Answer

Verified

384.6k+ views

**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.

Recently Updated Pages

The base of a right prism is a pentagon whose sides class 10 maths CBSE

A die is thrown Find the probability that the number class 10 maths CBSE

A mans age is six times the age of his son In six years class 10 maths CBSE

A started a business with Rs 21000 and is joined afterwards class 10 maths CBSE

Aasifbhai bought a refrigerator at Rs 10000 After some class 10 maths CBSE

Give a brief history of the mathematician Pythagoras class 10 maths CBSE

Trending doubts

Difference Between Plant Cell and Animal Cell

Give 10 examples for herbs , shrubs , climbers , creepers

Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE

Fill the blanks with the suitable prepositions 1 The class 9 english CBSE

Name 10 Living and Non living things class 9 biology CBSE

Change the following sentences into negative and interrogative class 10 english CBSE

Write a letter to the principal requesting him to grant class 10 english CBSE

Select the word that is correctly spelled a Twelveth class 10 english CBSE

Fill the blanks with proper collective nouns 1 A of class 10 english CBSE