Question

Please answer these questions(please include work/explanation)(3)

Answer 1

Answer:

The Answer for number 3.

$\sf \frac{k + 4}{4} + \frac{k - 1}{4} = \frac{k + 4}{4k} \\ \sf \frac{2k + 3}{4} = \frac{k + 4}{4k} \\ \sf \frac{k(2k + 3)}{4k} = \frac{k + 4}{4k} \\ \sf k(2k + 3) = k + 4 \\ \sf {2k}^{2} + 3k = k + 4 \\ \sf {2k}^{2} + 3k - k - 4 = 0 \\ \sf {2k}^{2} + 2k - 4 = 0 \\ \sf {k}^{2} + k - 2 = 0 \\ \sf (k + 2)(k - 1) = 0 \\ \sf k + 2 = 0 \to k = - 2 \\ \sf k - 1 = 0 \to k = 1 \\ \tt the \: answer \: is \\ \boxed{ \bold{ \red{ - 2 \: and \: 1}}}$

Hope it helps!

Answer 2

Answer:

See below.

Step-by-step explanation:

(k + 4)/4 + (k - 1)/4 = (k + 4)/4k

To solve these fractional equations the first step is to eliminate the fractions by multiplying through by the Lowest Common Multiple.

Here the LCM of 4, 4 and 4k  is 4k:

4k* (k + 4)/4 + 4k*(k - 1)/4 = 4k*(k + 4)/4k

k(k + 4) + k(k - 1) = k + 4

k^2 + 4k + k^2 - k = k + 4

2k^2 + 2k - 4 = 0

k^2 + k - 2 = 0

(k + 2)(k - 1) = 0

k = -2, 1   Answer.

The other equations are solved in a similar way.