Chapter 4.4, Problem 47PPS

To determine

Find the values of the variables that make the given equation true.

Answer to Problem 47PPS

The values of $a=25$ and $b=-2$ .

Explanation of Solution

Given:

The given equation is $\left(2a-4b\right)i+a+5b=15+58i$ .

Calculation:

Set the real parts equal to each other and the imaginary parts equal to each other.

The real parts

  $a+5b=15---\left(i\right)$

The imaginary parts.

  $2a-4b=58---\left(ii\right)$

Subtract $\left(i\right)\times 2$ from $\left(ii\right)$ .

  $\begin{array}{l}\frac{\begin{array}{l}2a-4b=58\\ \left[a+5b=15\right]\times 2\end{array}}{}\\ \frac{\begin{array}{l}2a-4b=58\\ 2a+10b=30\\ ---\end{array}}{\begin{array}{l}0-14b=28\\ \Rightarrow \frac{-14}{-14}b=\frac{28}{-14}\\ \Rightarrow b=-2\end{array}}\end{array}$

Substitute the value of $b=-2$ in equation $\left(i\right)$ .

  $\begin{array}{l}a+5b=15\\ \Rightarrow a+5\left(-2\right)=15\\ \Rightarrow a-10=15\\ \Rightarrow a-10+10=15+10\\ \Rightarrow a=25\end{array}$

Hence the values of $a=25$ and

  $b=-2$ .