Question

Solve the following systems of equations: $\dfrac{2}{x} + \dfrac{3}{y} = \dfrac{9}{{xy}}$, \[\dfrac{4}{x} + \dfrac{9}{y} = \dfrac{{21}}{{xy}}\], \[x \ne 0,y \ne 0\]

Answer 1

Hint: In these types of questions remember to multiply the equations by xy and simplify them to find the values of x and y.

Complete step-by-step answer:
Let $\dfrac{2}{x} + \dfrac{3}{y} = \dfrac{9}{{xy}}$ be the equation 1 and \[\dfrac{4}{x} + \dfrac{9}{y} = \dfrac{{21}}{{xy}}\]be the equation 2
Multiplying equation 1 by xy and equation 2 by xy
So we get $xy \times \dfrac{2}{x} + xy \times \dfrac{3}{y} = xy \times \dfrac{9}{{xy}}$, \[xy \times \dfrac{4}{x} + xy \times \dfrac{9}{y} = xy \times \dfrac{{21}}{{xy}}\]
\[ \Rightarrow \]$ 2y + 3x = 9$, \[4y + 9x = 21\]
Now le t$2y + 3x = 9$ equation 3 and \[4y + 9x = 21\] equation 4
Multiplying equation 3 by 3 and equation 4 by 1
\[ \Rightarrow \]$3 \times (2y) + 3 \times (3x) = 3 \times (9)$, \[1 \times (4y) + 1 \times (9x) = 1 \times (21)\]
\[ \Rightarrow \]$ 6y + 9x = 27$, \[4y + 9x = 21\] now subtracting both the equation
We get $6y + 9x - 4y - 9x = 27 - 21$
\[ \Rightarrow \]\[2y = 6\]
\[ \Rightarrow \]\[y = 3\]
Now using the value of y to find the value of x
Putting the value of y in equation 4
\[ \Rightarrow \]\[4 \times 3 + 9x = 21\]
\[ \Rightarrow \]\[x = 1\]
Therefore the value of x and y are $1,3$

Note: In this question first let $\dfrac{2}{x} + \dfrac{3}{y} = \dfrac{9}{{xy}}$ be the equation 1 and \[\dfrac{4}{x} + \dfrac{9}{y} = \dfrac{{21}}{{xy}}\]be the equation 2 then multiply equation 1 by xy and equation 2 by xy then let $2y + 3x = 9$equation 3 and \[4y + 9x = 21\]equation 4 and then multiply equation 3 by 3 and equation 4 by 1 subtract both the equation and find the value of y then with the help of the value of y find the value of x.