Question

1. \\(\frac{k + 4}{4} + \frac{k - 1}{4} = \frac{k + 4}{4k}\\)

Explanation:

Step1: Combine left - hand side fractions

Since the two fractions on the left - hand side have the same denominator (\(4\)), we can add their numerators:
\(\frac{k + 4}{4}+\frac{k-1}{4}=\frac{(k + 4)+(k - 1)}{4}=\frac{k+4+k - 1}{4}=\frac{2k + 3}{4}\)
So the equation becomes \(\frac{2k + 3}{4}=\frac{k + 4}{4k}\)

Step2: Cross - multiply to eliminate denominators

Cross - multiplying gives us \((2k + 3)\times4k=4\times(k + 4)\)
Simplify both sides:
Left - hand side: \(4k(2k + 3)=8k^{2}+12k\)
Right - hand side: \(4(k + 4)=4k+16\)
So the equation is \(8k^{2}+12k=4k + 16\)

Step3: Rearrange into standard quadratic form

Subtract \(4k\) and \(16\) from both sides to get \(8k^{2}+12k-4k - 16 = 0\)
Simplify the left - hand side: \(8k^{2}+8k-16 = 0\)
We can divide the entire equation by \(8\) to simplify: \(k^{2}+k - 2=0\)

Step4: Factor the quadratic equation

Factor \(k^{2}+k - 2\). We need two numbers that multiply to \(-2\) and add to \(1\). The numbers are \(2\) and \(-1\).
So \(k^{2}+k - 2=(k + 2)(k - 1)=0\)

Step5: Solve for \(k\)

Set each factor equal to zero:

• If \(k + 2=0\), then \(k=-2\)
• If \(k - 1=0\), then \(k = 1\)

But we need to check for extraneous solutions because we had a denominator \(4k\) (so \(k
eq0\)) and also when we cross - multiplied, we need to ensure that the original equation is defined.
For \(k = 1\):
Left - hand side: \(\frac{1 + 4}{4}+\frac{1-1}{4}=\frac{5}{4}+0=\frac{5}{4}\)
Right - hand side: \(\frac{1 + 4}{4\times1}=\frac{5}{4}\)
For \(k=-2\):
Left - hand side: \(\frac{-2 + 4}{4}+\frac{-2-1}{4}=\frac{2}{4}+\frac{-3}{4}=\frac{2-3}{4}=-\frac{1}{4}\)
Right - hand side: \(\frac{-2 + 4}{4\times(-2)}=\frac{2}{-8}=-\frac{1}{4}\)
Both solutions are valid.

Answer:

\(k = 1\) or \(k=-2\)