
Solve the following systems of equations by method of cross-multiplication
$
2ax + 3by = a + 2b \\
3ax + 2by = 2a + b \\
$
Answer
612.6k+ views
Hint – In this question use the concept that if we have two general question of the form $ax + by + c = 0{\text{ and }}px + qy + r = 0$, then using the cross-multiplication rule the relative between x, y and the coefficients of the two equations can be written as $\dfrac{x}{{br - cq}} = \dfrac{y}{{cp - ar}} = \dfrac{1}{{aq - bp}}$. This will help getting the values of x and y.
Complete Step-by-Step solution:
Let us consider two equations
$ax + by + c = 0$
And
$px + qy + r = 0$
So according to cross-multiplication rule the solution of these equation is
$\dfrac{x}{{br - cq}} = \dfrac{y}{{cp - ar}} = \dfrac{1}{{aq - bp}}$
$ \Rightarrow x = \dfrac{{br - cq}}{{aq - bp}},y = \dfrac{{cp - ar}}{{aq - bp}}$
So this is the rule of cross-multiplication.
Now given system of equation are
$2ax + 3by = a + 2b$
$3ax + 2by = 2a + b$
Now convert these system of equation in standard form we have,
$ \Rightarrow 2ax + 3by - \left( {a + 2b} \right) = 0$
And
$ \Rightarrow 3ax + 2by - \left( {2a + b} \right) = 0$
Now solve these system of equation using cross-multiplication rule which is described above,
$ \Rightarrow \dfrac{x}{{3b\left( { - \left( {2a + b} \right)} \right) - \left( { - \left( {a + 2b} \right)} \right)\left( {2b} \right)}} = \dfrac{y}{{ - \left( {a + 2b} \right)\left( {3a} \right) - 2a\left( { - \left( {2a + b} \right)} \right)}} = \dfrac{1}{{2a\left( {2b} \right) - \left( {3b} \right)\left( {3a} \right)}}$
Now simplify the above equation we have,
$ \Rightarrow \dfrac{x}{{ - 6ab - 3{b^2} + 2ab + 4{b^2}}} = \dfrac{y}{{ - 3{a^2} - 6ab + 4{a^2} + 2ab}} = \dfrac{1}{{4ab - 9ab}}$
$ \Rightarrow \dfrac{x}{{{b^2} - 4ab}} = \dfrac{y}{{{a^2} - 4ab}} = \dfrac{1}{{ - 5ab}}$
$ \Rightarrow x = \dfrac{{{b^2} - 4ab}}{{ - 5ab}} = \dfrac{{4a - b}}{{5a}}$
And
$ \Rightarrow y = \dfrac{{{a^2} - 4ab}}{{ - 5ab}} = \dfrac{{4b - a}}{{5b}}$
So the required solution of the given system of equation is
$ \Rightarrow \left( {x,y} \right) = \left( {\dfrac{{4a - b}}{{5a}},\dfrac{{4b - a}}{{5b}}} \right)$
So this is the required answer.
Note – In this question it is being specifically asked to use cross-multiplication, if this would not have been the case then we could have used two other methods to solve, first one is substitution. In this one variable is evaluated in terms of another variable and then substituted into another equation. The other method of elimination in this one variable of both the equations is eliminated by making their coefficients same in both the equations and then performing basic operations of addition or subtraction.
Complete Step-by-Step solution:
Let us consider two equations
$ax + by + c = 0$
And
$px + qy + r = 0$
So according to cross-multiplication rule the solution of these equation is
$\dfrac{x}{{br - cq}} = \dfrac{y}{{cp - ar}} = \dfrac{1}{{aq - bp}}$
$ \Rightarrow x = \dfrac{{br - cq}}{{aq - bp}},y = \dfrac{{cp - ar}}{{aq - bp}}$
So this is the rule of cross-multiplication.
Now given system of equation are
$2ax + 3by = a + 2b$
$3ax + 2by = 2a + b$
Now convert these system of equation in standard form we have,
$ \Rightarrow 2ax + 3by - \left( {a + 2b} \right) = 0$
And
$ \Rightarrow 3ax + 2by - \left( {2a + b} \right) = 0$
Now solve these system of equation using cross-multiplication rule which is described above,
$ \Rightarrow \dfrac{x}{{3b\left( { - \left( {2a + b} \right)} \right) - \left( { - \left( {a + 2b} \right)} \right)\left( {2b} \right)}} = \dfrac{y}{{ - \left( {a + 2b} \right)\left( {3a} \right) - 2a\left( { - \left( {2a + b} \right)} \right)}} = \dfrac{1}{{2a\left( {2b} \right) - \left( {3b} \right)\left( {3a} \right)}}$
Now simplify the above equation we have,
$ \Rightarrow \dfrac{x}{{ - 6ab - 3{b^2} + 2ab + 4{b^2}}} = \dfrac{y}{{ - 3{a^2} - 6ab + 4{a^2} + 2ab}} = \dfrac{1}{{4ab - 9ab}}$
$ \Rightarrow \dfrac{x}{{{b^2} - 4ab}} = \dfrac{y}{{{a^2} - 4ab}} = \dfrac{1}{{ - 5ab}}$
$ \Rightarrow x = \dfrac{{{b^2} - 4ab}}{{ - 5ab}} = \dfrac{{4a - b}}{{5a}}$
And
$ \Rightarrow y = \dfrac{{{a^2} - 4ab}}{{ - 5ab}} = \dfrac{{4b - a}}{{5b}}$
So the required solution of the given system of equation is
$ \Rightarrow \left( {x,y} \right) = \left( {\dfrac{{4a - b}}{{5a}},\dfrac{{4b - a}}{{5b}}} \right)$
So this is the required answer.
Note – In this question it is being specifically asked to use cross-multiplication, if this would not have been the case then we could have used two other methods to solve, first one is substitution. In this one variable is evaluated in terms of another variable and then substituted into another equation. The other method of elimination in this one variable of both the equations is eliminated by making their coefficients same in both the equations and then performing basic operations of addition or subtraction.
Recently Updated Pages
Master Class 11 English: Engaging Questions & Answers for Success

Master Class 11 Maths: Engaging Questions & Answers for Success

Master Class 11 Biology: Engaging Questions & Answers for Success

Master Class 11 Physics: Engaging Questions & Answers for Success

Master Class 11 Accountancy: Engaging Questions & Answers for Success

Class 11 Question and Answer - Your Ultimate Solutions Guide

Trending doubts
Why is there a time difference of about 5 hours between class 10 social science CBSE

What is the median of the first 10 natural numbers class 10 maths CBSE

The draft of the Preamble of the Indian Constitution class 10 social science CBSE

Who gave "Inqilab Zindabad" slogan?

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

Who was Subhash Chandra Bose Why was he called Net class 10 english CBSE

