Chapter 45, Problem 1A

Solve $\frac{3}{4}E+\left(18-E\right)=2\frac{1}{4}E+12$ .

To determine

To solve the given equation.

Answer to Problem 1A

  $H=\pm 2.885$.

Explanation of Solution

Given information:

The given equation is $\frac{3}{4}E+(18-E)=2\frac{1}{4}E+12$.

Calculation:

We have been given the equation as $\frac{3}{4}E+(18-E)=2\frac{1}{4}E+12$.

We will perform some basic mathematical operation to solve the given equation as below.

  $\begin{array}{l}\frac{3}{4}E+(18-E)=2\frac{1}{4}E+12\\ \Rightarrow \frac{3}{4}E+18-E=\frac{9}{4}E+12\left\{\text{Simplify}\right\}\\ \Rightarrow \frac{3}{4}E-E-\frac{9}{4}E=12-18\text{ {Bringing the variables on LHS} }\\ \Rightarrow \frac{3E-4E-9E}{4}=-6\text{ {Taking LCD }}\\ \Rightarrow \frac{-10E}{4}=-6\left\{\text{Simplify}\right\}\\ \Rightarrow \frac{10E}{4}\times 4=6\times 4\text{{Multiplying}\text{by}4\text{on}\text{both}\text{sides}\}\\ \Rightarrow 10E=24\\ \Rightarrow E=\frac{24}{10}\\ \Rightarrow E=\frac{12}{5}\\ \Rightarrow E=2\frac{2}{5}\text{ {converting into mixed fraction} }\end{array}$

Hence, solution of the equation will be given by-

  $E=2\frac{2}{5}$.