Q. 40 - Precalculus Enhanced with Graphing Utilities

Question

Solve each system of equations. If the system has no solution, say that it is inconsistent.
$\left\{\begin{array}{c}\frac{4}{x}-\frac{3}{y}=0\\ \frac{6}{x}+\frac{3}{2y}=2\end{array}\right.$

Step-by-Step Solution

Answer

The solution of the given system of equation is $x=4, y=3; \left(4,3\right)$.

Step 1. Given information.

The given equations are:

$\begin{array}{rcl}\frac{4}{x}-\frac{3}{y}& =& 0\\ \frac{6}{x}+\frac{3}{2y}& =& 2\end{array}$

Step 2. Rewrite the equations.

Let $u=\frac{1}{x}$ and $v=\frac{1}{y}$. Then, $x=\frac{1}{u}$ and $y=\frac{1}{v}$.

The given equations can be written as:

$\begin{array}{rcl}4u-3v& =& 0 ...\left(i\right)\\ 6u+\frac{3}{2}v& =& 2 ...\left(ii\right)\end{array}$

Equation (i) can be written as:

$\begin{array}{rcl}4u& =& 3v\\ u& =& \frac{3v}{4} ...\left(iii\right)\end{array}$

Step 3. Determine the value of one variable.

Substitute the above value in equation (ii)

$\begin{array}{rcl}6\left(\frac{3v}{4}\right)+\frac{3}{2}v& =& 2\\ \frac{9v}{2}+\frac{3v}{2}& =& 2\\ \frac{12v}{2}& =& 2\\ 6v& =& 2\\ v& =& \frac{2}{6}\\ v& =& \frac{1}{3}\end{array}$

Step 4. Determine the value of another variable.

Substitute $v=\frac{1}{3}$ in equation (iii).

$$\begin{gathered} u=\frac{3\times \frac{1}{3}}{4} \\ u=\frac{1}{4} \end{gathered}$$

We know that $x=\frac{1}{u}$ and $y=\frac{1}{v}$. So, $x=4$ and $y=3$.

The solution of the given system of equation is $x=4, y=3; \left(4,3\right)$.