Question

Solve the following system of equations

$\frac{4}{x}+5y=7;\frac{3}{x}+4y=5$

Answer 1

Step1: Using Reciprocal Equations

$\frac{4}{x}+5y=7$; $\frac{3}{x}+4y=5$

Let $\frac{1}{x}=A$

These two equations becomes

$$\begin{gathered} 4\times \frac{1}{x}+5\times y=7 \\ \Rightarrow 4A+5y=7.......................................\left(1\right) \\ 3\times \frac{1}{x}+4\times y=5 \\ \Rightarrow 3A+4y=5.......................................\left(2\right) \end{gathered}$$

Step 2:Using elimination method

Multiplying equation $\left(1\right)$ by $3$ and equation $\left(2\right)$ by $4$ ,we get

$$\begin{gathered} \left(1\right)\times 3\Rightarrow 12A+15y=21...........................................\left(3\right) \\ \left(2\right)\times 4\Rightarrow 12A+16y=20...........................................\left(4\right) \end{gathered}$$

Subtracting (3) and (4),we get,

$$\begin{gathered} \left(3\right)-\left(4\right)\Rightarrow -y=1 \\ y=-1 \end{gathered}$$

Step :3 Substituting $y=-1$ in $\left(1\right)$

we get,

$$\begin{gathered} 4A+5\times -1=7 \\ 4A=7+5 \\ =12 \\ A=\frac{12}{4}=3 \end{gathered}$$

Now,

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

Hence the value of $x=\frac{1}{3}$ and $y=-1$ is the required solution.