Question

How do you solve `8- 6k = 6( 1+ 3k )`?

Answer 1

`k = 1/12`

Explanation:

Step 1) Expand the terms within parenthesis:

`8 - 6k = 6 + 18k`

Step 2) Isolate the `x` terms on one side of the equation and the constants on the other side of the equation while keeping the equation balanced:

`8 - 6k + 6k - 6 = 6 + 18k + 6k - 6`

`8 - 6 - 0 = 0 + 24k`

`2 = 24k`

Step 3) Solve for `k` while keeping the equation balanced:

`2/24 = (24k)/24`

`1/12 = k`

Answer 2

`k=1/12`

Explanation:

`"distribute the bracket on the right side"`

`rArr8-6k=6+18k`

`"add 6k to both sides"`

`8cancel(-6k)cancel(+6k)=6+18k+6k`

`rArr8=6+24k`

`"subtract 6 from both sides"`

`8-6=cancel(6)cancel(-6)+24k`

`rArr24k=2`

`"divide both sides by 24"`

`(cancel(24) k)/cancel(24)=2/24`

`rArrk=1/12`

`color(blue)"as a check"`

Substitute this value into the equation and if the left side equals the right side then it is the solution.

`"left side "=8-6/12=8-1/2=7 1/2`

`"right side "=6(1+3/12)=6xx1 1/4=7 1/2`

`rArrk=1/12" is the solution"`