4. Solution by cross Multiplication:
Let us consider a system of equations:
From equation (1) we have
Putting this value of y in eq (2) we get
Or,
Similarly by eliminating x, we get
Here we see that (3) and (4) are linear equation in one variable only.
Provided .
The above results can be written as:
and
and combining these two we get
The above result can be written in a picture form as:
When , then we can not divide equation’s (3) and (4) by to fine the value of x and y. if
then
and
If , then equation (2) reduces to
ka1x + kb1y + kc1 = 0
Or,
Or,
Which is equation (1). Here we see that every solution of equation (1) is a solution of equation (2) hence the system has infinite solutions.
If , equation on (2) reduces to
Or,
Or, k (-c1) + c2 = 0
Or, C2 = KC1
But this is not true. Therefore, no solution exists.
The system of equations
has exactly one or, unique solution if
i. e; if
has no solution if
and has infinite solution if
Subjects | Maths (Part-1) by Mr. M. P. Keshari |
Chapter 1 | Linear Equations in Two Variables |
Chapter 2 | HCF and LCM |
Chapter 3 | Rational Expression |
Chapter 4 | Quadratic Equations |
Chapter 5 | Arithmetic Progressions |
Chapter 6 | Instalments |
Chapter 7 | Income Tax |
Chapter 8 | Similar Triangles |