Question

Solve $\dfrac{2}{x} + \dfrac{3}{y} = 13$ and $\dfrac{5}{x} - \dfrac{4}{y} = - 2$.

Answer 1

Hint: A linear equation is any equation that can be written in the form.
$ax + by + c = 0$
Where $a$ and $b$ are real numbers and $x,y$ are variables.
This form is sometimes called the standard form of a linear equation in two-variable.
If the equation contains any fractions use the least common denominator to clear the fractions we will do this by multiplying both sides of the equation by the L.C.D.
We shall draw the solution of such pairs of equations which are not linear but can be reduced to linear form by making some suitable substitution.

Complete step-by-step answer:
Given equation $\dfrac{2}{x} + \dfrac{3}{y} = 13$ and $\dfrac{5}{x} - \dfrac{4}{y} = - 2$ .
Here in given equations which are not is form of \[ax + by + c = 0\]. Firstly reduce it and form in \[ax + by + c = 0\]. So, we substitute $\dfrac{1}{x} = b$ and $\dfrac{1}{y} = q$ in both equations we get.
\[ \Rightarrow \]$2b + 3q = 13$ --------(i) and
\[ \Rightarrow \]$5b + 4q = - 2$ --------(ii)
Now, these equations is the form of $ab + 2q + c = 0$
where $p,q$ are variables.
We will use elimination method to solve the equation (i) and (ii)
Firstly, we should multiply (i) equation by $5$ and (ii) equation by $2$,
we get,
\[ \Rightarrow \]$10p + 15q = 65$ --------(iii) equation
And $10p - 8q = - 4$ --------(iv)
Now (iv) equation subtract form (iii) equation
\[ \Rightarrow \]$10p + 15q = 65 $
\[ \Rightarrow \]$10p - 8q = - 4$
\[ \Rightarrow \]$239 = 69$
After subtraction we get on the left hand side $23q$ and on the right hand side $69$.
After solving we get $q = \dfrac{{69}}{{23}}$
\[ \Rightarrow \]$q = 3$
Further but the value of $q$ in (ii) equation we get the value of $p$.
\[ \Rightarrow \]$2b + 3(3) = 13$
\[ \Rightarrow \]$2b + 9 = 13$
\[ \Rightarrow \]\[2b = 13 - 9\]
\[ \Rightarrow \]\[2b = 4\]
\[ \Rightarrow \]\[b = \dfrac{4}{2}\]
\[ \Rightarrow \]\[b = 2\]
Substitute the value of $p$ and $q$ we get,
$\dfrac{1}{x} = p$ and $\dfrac{1}{y} = q$
$i.e.$$\dfrac{1}{x} = 2$ $i.e.$$\dfrac{1}{2} = x$
$i.e.$ $\dfrac{1}{y} = 3$ $i.e.$ $\dfrac{1}{3} = y$

Therefore the value of $x = \dfrac{1}{2}$ and value of $x = \dfrac{1}{3}$.

Note: Be careful while processing simultaneous linear equations from mathematical problems.
- Make sure to remember how to solve simultaneous equations by the method of comparison and method of cross-multiplication.