MATHEMATICS : MATHEMATICS Linear Equations In Two Variables
Linear Equation means -: : Linear Equation means -: Degree (Highest power ) is 1.
It has “=“ sign in between. Examples 2x = 9
7z -5 =0
In two variables means : In two variables means It has two unknown quantities. E.g. x and y or t and z etc. Examples 2x - 6y = 23
7z -5p = -5/3
3t/2 + 8s = 6
No of equations required -: : No of equations required -: To be able to solve Linear Equations In Two Variables, we need two equations Note : Graph of a Linear equation is a Line.
Standard Forms : Standard Forms a1x+ b1y + c1 =0 (1)
a2x +b2y + c2 =0 (2)
Where a1,b1 and c1 are coefficients of x,y and constant in 1st equation. a2,b2 and c2 are coefficients of x,y and constant in 2nd equation.
Two lines can be drawn as -: : Two lines can be drawn as -: Intersecting Lines – They meet in one point. Unique solution.
Coincident Lines – They overlap each other. Infinitely many solution.
Parallel Lines – They never meet. No solution
Slide 7 :
Methods to solve -: : Methods to solve -: Elimination
Substitution
Cross Multiplication
Graphically
Elimination Method : Elimination Method Make coefficients of variable( to be eliminated) equal in both equations by multiplying/dividing the equation (s) with a number.
Add/subtract the new equations.
Example of Elimination : Example of Elimination 2x + 3y = 13 (1)
5x + 4y = 22 (2)
To eliminate y, take L.C.M of 3 & 4 =12. so multiply (1) by 4 & (2) by 3
{2x + 3y = 13 } x 4
=> 8x + 12y = 52 (3)
{5x + 4y = 22} x 3
15x + 12y = 66 (4)
Subtract (3) from (4),
Slide 11 : 15x + 12y = 66 (4)
8x + 12y = 52 (3)
(-) (-) (-)
( in case of subtract, change sign)
7x + 0y = 14
x = 14 / 7 =2
Put x= 2 in (1),
2 x 2 + 3y =13
3y = 13 - 4 = 9
y = 9 / 3 =3
Substitution Example : Substitution Example 2x + y = 5 (1)
5x - 3y = - 4 (2)
From (1) ,
y= 5 -2x (3)
Put value of y in (2), ( remember not in (1) )
5x – 3( 5 - 2x) = - 4
5x – 15 + 6x = - 4
11x – 15 = - 4
11x = 11
x = 11 / 11 = 1 put it in (3), y= 5 – 2 x 1 = 3
Cross Multiplication Example : Cross Multiplication Example 7t – 3p = - 4 (1)
3t – 5p = 2 (2)
(-3) (-4) (7) (-3)
b1 c1 a1 b1
b2 c2 a2 b2
(-5) (2) (3) (-5)
t p 1
b1 c2 - b2 c1 = c1 a2 - c2 a1 = a1 b2 - a2 b1
Slide 14 : t p 1
(-3)(2) - (-5)(-4) = (-4)(3) – (2)(7) = (7)(-5) – (3)(-3)
t p 1
- 26 = -26 = -26
So,
t 1 p 1
26 = - 26 AND -26 = -26
> t = -1 and p = -1