Question

solve:
\\(\frac{k}{k + 5} + \frac{5}{7 - k} = 1\\)
\\(k = ?\\)

Explanation:

Step1: Find the common denominator

The denominators are \(k + 5\) and \(7 - k\), so the common denominator is \((k + 5)(7 - k)\). Multiply each term by the common denominator to eliminate the fractions:

$$ \frac{k}{k + 5} \times (k + 5)(7 - k) + \frac{5}{7 - k} \times (k + 5)(7 - k) = 1 \times (k + 5)(7 - k) $$

Simplify each term:

$$ k(7 - k) + 5(k + 5) = (k + 5)(7 - k) $$

Step2: Expand both sides

Expand the left - hand side:

$$ 7k - k^{2}+5k + 25=-k^{2}+2k + 35 $$

Combine like terms on the left - hand side:

$$ (7k + 5k)-k^{2}+25=-k^{2}+2k + 35 $$

$$ 12k - k^{2}+25=-k^{2}+2k + 35 $$

Step3: Solve for k

Add \(k^{2}\) to both sides of the equation:

$$ 12k+25 = 2k + 35 $$

Subtract \(2k\) from both sides:

$$ 12k-2k + 25=35 $$

$$ 10k+25 = 35 $$

Subtract 25 from both sides:

$$ 10k=35 - 25 $$

$$ 10k = 10 $$

Divide both sides by 10:

$$ k=\frac{10}{10}=1 $$

Step4: Check for extraneous solutions

We need to check if \(k = 1\) makes the original denominators zero.
For \(k+5\), when \(k = 1\), \(k + 5=1 + 5=6
eq0\).
For \(7 - k\), when \(k = 1\), \(7-1 = 6
eq0\). So \(k = 1\) is a valid solution.

Answer:

\(k = 1\)