Chapter 7.CT, Problem 1CT

Solve the following equations:

a. $\frac{3}{4}x+5=\frac{2}{3}x+4$

b. $0.25x+4=1.5x-0.2$

To determine

(a)

To solve:

The following equation $\frac{3}{4}x+5=\frac{2}{3}x+4$.

Answer to Problem 1CT

Solution:

$x=-12$

Explanation of Solution

Given:

The equation is $\frac{3}{4}x+5=\frac{2}{3}x+4$

Calculation:

Consider the equation $\frac{3}{4}x+5=\frac{2}{3}x+4$

Subtract 5 from both sides,

$\begin{array}{rll}\frac{3}{4}x+5-5& =& \frac{2}{3}x+4-5\\ \frac{3}{4}x& =& \frac{2}{3}x-1\end{array}$

Subtract $\frac{2}{3}x$ from both sides,

$\begin{array}{rll}\frac{3}{4}x-\frac{2}{3}x& =& \frac{2}{3}x-1-\frac{2}{3}x\\ \frac{3}{4}x-\frac{2}{3}x& =& -1\\ \frac{9-8}{12}x& =& -1\\ \frac{1}{12}x& =& -1\end{array}$

Multiply both sides by 12,

$\begin{array}{rll}\frac{x}{12}\cdot 12& =& -1\cdot 12\\ x& =& -12\end{array}$

Final statement:

Therefore, the value of $x$ is $-12$.

To determine

(b)

To solve:

The following equation $0.25x+4=1.5x-0.2$.

Answer to Problem 1CT

Solution:

$x=3.36$

Explanation of Solution

Given:

The equation is $0.25x+4=1.5x-0.2$

Calculation:

Consider the equation $0.25x+4=1.5x-0.2$

Multiply both sides by 100,

$\begin{array}{rll}0.25x\cdot 100+4\cdot 100& =& 1.5x\cdot 100-0.2\cdot 100\\ 25x+400& =& 150x-20\\ 25x+400& =& 150x-20\end{array}$

Subtract $25x$ from both sides,

$\begin{array}{rll}25x+400-25x& =& 150x-20-25x\\ 400& =& 125x-20\end{array}$

Add $20$ on both sides,

$\begin{array}{rll}400+20& =& 125x-20+20\\ 420& =& 125x\end{array}$

Multiply both sides by 125,

$\begin{array}{rll}\frac{420}{125}& =& \frac{125}{125}x\\ x& =& \frac{420}{125}\\ x& =& \frac{84}{25}& =& 3.36\end{array}$

Final statement:

Therefore, the value of $x$ is $3.36$