Question

*Solve*
$\left.\begin{array}{l}{\frac{1}{2(2x+3y)}+\frac{12}{7(3x-2y)}=\frac{1}{2}}\\{\frac{7}{2x+3y}+\frac{4}{3x-2y}=2}\end{array}\right.$
please help me with this question ​

Answer 1

Answer:

Solve*

$\left.\begin{array}{l}{\frac{1}{2(2x+3y)}+\frac{12}{7(3x-2y)}=\frac{1}{2}}\\{\frac{7}{2x+3y}+\frac{4}{3x-2y}=2}\end{array}\right.$

Answer 2

$\sf\huge\bold\pink{Answer:}$

$\left.\begin{array}{l}{\frac{1}{2(2x+3y)}+\frac{12}{7(3x-2y)}=\frac{1}{2}}\\{\frac{7}{2x+3y}+\frac{4}{3x-2y}=2}\end{array}\right. \\ taking \: \frac{1}{2x + 3y} = u \: and \: \frac{1}{3x - 2y} = v$

$\left.\begin{array}{l}{\frac{u}{2}+\frac{12v}{7}=\frac{1}{2}}..i\\{7u+4v=2}\end{array}\right...ii \\7u + 24v = 1 \\ 7u + 4v = 2$