Question

How do you rewrite `w=3(2a + b) - 4` to solve for `a`?

Answer 1

`color(blue)(a=(w+4-3b)/6)`

Explanation:

If `w=3(2a+b)-4`
then
`color(white)("XXX")w+4=3(2a+b)`

`color(white)("XXX")(w+4)/3=2a+b`

`color(white)("XXX")(w+4)/3-b=2acolor(white)("xx")rarrcolor(white)("xx")(w+4-3b)/3=2a`

`color(white)("XXX")(w+4-3b)/6=acolor(white)("xxx")"or"color(white)("xxx")a=(w+4-3b)/6`