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# Class 10 maths

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### Class 10 maths

1. 1. SUBMITTED BY:- BHASKAR GUPTA CLASS:- 10th A In the guidance of:- MEERA MEENA
2. 2. Linear Equations Definition of aLinearEquation • Alinear equation in two variable xis an equation that canbe written inthe form ax+ by+c=0, where a ,b and care real numbers and a and b is not equal to0. For example-: 3x+4y+1=0 .
3. 3. Let ax + by + c = 0, where a ,b & c are real numbers such that a and b ≠ O. Then, any pair of values of x and y which satisfies the equation ax +by +c=O,is called asolution of it. APair of Linear Equation In Two Variables canbesolved –: i. ByGraphical Method ii. ByAlgebraic Method SOLUTIONS OF A LINEAR EQUATION
4. 4. GRAPHICAL SOLUTIONS OF A LINEAR EQUATION Let usconsider the following system of linear equations in two variable I. 2x-y=-1 II. 3x+2y=9 Now solution of eachequation are tobe taken by- i. Firstly expressone variable in terms ofother. ii. Now assignany value to one variable andthen determine the value of othervariable. Then plot the equations and determine thesolutions Of system of linear equation in twovariables
5. 5. For Eq.1 2x - y=-1 y =2x +1 Put x=0 y =2(0) +1 y =1 Put x= 2 y =2(2) + 1 y =5 For Eq.2 3x +2y = 9 y =9 - 3x/2 Put x=3 y =9 - 3(3) /2 y =0 Put x=-1 y =9 - 3(-1) /2 y =6 X 0 2 y 1 5 X 3 -1 y 0 6
6. 6. 2 3 4-4 -3 -2 -1 1 -1 -2 -3 -4 -5 -6 ( 2,5) ( -1,6) ( 3,0) 5 4 3 2 ( 0,1)1 X=1 Y=3 Isthe solution for system of equations
7. 7. TYPES OF SOLUTIONS of system of equations UNIQUE SOLUTI ON consistent a1/a2 ≠ b1/b2 Linesintersecting at apoint INFINITE SOLUTIO NS consistent a1/a2 =b1/b2 =c1/c2 Coincident Lines NO SOLUTI ON Non consisten t a1/a2 =b1/b2 ≠ c1/c2 Parallel Lines
8. 8. TYPESOFMETHOD:- Tosolveapairoflinearequationintwo variablebyALGEABRICMETHOD. I. Elimination Method II. Substitution Method III.Cross-Multiplication Method
9. 9. • Make one variable equal in both theequations and then either add or subtract the equations to eliminate thevariable. • Theresulting equation in one variable is solved and then by substituting this valueof the variable in either of the given equation, the value of the other variable is also obtained. ELIMINATIONMETHOD
10. 10. Q.Solveusing the method of Elimination I. 2x +y = 8 II. 2(x +6y = 15) For making the coefficients of xin eq.(1) and eq.(2)equal, we multiply eq.(2) by 2 andget -(3) -(4) 2x +y = 8 2x +12y = 30 Subtracting Eq.(3) from eq.(4), we have 11y =22 y =2 Put this value of y ineq.(1) 2x +2 = 8 2x =6 x=3
11. 11. •Find the value of one variable in the terms of other variable. Substitute it in other equation and we will get value of one of thevariable. Then put the value of variable in any one of the equation and the value of other variable is also obtained. SUBSTITUTIONMETHOD
12. 12. Q.Solveusing the method of Substitution I. x+2y = -1 II. 2x - 3y =12 y =-2 From Eq.(1), we have x=-1 – 2y - (3) Substituting this value of xin eq.(2), wehave 2(-1 – 2y) – 3y =12 - 2 - 4y – 3y =12 - 7y =14 Put this value of y in Eq.(3), wehave x=-1 -2(-2) x=-1 + 4 x=3
13. 13. CROSS-MULTIPLICATIONMETHOD
14. 14. Q.Solveusing the method of cross-Multiplication 5x +3y = 35 2x +4y = 28 5x +3y - 35 = 0 2x +4y - 28 = 0 a1 =5 a2 =2 b1 =3 b2 =4 c1=-35 c2=-28 x = y = 1 (3)(-28)-(4)(-35) (-35)(2)-(-28)(5) (5)(4)-(2)(3) -84 +140 -70 +140 20 – 6 56/14 70/14 14 x=4and y =5