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# Linear equation in two variable for class X(TEN) by G R Ahmed

A presentation on Linear equation for class 10

A presentation on Linear equation for class 10

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### Linear equation in two variable for class X(TEN) by G R Ahmed

1. 1. Standard form of Linear Equation in two variable ax + by +c= 0 ● The degree of each variable is 1. ● The variables are added or subtracted. ● a or b are not equal to zero. ● Besides x and y, other commonly used variables are m and n, a and b, and r and s. ● There are no radicals(√) in the equation. ● Every linear equation graphs as a line.
2. 2. Examples of linear equations 2x + 4y =8 6y = 3 – x x = 1 -2a + b = 5 4 7 3 x y   
3. 3. Examples of Nonlinear Equations 4x2 + y = 5 xy + x = 5 s/r + r = 3 4x  The exponent is 2 There is a radical in the equation Variables are multiplied Variables are divided The following equations are NOT in the standard form of ax + by +c=0:
4. 4. x and y -intercepts ● The x-intercept is the point where a line crosses the x- axis. The general form of the x-intercept is (x, 0). The y- coordinate will always be zero. ● The y-intercept is the point where a line crosses the y- axis. The general form of the y-intercept is (0, y). The x- coordinate will always be zero.
5. 5. Finding the x-intercept ● For the equation 2x + y = 6, we know that y must equal 0. What must x equal? ● Plug in 0 for y and simplify. 2x + 0 = 6 2x = 6 x = 3 ● So (3, 0) is the x-intercept of the line.
6. 6. Finding the y-intercept ● For the equation 2x + y = 6, we know that x must equal 0. What must y equal? ● Plug in 0 for x and simplify. 2(0) + y = 6 0 + y = 6 y = 6 ● So (0, 6) is the y-intercept of the line.
7. 7. Find the x and y- intercepts of x = 4y – 5 ● x-intercept: ● Plug in y = 0 x = 4y - 5 x = 4(0) - 5 x = 0 - 5 x = -5 ● (-5, 0) is the x-intercept ● y-intercept: ● Plug in x = 0 x = 4y - 5 0 = 4y - 5 5 = 4y = y ● (0, )is the y-intercept 5 4 5 4
8. 8. Solution: Substitution Method • Given: y +2x= 4…….(i) 7x – 2y = 3………(ii) From (i) y= -2x+4 Putting y in equation(ii) 7x – 2(-2x + 4 = 3 7x + 4x – 8 = 3 11 x = 11 x = 1 Now again putting the value of x in equation (i) y = -2(1) + 4 = 2 • Solution (1, 2)
9. 9. Solution: Eliminating One Variable • Given: 3x + 4y = -10 ……..(i) 5x – 2y = 18……….(ii) (i) x 1= 3x + 4y = -10 (ii) x 2 =(10x – 4y) = 36 Adding these two equation 13x = 26 x = 2 Putting value of x in equation (i) 3(2) + 4y = -10 4y = -16 y = -4 Therefore the solution is (2, -4)
10. 10. Compare the Ratio Graphic representation Algebraic interpretation Linear equations Intersection lines at one point Exactly one solution or unique solution Consistent(Unique solution) Coincident line Infinity solution or many solutions Dependent and consistent Parallel lines No solution Inconsistent
11. 11. A chemist has two solutions which is 50% acid and another which is 25% acid. How much of each should be mixed to make 10 liters of 40% acid solution. Ans: Let 50 % acids in the solution be x Let 25 % of other solution be y Total Volume in the mixture = x + y A.P.Q: x + y = 10 --------(1) A.P.Q: 50%of x + 25%of y = 40% ×10 2x + y = 16 --------(2) So x = 6 & y = 4
12. 12. Q.In an election contested between A and B, A obtained votes equal to twice the no. of persons on the electoral roll who did not cast their votes & the number of person who did not caste vote was equal to twice the majority of A over B. If there were 18000 persons on the electoral roll. How many voted for B.Ans: Let x and y be the no. of votes for A & B respectively. The no. of persons who did not vote = (18000 – x – y) APQ: x = 2(18000 – x – y) => 3x + 2y = 36000 ---------------(1) Again APQ (18000 – x – y) = 2 (x – y) 3x – y = 18000 ----------------(2) On solving we get, y = 6000 and x = 8000 Vote for B = 6000
13. 13. Q. The population of the village is 5000. If in a year, the number of males were to increase by 5% and that of a female by 3% annually, the population would grow to 5202 at the end of the year. Find the number of males and females in the village. APQ, X + Y = 5000………………..(i) APQ X+(5% of x) + y +( 3% of y)=5202 X +x/20 + y +3y/100=5202……………………….(ii) On solving 1 & 2 we get x= 2600 y=2400 No. of males = 2600 No. of females = 2400
14. 14. Q. The income of a and b are in the ratio of 3:2 and their expenditures are in the ratio of 5:3 . If each saves rs 1000 the income of B is? let income of A is 3x and income of B is 2x same way expenditure ratio is 5:3 let expenditure of A is 5y and expenditure of B is 3y saving of A will be 3x-5y =1000 (i) multiplied by 3 saving of B will be 2x-3y =1000 (ii) multiplied by 5 we will get 9x-15y =3000 10x-15y =5000 by elimination method -x =-2000 x =2000 income of A will be 3x=6000 income of A = 6000 income of B will be 2x =2x2000 =4000
15. 15. Q. When the son will be as old as the father today their ages will add up to 126 years. When the father was old as the son is today, their ages add upto 38 years. Find their present ages let the son’s present age be x Father’s age be y Difference in age (y – x) Of this difference is added to the present age of son, then son will be as old as the father now and at that time, the father’s age will be [ y + (y – x)] APQ: [x + (y – x)] + [y (y – x)] = 126 [y + (x – y)] + [x + (x – y)] = 38 Solving we get the value of x and y Q.A says to B "my present age is five times your that age when I was as old as you are now." If the sum of their present ages is 48 years. Find their present ages Let Age of A=x years and age of B= y years APQ-1 x + y =48………………….(i) APQ-2 X = 5{ y –(x –y)}………………….(ii) On solving We get A=30 years and B= 18 years
16. 16. Q.A train covers a certain distance at a uniform speed. if the train had been 10 km/hr faster, it would have taken 2 hour less than the scheduled time. if the train was slower by 10 km/hr, it would have taken 3 hours more than the scheduled time. Write the two equation Let the speed=x and the time taken=y therefore Distance=speed . Time=xy First statement if speed increased by 10 km per hour then distance covered in 2 hours. Distance= speed ×time (X+10)(y-2)=xy Xy -2x +10y -20=xy -2x +10y=20 x -5y=-10………………(i) And second statement if speed decreased by 10 km per hour them distance covered in 03 hours (x-10)(y+3)=xy 3x -10y=30………………….(ii) On Solving we get Distance= 50×12 Distance= 600 kms.