Question

How do you solve `3-4(2k-5)=71` using the distributive property?

Answer 1

See the entire solution process below:

Explanation:

First, expand the term in parenthesis on the left side of the equation by multiplying each term within the parenthesis by `color(red)(4)` - the term outside the parenthesis:

`3 - color(red)(4)(2k - 5) = 71`

`3 - (color(red)(4)xx2k) + (color(red)(4)xx5) = 71`

`3 - 8k + 20 = 71`

`-8k + 20 + 3 = 71`

`-8k + 23 = 71`

Next, subtract `color(red)(23)` from each side of the equation to isolate the `k` term while keeping the equation balanced:

`-8k + 23 - color(red)(23) = 71 - color(red)(23)`

`-8k + 0 = 48`

`-8k = 48`

Now, divide each side of the equation by `color(red)(-8)` to solve for `k` while keeping the equation balanced:

`(-8k)/color(red)(-8) = 48/color(red)(-8)`

`(color(red)(cancel(color(black)(-8)))k)/cancel(color(red)(-8)) = -6`

`k = -6`

Answer 2

`k=-6`

Explanation:

distribute the bracket to begin with.

`3-8k+20=71`

`rArr-8k+23=71`

subtract 23 from both sides.

`-8kcancel(+23)cancel(-23)=71-23`

`rArr-8k=48`

divide both sides by - 8

`(cancel(-8) k)/cancel(-8)=48/(-8)`

`rArrk=-6`

`color(blue)"As a check"`

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

`"left side "=3-4((2xx-6)-5)`

`color(white)(left side)=3-(4xx(-17))`

`color(white)(left side)=3-(-68)=3+68=71`

`rArrk=-6" is the solution"`