Simplifying Algebraic Expressions

The question asks us to simplify the expression \( \frac{1}{2+2p} + \frac{1}{2+2q}+\frac{1}{2+2r} \) given the values of p, q, and r in terms of x, y, and z.

We are given:

• \( p = \frac{x}{y+z} \)
• \( q = \frac{y}{z+x} \)
• \( r = \frac{z}{x+y} \)

Let's substitute the given values of p, q, and r into the expression we need to simplify.

Step-by-Step Simplification

First, consider the first term \( \frac{1}{2+2p} \). Substitute the value of p:

\( \frac{1}{2+2p} = \frac{1}{2+2\left(\frac{x}{y+z}\right)} \)

Factor out 2 from the denominator:

\( = \frac{1}{2\left(1+\frac{x}{y+z}\right)} \)

Combine the terms inside the parenthesis by finding a common denominator:

\( = \frac{1}{2\left(\frac{y+z}{y+z}+\frac{x}{y+z}\right)} \)

\( = \frac{1}{2\left(\frac{y+z+x}{y+z}\right)} \)

Now, invert the fraction in the denominator and multiply:

\( = \frac{1}{2} \times \frac{y+z}{x+y+z} \)

\( = \frac{y+z}{2(x+y+z)} \)

Next, consider the second term \( \frac{1}{2+2q} \). Substitute the value of q:

\( \frac{1}{2+2q} = \frac{1}{2+2\left(\frac{y}{z+x}\right)} \)

Following the same steps as above:

\( = \frac{1}{2\left(1+\frac{y}{z+x}\right)} \)

\( = \frac{1}{2\left(\frac{z+x+y}{z+x}\right)} \)

\( = \frac{z+x}{2(x+y+z)} \)

Finally, consider the third term \( \frac{1}{2+2r} \). Substitute the value of r:

\( \frac{1}{2+2r} = \frac{1}{2+2\left(\frac{z}{x+y}\right)} \)

Following the same steps:

\( = \frac{1}{2\left(1+\frac{z}{x+y}\right)} \)

\( = \frac{1}{2\left(\frac{x+y+z}{x+y}\right)} \)

\( = \frac{x+y}{2(x+y+z)} \)

Combining the Terms

Now, add the three simplified terms together:

\( \frac{y+z}{2(x+y+z)} + \frac{z+x}{2(x+y+z)} + \frac{x+y}{2(x+y+z)} \)

Since all terms have the same denominator \( 2(x+y+z) \), we can add the numerators directly:

\( = \frac{(y+z) + (z+x) + (x+y)}{2(x+y+z)} \)

Combine like terms in the numerator:

\( = \frac{y+z+z+x+x+y}{2(x+y+z)} \)

\( = \frac{2x+2y+2z}{2(x+y+z)} \)

Factor out 2 from the numerator:

\( = \frac{2(x+y+z)}{2(x+y+z)} \)

Cancel the common factor \( 2(x+y+z) \) from the numerator and the denominator. Assuming \( x+y+z \neq 0 \), the expression simplifies to:

\( = 1 \)

Therefore, the simplified value of the expression \( \frac{1}{2+2p} + \frac{1}{2+2q}+\frac{1}{2+2r} \) is 1.

Revision Table - Algebraic Simplification

Additional Information - Related Algebraic Concepts

This problem involves algebraic manipulation and working with rational expressions (fractions involving variables). Key concepts used include:

• Rational Expressions: Fractions where the numerator and denominator are polynomials. Simplifying them often involves factoring and cancelling common terms.
• Least Common Denominator (LCD): When adding or subtracting fractions with different denominators, finding the LCD is crucial. In this problem, after substitution, all terms naturally had the same denominator \( 2(x+y+z) \), making addition straightforward.
• Algebraic Identities: While not directly used for simplification here, recognizing patterns and identities can sometimes simplify complex expressions more quickly.

Problems like this test your ability to systematically substitute expressions and perform algebraic operations accurately. Always simplify each part of the expression before combining them if possible, as shown in the step-by-step solution above.