \frac{p + 1}{2p + 5} \times \frac{2p + 5}{4(p + 1)} \][/tex]
4. More cancellation:
- The terms [tex]\(2p + 5\)[/tex] in the numerator of the second fraction and the denominator of the first fraction cancel each other out:
[tex]\[ \frac{p + 1}{2p + 5} \times \frac{2p + 5}{4(p + 1)} = \frac{\cancel{(p + 1)}}{\cancel{(2p + 5)}} \times \frac{\cancel{(2p + 5)}}{4\cancel{(p + 1)}} \][/tex]
5. Rewriting the remaining terms:
- The simplified fraction becomes:
[tex]\[ \frac{1}{4} \][/tex]
Thus, the simplified result of [tex]\(\frac{2p+2}{4p+10} \div \frac{4p+4}{2p+5}\)[/tex] is:
[tex]\[ \boxed{\frac{1}{4}} \][/tex]
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