Chapter 4.4, Problem 60SR

To determine

To solve: equation and check the solution

Answer to Problem 60SR

The required solution is

Explanation of Solution

Given: $p+\frac{1}{10}=-\frac{3}{4}$

Calculation:

Add both sides of equation with $-\frac{1}{10}$

  $p=-\frac{3}{4}-\frac{1}{10}$

LCD is 20. Get the denominators as 20

  $\begin{array}{l}p=-\frac{3.5}{4.5}-\frac{1.2}{10.2}\\ p=-\frac{15}{20}-\frac{2}{20}\\ p=\frac{-15-2}{20}\\ p=-\frac{17}{20}\end{array}$

Solution:- $p=-\frac{17}{20}$

Check

Substitute $p=-\frac{17}{20}$ in $p+\frac{1}{10}=-\frac{3}{4}$ .

  $-\frac{17}{20}+\frac{1}{10}=-\frac{3}{4}$

LDR is 20. Make 20 as common denominator

  $\begin{array}{l}-\frac{17}{20}+\frac{1.2}{10.2}=-\frac{3}{4}\\ -\frac{17}{20}+\frac{2}{20}=-\frac{3}{4}\\ \frac{-17+2}{20}=-\frac{3}{4}\\ \frac{-15}{20}=-\frac{3}{4}\\ \frac{-3}{4}=-\frac{3}{4}\end{array}$

Left hand side= Right hand side

Conclusion:

Hence, the solution is $p=-\frac{17}{20}$