How do you solve $k + 1 = 3 k - 1$ $k+ 1= 3k - 1$ ?

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Answer 1 · 2018-05-09T18:01:20

$k = 1$ $k = 1$

As per the question, we have

$k + 1 = 3 k - 1$ $k + 1 = 3k -1$

$:. 1 + 1 = 3k - k$

$:. 2k = 2$

$:. k = 2/2$

$:. k =1$

$:. k = 1$ is the solution to this problem.

Answer 2 · 2018-05-09T18:14:20

$k = 1$

$k + 1 = 3k - 1$

Subtract 1k from each side, remember what you do on one side you do on the other.
$1 = 2k - 1$

Add 1 to each side.
$2 = 2k$

You can switch them:
$2k = 2$

Now divide each by 2.
$(2k)/2 = 2/2$

$2/2$ is the same thing as $1/1$ which is equal to one:

$k = 1$

How do you solve k+ 1= 3k - 1k+1=3k−1k+ 1= 3k - 1?