Chapter 8.4, Problem 27IP

To determine

What is the distance to and from work each day.

Answer to Problem 27IP

  $12.75mi$

Explanation of Solution

Given:

Brody drives the same distance to and from work each day. He also drives an additional 1.5 miles each day to go to the gym, 5 day workweek brody drives a total of 71.25.

Concept Used:

Distributive property:

Left Distributive property: $a\left(b\pm c\right)=a\cdot b\pm a\cdot c$

Right Distributive property: $\left(a\pm b\right)c=a\cdot c\pm b\cdot c$

• To get rid of a number in addition from one side, subtract the same number from both sides of equal sign.
• To get rid of a number in subtraction from one side, add the same number both sides of equal sign.
• To get rid of a number in multiplication from one side, divide the same number from both sides of equal sign.
• To get rid of a number in division from one side, multiply the same number both sides of equal sign.

Rules of Addition/ Subtraction:

• Two numbers with similar sign always get added and the resulting number will carry the similar sign.
• Two numbers with opposite signs always get subtracted and the resulting number will carry the sign of larger number.

Rules of Multiplication/ Division:

• The product/quotient of two similar sign numbers is always positive.
• The product/quotient of two numbers with opposite signs is always negative.

Calculation:

In order to find the distance to and from work, let d be the distance to and from work each day, then adding the actual distance and additional distance and then multiplying it by the number of days, is equal to 71.25 the total distance, so the equation is :

  $5\left(d+1.5\right)=71.25$

Now, to solving the equation first using distributive property on left side of the equation and then isolate the variable term d on one side by performing some basic algebraic operations to get rid of the other numbers and terms associated with it.

Here to isolate d on left side, first subtract 7.5 from both sides and then divide both sides by 5 and then simplify further as shown below,

  $\begin{array}{l}5\left(d+1.5\right)=71.25\\ 5\cdot d+5\cdot 1.5=71.25\\ 5d+7.5=71.25\\ 5d+7.5-7.5=71.25-7.5\\ 5d=63.75\\ \frac{5d}{5}=\frac{63.75}{5}\\ d=12.75\end{array}$

So, the distance is 12.75mi .