Chapter 1.2, Problem 47E

To determine

Tofind:the whether the number of quarters is eight more than the number of dimes.

Answer to Problem 47E

It is not possible to have $8$ more quarters than dimes when the total is $2.80$ , because it is not possible to have only part of a dime.

Explanation of Solution

Given:

Total cost of quarter and dimes is $\$2.80$ .

Concept used:

Distributive law:

In mathematics the law relating the operations of multiplication and addition, stated symbolically:

  $a\left(b+c\right)=ab+ac$ .

Here monomial factor $a$ is distributed, or separately applied, to each term of the binomial factor $\left(b+c\right)$ resulting in the product $ab+ac$ .

Probably the most common thing it will be doing with polynomials is combining like terms.

This is the process of adding together whatever terms it can, but not overdoing it by trying to add together terms that can’t actually be combined and term can combined only if there has the exact same variable portion and exact same variable portion and by exact same.

Calculation:

To find the number of dimes used first define a variable as:

Let $d=$ number of dimes.

The equation can be written as:

  $2.80=0.1d+0.25\left(d+8\right).$

By distributing and combining the like terms:

  $\begin{array}{l}2.80=0.1d+0.25d+2.\\ 2.80=0.35d+2.\\ 0.8=0.35d.\\ d\approx \mathrm{2.29.}\end{array}$

Hence, it is not possible to have $8$ more quarters than dimes when the total is $2.80$ , because it is not possible to have only part of a dime.