Chapter 6.6, Problem 4E

a.

To determine

To Solve: each system of equations by letting $X=\frac{1}{x},Y=\frac{1}{y}$

Answer to Problem 4E

The solutions are $x=\frac{1}{3},y=\frac{-1}{4}$

Explanation of Solution

Given:

The given system of equations is $\{\begin{array}{c}\frac{12}{x}-\frac{12}{y}=7\\ \frac{3}{x}+\frac{4}{y}=0\end{array}$

  $X=\frac{1}{x},Y=\frac{1}{y},Z=\frac{1}{z}$ is to be used.

Concept Used:

Substitution method is used in this problem.

Calculation:

Let us substitute $X=\frac{1}{x},Y=\frac{1}{y}$ .

The equations changes to

  $\begin{array}{l}12X-12Y=7\cdots \left(1\right)\\ 3X+4Y=0\cdots \left(2\right)\end{array}$

From equation (2) we get

  $\begin{array}{l}3X+4Y=0\\ X=\frac{-4Y}{3}\end{array}$

Substitute this in equation (1)

  $\begin{array}{l}12\left(\frac{-4Y}{3}\right)-12Y=7\\ -16Y-12Y=7\\ -28Y=7\\ Y=\frac{-1}{4}\end{array}$

The value of X is

  $\begin{array}{l}X=\frac{-4Y}{3}\\ X=\frac{-4\left(\frac{-1}{4}\right)}{3}\\ X=\frac{1}{3}\end{array}$

The value of x is $x=\frac{1}{X}=\frac{1}{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}=3$

The value of y is $y=\frac{1}{Y}=\frac{1}{\raisebox{1ex}{$-1$}\!\left/ \!\raisebox{-1ex}{$4$}\right.}=-4$

Conclusion:

The solutions are $x=3,y=-4$

b.

To determine

To Solve: each system of equations by letting $X=\frac{1}{x},Y=\frac{1}{y},Z=\frac{1}{z}$ .

Answer to Problem 4E

For any arbitrary constant a, there exist infinitely many solutions $X=5-\frac{a+1}{4},Y=a$ and $Z=\frac{a+1}{4}$

Explanation of Solution

Given:

The given system of equations is $\{\begin{array}{c}\frac{2}{x}+\frac{1}{y}-\frac{3}{z}=4\\ \frac{4}{x}+\frac{2}{z}=10\\ \frac{-2}{x}+\frac{3}{y}-\frac{13}{z}=-8\end{array}$

  $X=\frac{1}{x},Y=\frac{1}{y},Z=\frac{1}{z}$ is to be used.

Concept Used:

Substitution method is used in this.

Calculation:

Let us substitute $X=\frac{1}{x},Y=\frac{1}{y},Z=\frac{1}{z}$ . The equations changes to

  $\begin{array}{l}2X+Y-3Z=4\cdots \left(1\right)\\ 4X+2Z=10\cdots \left(2\right)\\ -2X+3Y-13Z=-8\cdots \left(3\right)\end{array}$

From equation (2), we get,

  $\begin{array}{l}X=\frac{10-2Z}{4}\\ X=\frac{5-Z}{2}\end{array}$

Substitute this value in other two equations.

  $\begin{array}{l}2\left(\frac{5-Z}{2}\right)+Y-3Z=4\\ \Rightarrow 5-Z+Y-3Z=4\\ \Rightarrow -4Z+Y=-1\\ \Rightarrow 4Z-Y=1\cdots \left(4\right)\end{array}$

And

  $\begin{array}{l}-2\left(\frac{5-Z}{2}\right)+3Y-13Z=-8\\ \Rightarrow -5+Z+3Y-13Z=-8\\ \Rightarrow -12Z+3Y=-3\\ \Rightarrow 4Z-Y=1\cdots \left(5\right)\end{array}$

Equation (4) and (5) are same so $Z=\frac{Y+1}{4}\cdots \left(6\right)$

Solve these equations.

Multiply equation (1) by 3 and subtract it from equation (3)

  $\begin{array}{l}6X-3Y-9Z-2X-3Y+13Z=12+8\\ 4X+4Z=20\\ X+Z=5\end{array}$

Put $Z=\frac{Y+1}{4}$

Then

  $\begin{array}{l}X+\frac{Y+1}{4}=5\\ X=5-\frac{Y+1}{4}\end{array}$

Let $Y=a$

Here a is an arbitrary constant

Then $X=5-\frac{a+1}{4}$ and $Z=\frac{a+1}{4}$

Hence there exist infinitely many solutions $X=5-\frac{a+1}{4},Y=a$ and $Z=\frac{a+1}{4}$

Conclusion:

There exist infinitely many solutions $X=5-\frac{a+1}{4},Y=a$ and $Z=\frac{a+1}{4}$