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

Mar 12, 2017

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 $\textcolor{red}{4}$ - the term outside the parenthesis:

$3 - \textcolor{red}{4} \left(2 k - 5\right) = 71$

$3 - \left(\textcolor{red}{4} \times 2 k\right) + \left(\textcolor{red}{4} \times 5\right) = 71$

$3 - 8 k + 20 = 71$

$- 8 k + 20 + 3 = 71$

$- 8 k + 23 = 71$

Next, subtract $\textcolor{red}{23}$ from each side of the equation to isolate the $k$ term while keeping the equation balanced:

$- 8 k + 23 - \textcolor{red}{23} = 71 - \textcolor{red}{23}$

$- 8 k + 0 = 48$

$- 8 k = 48$

Now, divide each side of the equation by $\textcolor{red}{- 8}$ to solve for $k$ while keeping the equation balanced:

$\frac{- 8 k}{\textcolor{red}{- 8}} = \frac{48}{\textcolor{red}{- 8}}$

$\frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{- 8}}} k}{\cancel{\textcolor{red}{- 8}}} = - 6$

$k = - 6$

Mar 12, 2017

$k = - 6$

#### Explanation:

distribute the bracket to begin with.

$3 - 8 k + 20 = 71$

$\Rightarrow - 8 k + 23 = 71$

subtract 23 from both sides.

$- 8 k \cancel{+ 23} \cancel{- 23} = 71 - 23$

$\Rightarrow - 8 k = 48$

divide both sides by - 8

$\frac{\cancel{- 8} k}{\cancel{- 8}} = \frac{48}{- 8}$

$\Rightarrow k = - 6$

$\textcolor{b l u e}{\text{As a check}}$

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

$\text{left side } = 3 - 4 \left(\left(2 \times - 6\right) - 5\right)$

$\textcolor{w h i t e}{\le f t s i \mathrm{de}} = 3 - \left(4 \times \left(- 17\right)\right)$

$\textcolor{w h i t e}{\le f t s i \mathrm{de}} = 3 - \left(- 68\right) = 3 + 68 = 71$

$\Rightarrow k = - 6 \text{ is the solution}$