Question

Solve the system analytically. If the equations are dependent, write the solution set in terms of the variable z. Hint, let 1/x=t,1/y=u and 1/z=w.

Answer 1

Answer:

x=4, y=8 and z=1

Explanation:

Given the system of equations:

$\begin{gathered} \frac{3}{x}+\frac{2}{y}-\frac{2}{z}=-1 \\ \frac{6}{x}-\frac{12}{y}+\frac{5}{z}=5 \\ \frac{7}{x}+\frac{2}{y}-\frac{1}{z}=1 \end{gathered}$

Making the substitutions: 1/x=t,1/y=u and 1/z=w., we have:

$\begin{gathered} 3t+2u-2w=-1\ldots(1) \\ 6t-12u+5w=5\ldots(2) \\ 7t+2u-w=1\ldots(3) \end{gathered}$

From the third equation:

$w=7t+2u-1$

Substitute w into the first and second equations:

First Equation

$\begin{gathered} 3t+2u-2w=-1 \\ 3t+2u-2(7t+2u-1)=-1 \\ 3t+2u-14t-4u+2=-1 \\ -11t-2u=-3 \\ 11t+2u=3\ldots(4) \end{gathered}$

Second Equation

$\begin{gathered} 6t-12u+5w=5 \\ 6t-12u+5(7t+2u-1)=5 \\ 6t-12u+35t+10u-5=5 \\ 41t-2u=10\ldots(5) \end{gathered}$

We then solve equations 4 and 5 simultaneously:

$\begin{gathered} 11t+2u=3 \\ 41t-2u=10 \\ \text{Add} \\ 52t=13 \\ t=\frac{13}{52} \\ t=\frac{1}{4} \end{gathered}$

Substitute t to solve for u.

$\begin{gathered} 11t+2u=3 \\ \frac{11}{4}+2u=3 \\ 2u=3-\frac{11}{4} \\ 2u=\frac{1}{4} \\ u=\frac{1}{8} \end{gathered}$

Recall that w=7t+2u-1:

$\begin{gathered} w=7(\frac{1}{4})+2(\frac{1}{8})-1 \\ =\frac{7}{4}+\frac{2}{8}-1 \\ w=1 \end{gathered}$

Therefore, we have that:

$\begin{gathered} \frac{1}{x}=\frac{1}{4}\implies x=4 \\ \frac{1}{y}=\frac{1}{8}\implies y=8 \\ \frac{1}{z}=1\implies z=1 \end{gathered}$

The solution to the system of equations is:

$x=4,y=8\text{ and z=1}$

There is only one solution. The solution set is (4,8,1).