Question

How do we do this system of equations?$1. \: \frac{3}{x - 2y} + \frac{2}{2x + y} = 3$$\frac{2}{x - 2y} - \frac{1}{4x + 2y} = \frac{1}{6}$

Answer 1

SOLUTION

Write out the equation given

$\begin{gathered} \frac{3}{x-2y}+\frac{2}{2x+y}=3\ldots\text{equation 1} \\ \frac{2}{x-2y}-\frac{1}{4x+2y}=\frac{1}{6}\ldots\text{equation 2} \end{gathered}$

Rewrite the equation as

$\begin{gathered} \text{let} \\ x-2y=p \\ 2x+y=q \end{gathered}$

Then, the equation 1 becomes

$\frac{3}{p}+\frac{2}{q}=3$

Then, equation 2 becomes

$\begin{gathered} \frac{2}{x-2y}-\frac{1}{4x+2y}=\frac{1}{6} \\ \frac{2}{x-2y}-\frac{1}{2(2x+y)}=\frac{1}{6} \\ \text{Then} \\ \frac{2}{p}-\frac{1}{2q}=\frac{1}{6} \end{gathered}$

Then, the new equation becomes

$\begin{gathered} \frac{3}{p}+\frac{2}{q}=3 \\ \text{and } \\ \frac{2}{p}-\frac{1}{2q}=\frac{1}{6} \end{gathered}$

Then we solve the system of equation above

$\begin{gathered} \frac{3}{p}=3-\frac{2}{q} \\ \text{The}n \\ p=-\frac{3q}{2-3q} \end{gathered}$

Then , substitute into the second equation, we have

$\begin{gathered} \begin{bmatrix}\frac{2}{-\frac{3q}{2-3q}}-\frac{1}{2q}=\frac{1}{6}\end{bmatrix} \\ \text{Hence },\text{ simplify we have } \\ \begin{bmatrix}\frac{12q-11}{6q}=\frac{1}{6}\end{bmatrix} \end{gathered}$

Then isolate q from the eequation above, we have

$\begin{gathered} \frac{12q-11}{6q}=\frac{1}{6} \\ 6q=6(12q-11) \\ 6q=72q-66 \\ 6q-72q=-66 \\ -66q=-66 \end{gathered}$

Hence

$q=-\frac{66}{-66}=1$

Hence q=1

Substitute to find the value of p, we have

$\begin{gathered} p=-\frac{3q}{2-3q} \\ p=-\frac{3(1)}{2-3(1)}=-\frac{3}{2-3}=-\frac{3}{-1}=3 \end{gathered}$

P = 3

Recall that

$\begin{gathered} p=x-2y \\ q=2x+y \\ \text{Then } \\ 3=x-2y \\ 1=2x+y \end{gathered}$

Hence

$\begin{gathered} x-2y=3\Rightarrow x=3+2y \\ 2(3+2y)+y=1\Rightarrow6+4y+y=1 \\ \text{Then } \\ 5y=1-6 \\ 5y=-5 \\ \text{Hence } \\ y=-1 \end{gathered}$

Thus,

$\begin{gathered} x=3+2y \\ x=3+2(-1)=3-2=1 \\ x=1 \end{gathered}$

Therefore

x=1, y= -1