Question

How do you solve `1/7 k -6 = 3/7k +4`?

Answer 1

`k=-35`

Explanation:

Solution:

`1/7k-6=3/7k+4`

`1/7k-3/7k=4+6`

`(-2)/7k=10`

`k=10*(7/(-2))`

`k=-35`

God bless....I hope the explanation is useful

Answer 2

In support of and expanding on Leland's solution showing first principle method ( in detail ).

`k=-35`

Explanation:

The shortcuts you have probably been shown are the final result of applying first principles.

`color(blue)("Using first principles")`

`color(brown)("Step 1")`
Have all the eterms with `k` on the left of =

`color(magenta)("Subtract "3/7 k" from both sides")`

`" "1/7kcolor(magenta)(-3/7k)-6" "=" "4+3/7 kcolor(magenta)( -3/7 k)`

`color(magenta)("But "3/7 k-3/7 k=0)`

`" "1/7 k-3/7 k-6" "=" "4color(magenta)(+0)`
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
`color(brown)("Step 2")`
Only have the terms with `k` on the left of =

`color(magenta)("Add 6 to both sides")`

`" "1/7 k-3/7 k-6color(magenta)(+6)" "=" "4color(magenta)(+6)`

`color(magenta)("But 6-6=0")`

`" "1/7 k-3/7 kcolor(magenta)(+0)" "=" "10`

`" "-2/7 k" "=" "10`
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
`color(brown)("Step 3")`
Only have `k` on the left

`color(magenta)("Multiply both sides by "7/2)`

`" "-(2/7color(magenta)(xx7/2))xx k" "=" "10color(magenta)(xx7/2)`

`color(magenta)("But "2/7xx7/2" is the same as "2/2xx7/7)`

`color(magenta)("and "10xx7/2" is the same as "10/2xx7)`

`" "-(2/2xx7/7)xx k" "=" "10/2xx7`

`" "-(1xx1)xx k" "=" "5xx7`

`" "-k" "=" "35`

`color(green)("Multiply both sides by "-1" to make "-k" become "+k)`

`" "bar(ul(|color(white)(2/2)k" "=" "-35color(white)(2/2)))|`