Question

Solve for $xandy$

$\frac{1}{5x}+\frac{9}{y}=4;\frac{3}{x}+\frac{27}{y}=24$

Answer 1

Step 1:Solve the equation in assumption of $\frac{\mathbf{1}}{\mathbf{x}}\mathbf{=}\mathit{A}\mathbf{}\mathit{a}\mathit{n}\mathit{d}\mathbf{}\frac{\mathbf{1}}{\mathbf{y}}\mathbf{=}\mathit{B}$

Given equations,

$$\begin{gathered} \frac{1}{5x}+\frac{9}{y}=4........................................\left(1\right) \\ \frac{3}{x}+\frac{27}{y}=24.......................................\left(2\right) \end{gathered}$$

Giving substitutions

$$\begin{gathered} Let\frac{1}{x}=Aand\frac{1}{y}=B \\ equation\left(1\right)\Rightarrow \frac{A}{5}+9B=4 \\ \Rightarrow A+45B=20.....................\left(3\right) \\ equation\left(2\right)\Rightarrow 3A+27B=24....................\left(4\right) \end{gathered}$$

Step 2: To find A and B

Equation $\left(3\right)$ $\times 3$

$$\begin{gathered} \Rightarrow 3A+135B=60...............................\left(5\right) \\ \Rightarrow 3A+27B=24.................\left(4\right) \end{gathered}$$

Using elimination method

equation (5)$-$(4)

$$\begin{gathered} \Rightarrow 108B=36 \\ B=\frac{36}{108}=\frac{1}{3} \end{gathered}$$

Now,

from equation $\left(3\right)$

$$\begin{gathered} \Rightarrow A=20-45B \\ =20-45\times \frac{1}{3} \\ =20-15=5 \end{gathered}$$

Step 3: To find $\mathit{x}\mathbf{}\mathit{a}\mathit{n}\mathit{d}\mathbf{}\mathit{y}$

We have,

$$\begin{gathered} A=\frac{1}{x}\Rightarrow 5=\frac{1}{x}\Rightarrow x=\frac{1}{5} \\ B=\frac{1}{y}\Rightarrow \frac{1}{3}=\frac{1}{y}\Rightarrow y=3 \end{gathered}$$

Hence the values of $x=\frac{1}{5}andy=3$