Question

can anyone explain me a pair of linear equations class 10th

Answer 1

I'll start with a practical answer, but what I believe is the true answer is down at the bottom.

One main point of linear algebra is that linear equations are easy to solve. Let me explain.

Equations arise naturally all the time, for example when we try to model something in the world mathematically. Sometimes the equations will involve special functions of the parameters (trigonometric, exponential, logarithm, etc). But more often than not the equations are polynomials -- they involve only performing addition, subtraction, multiplication, and division operations upon the parameters, together with scaling various quantities by constants. This is quite reasonable, as a big part of why mathematics is a useful tool for describing the world is that the basic operations of arithmetic already yield enormous versatility.could think of it either geometrically or analytically:-
Geometrical Proof:-
a1x + b1y + c1 = 0 ; a2x + b2y + c2 = 0 both represent equation of straight lines. So for 2 lines in 2-D space:-
1.) They could be either co-incident (The case of infinite solutions) OR
2.) They could be parallel (The case of no solution) OR
3.) They could be intersecting (The case of unique solution)

In the case 1.) : The equations of the two lines must look alike (For ex. 2x+3y=8 ; 4x+6y=16) and one of them may be multiplied(or divided) by a non-zero number to get the other(Since an equation is not altered when multiplied or divided by a non-zero number).In the case 2.) : The equations must have same slope (Since they are parallel) and differ only by the term C (c1 or c2) (Slope of Ax+By+C=0 is (-A)/B) (For ex. 3x+4y=7 ; 6x+8y=5 <=> 3x+4y=2.5 )

In the case 3.) : The equations must be non-parallel , thus having different slopes (For ex. x+2y=5 ; 6x+9y=7).Analytical Proof:-
The solution of a1x + b1y + c1 = 0 ; a2x + b2y + c2 = 0 is given as :-
(a1*b2 - a2*b1)*x = (b1*c2 - b2*c1) ==>(1)
(a1*b2 - a2*b1)*y = (a2*c1 - a1*c2) ==>(2)

Let D = (a1*b2 - a2*b1) , D1 = (b1*c2 - b2*c1) , D2 = (a2*c1 - a1*c2)