How do you solve $\frac { k + 1} { 3} = \frac { k - 2} { 6}$ ?

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Answer 1 · 2016-10-21T08:42:57

$k=-4$

$(k+1)/3=(k-2)/6$

Multiply both sides by 6 and simplify.

$6xx(k+1)/3=6xx(k-2)/6$

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

$2k+2=k-2$

Subtract $k$ from both sides.

$k+2=-2$

Subtract $2$ from both sides.

$k=-2-2=-4$

$:.k=-4$

How do you solve \frac { k + 1} { 3} = \frac { k - 2} { 6}\frac { k + 1} { 3} = \frac { k - 2} { 6}?