Chapter 8.5, Problem 23IP

To determine

The solution of the equation $\frac{4}{5}c-\frac{1}{10}c=\frac{9}{10}c$ ,

Answer to Problem 23IP

  $c=0$

Explanation of Solution

Given:

The equation, $\frac{4}{5}c-\frac{1}{10}c=\frac{9}{10}c$

Concept Used:

• 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 solve the given equation $\frac{4}{5}c-\frac{1}{10}c=\frac{9}{10}c$ for the variable, isolate the variable term c on one side by performing some basic algebraic operations to get rid of the other numbers and terms associated with it.

Here to isolate c on left side, first simplify the like terms and then subtract both sides by $\frac{9}{10}c$ and then multiply both sides by $10$ and then simplify further as shown below,

  $\begin{array}{c}\frac{4}{5}c-\frac{1}{10}c=\frac{9}{10}c\\ \frac{4}{5}c\times \frac{2}{2}-\frac{1}{10}c=\frac{9}{10}c\\ \frac{8}{10}c-\frac{1}{10}c=\frac{9}{10}c\\ \frac{8c-1c}{10}=\frac{9}{10}c\\ \frac{7c}{10}=\frac{9}{10}c\\ \frac{7c}{10}-\frac{9}{10}c=\frac{9}{10}c-\frac{9}{10}c\\ \frac{7c-9c}{10}=0\\ \frac{-2c}{10}\times 10=0\times 10\\ 2c=0\\ c=0\end{array}$

Thus, the solution of the given equation is $c=0$ .