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solve for k.\\(\frac{1}{4}k - 3 + \frac{5}{4}k - 1 = \frac{3}{8}k + 2\\…

Question

solve for k.\\(\frac{1}{4}k - 3 + \frac{5}{4}k - 1 = \frac{3}{8}k + 2\\)\\(k =\\)

Explanation:

Step1: Combine like terms on the left

$\frac{1}{4}k + \frac{5}{4}k - 3 - 1 = \frac{6}{4}k - 4 = \frac{3}{2}k - 4$

Step2: Rewrite the equation

$\frac{3}{2}k - 4 = \frac{3}{8}k + 2$

Step3: Subtract $\frac{3}{8}k$ from both sides

$\frac{3}{2}k - \frac{3}{8}k - 4 = 2$
$\frac{12}{8}k - \frac{3}{8}k - 4 = 2$
$\frac{9}{8}k - 4 = 2$

Step4: Add 4 to both sides

$\frac{9}{8}k = 6$

Step5: Multiply both sides by $\frac{8}{9}$

$k = 6 \times \frac{8}{9}$
$k = \frac{48}{9} = \frac{16}{3}$

Answer:

$\frac{16}{3}$