Question

Solve the following pairs of equations:

${\displaystyle \frac{2}{x}+\frac{3}{y}=\frac{9}{xy}}$

${\displaystyle \frac{4}{x}+\frac{9}{y}=\frac{21}{xy}}$
Where x ≠ 0, y ≠ 0

Answer 1

The given equations are ${\displaystyle \frac{2}{x}+\frac{3}{y}=\frac{9}{xy}}$ and ${\displaystyle \frac{4}{x}+\frac{9}{y}=\frac{21}{xy}}$

Let ${\displaystyle \frac{1}{x}=\text{a}\text{and}\frac{1}{y}=\text{b}}$
Then, we have
2a + 3b = 9ab    ....(i)
4a + 9b = 21ab  ....(ii)
Multiplying eqn. (i) by 2, we get
4a + 6b = 18ab   ....(iii)
Subtracting eqn. (iii) from eqn. (ii), we get
3b = 3ab
⇒ a = 1
⇒ ${\displaystyle \frac{1}{x}}$ = 1
⇒ x = 1
Substituting the value of a in (i), we get
2(1) + 3b = 9(1)b
⇒ 2 + 3b = 9b
⇒ 6b = 2
⇒ b = ${\displaystyle \frac{1}{3}}$
⇒ ${\displaystyle \frac{1}{y}=\frac{1}{3}}$
⇒ y = 3
Thus, the solution set is (1, 3).