$\frac{3}{\mathrm{x}}$ - $\frac{4}{\mathrm{y}}$ = $\frac{1}{3}$ --- eqn 1

$\frac{2}{\mathrm{x}}$ - $\frac{5}{\mathrm{y}}$ = 1 --- eqn 2

Method I

Using elimination to eliminate y

--- eqn 1 x 5

5 x $\frac{3}{\mathrm{x}}$ - 5 x $\frac{4}{\mathrm{y}}$ = 5 x $\frac{1}{3}$

$\frac{15}{\mathrm{x}}$ - $\frac{20}{\mathrm{y}}$ = $\frac{5}{3}$ --- eqn 3

--- eqn 2 x 4

4 x $\frac{2}{\mathrm{x}}$ - 4 x $\frac{5}{\mathrm{y}}$ = 1 x 4

$\frac{8}{\mathrm{x}}$ - $\frac{20}{\mathrm{y}}$ = 4 --- eqn 4

eqn 4 - eqn 3

$\frac{8}{\mathrm{x}}$ - $\frac{15}{\mathrm{x}}$ - $\frac{20}{\mathrm{y}}$ - - $\frac{20}{\mathrm{y}}$ = 4 - $\frac{5}{3}$

Note: -- = +

$\frac{\mathrm{8\; -\; 15}}{\mathrm{x}}$ - $\frac{20}{\mathrm{y}}$ + $\frac{20}{\mathrm{y}}$ = 4 - $\frac{5}{3}$

Notes:

1. -$\frac{20}{\mathrm{y}}$ + $\frac{20}{\mathrm{y}}$ = 0

2. Every number is being divided by 1

4 = $\frac{4}{1}$

$\frac{-7}{\mathrm{x}}$ = $\frac{4}{1}$ - $\frac{5}{3}$

-$\frac{7}{\mathrm{x}}$ = $\frac{\mathrm{3\; x\; 4\; -\; 1\; x\; 5}}{3}$

-$\frac{7}{\mathrm{x}}$ = $\frac{\mathrm{12\; -\; 5}}{3}$

$\frac{-7}{\mathrm{x}}$ = $\frac{7}{3}$

Cross multiply.

x x 7 = -7 x 3

Divide both sides by 7

x = $\frac{\mathrm{-7\; x\; 3}}{7}$

Note: 7 divides itself 1 time.

x = -1 x 3

x = -3

Substitute x = -3 into --- eqn 1

$\frac{3}{\mathrm{x}}$ - $\frac{4}{\mathrm{y}}$ = $\frac{1}{3}$ --- eqn 1

$\frac{3}{-3}$ - $\frac{4}{\mathrm{y}}$ = $\frac{1}{3}$

-1 - $\frac{4}{\mathrm{y}}$ = $\frac{1}{3}$

Get rid of the fractions by multiplying both sides by the L.C.M of the denominators y and 3

The L.C.M is 3y.

-1 x 3y - $\frac{4}{\mathrm{y}}$ x 3y = $\frac{1}{3}$ x 3y

-3y -12 = y

-12 = y + 3y

-12 = 4y

Divide both sides by 4

y = $\frac{-12}{4}$

y = -3

Method 2

Using substitution

$\frac{3}{\mathrm{x}}$ - $\frac{4}{\mathrm{y}}$ = $\frac{1}{3}$ --- eqn 1

Make y the subject.

Get rid of the fractions by multiplying both sides by the L.C.M of the denominators x, y and 3.

The L.C.M is x x y x 3 = 3xy

3xy x $\frac{3}{\mathrm{x}}$ - $\frac{4}{\mathrm{y}}$ x 3xy = $\frac{1}{3}$ x 3xy

9y - 12x = xy

9y - xy = 12x

y(9 - x) = 12x

Divide both sides by 9 - x

y = $\frac{\mathrm{12x}}{\mathrm{9\; -\; x}}$

Substite y = $\frac{\mathrm{12x}}{\mathrm{9\; -\; x}}$ into --- eqn 2

$\frac{2}{\mathrm{x}}$ - $\frac{5}{\mathrm{y}}$ = 1 --- eqn 2

$\frac{2}{\mathrm{x}}$ - 5 ÷ $\frac{\mathrm{12x}}{\mathrm{9\; -\; x}}$ = 1

Reciprocate the fraction at the right of the ÷ and change ÷ to multiplication (x)

Note: reciprocate means the numerator becomes the denominator and the denominator becomes the numerator.

$\frac{2}{\mathrm{x}}$ - 5 x $\frac{\mathrm{9\; -\; x}}{\mathrm{12x}}$ = 1

$\frac{2}{\mathrm{x}}$ - $\frac{\mathrm{45\; -\; 5x}}{\mathrm{12x}}$ = 1

Get rid of the fractions by multiplying both sides by the L.C.M of the denominators x and 12x

The L.C.M is 12x

12x x $\frac{2}{\mathrm{x}}$ - $\frac{\mathrm{45\; -\; 5x}}{\mathrm{12x}}$ x 12x = 1 x 12x

24 - (45 - 5x) = 12x

24 - 45 + 5x = 12x

-21 = 12x - 5x

-21 = 7x

Divide both sides by 7

x = $\frac{-21}{7}$ = -3

Substitute x = -3 into --- eqn 1

$\frac{3}{\mathrm{x}}$ - $\frac{4}{\mathrm{y}}$ = $\frac{1}{3}$ --- eqn 1

$\frac{3}{-3}$ - $\frac{4}{\mathrm{y}}$ = $\frac{1}{3}$

-1 - $\frac{4}{\mathrm{y}}$ = $\frac{1}{3}$

Get rid of the fractions by multiplying both sides by the L.C.M of the denominators y and 3

The L.C.M is 3y.

-1 x 3y - $\frac{4}{\mathrm{y}}$ x 3y = $\frac{1}{3}$ x 3y

-3y -12 = y

-12 = y + 3y

-12 = 4y

Divide both sides by 4

y = $\frac{-12}{4}$

y = -3