logging in or signing up LINEAR EQUATIONS aSGuest37528 Download Post to : URL : Related Presentations : Share Add to Flag Embed Email Send to Blogs and Networks Add to Channel Uploaded from authorPOINT lite Insert YouTube videos in PowerPont slides with aS Desktop Copy embed code: Embed: Flash iPad Dynamic Copy Does not support media & animations Automatically changes to Flash or non-Flash embed WordPress Embed Customize Embed URL: Copy Thumbnail: Copy The presentation is successfully added In Your Favorites. Views: 3999 Category: Education License: All Rights Reserved Like it (1) Dislike it (1) Added: March 03, 2010 This Presentation is Public Favorites: 0 Presentation Description This is an introduction to linear equations in one variable Comments Posting comment... Premium member Presentation Transcript LINEAR EQUATIONS : LINEAR EQUATIONS IN ONE VARIABLE Slide 2: EQUATION A statement which states that two algebraic expressions are equal is called an equation. LINEAR EQUATIONS IN ONE VARIABLE The equation involving only one variable in first order is called a linear equation in one variable. Slide 3: PROPERTIES OF AN EQUATION If same quantity is added to both sides of the equation, the sums are equal. Thus: x=7 => x+a=7+a If same quantity is subtracted from both sides of an equation, the differences are equal Thus: x=7 => x-a=7-a If both the sides of an equation are multiplied by the same quantity, the products are equal. Thus: x=7 => ax=7a If both the sides of an equation are divided by the same quantity, the quotients are equal. Thus: x=7 => x÷a=7÷a Slide 4: TO SOLVE AN EQUATION 1.To solve an equation of the form x+a=b E.g.: Solve x+4=10 Solution: x+4=10 => x+4-4=10-4 (subtracting 4 from both the sides) => x=6 2.To solve an equation of the form x-a=b E.g.: Solve y-6=5 equal. Solution: y-6=5 => y-6+6=5+6 (adding 6 to both sides) => y=11 1.To Slide 5: 3.To solve an equation of the form ax=b E.g.: Solve 3x=9 Solution: 3x=9 => => x = 3 4. To solve an equation of the form x/a=b E.g.: Solve = 6 Solution: =6 => ×2=6×2 => x=12 Slide 6: SHORT- CUT METHOD (SOLVING AN EQUATION BY TRANSPOSING TERMS) 1. In an equation, an added term is transposed (taken) from one side to the other, it is subtracted. i.e., x+4=10 => x=10-4=6 (4 is transposed) 2. In an equation, a subtracted term is transposed to the other side, it is added. i.e., y-6=5 =>y=5+6=11 (6 is transposed) 3. In an equation, a term in multiplication is transposed to the other side, it is divided. i.e., 3x=12 4. In an equation a term in division is taken to the other side it is multiplied. i.e => y=6×4=24 (4 is transposed) (3 is transposed) Slide 7: TO SOLVE EQUATIONS USING MORE THAN ONE PROPERTY Solve: (1) 3x+8=14 Solution: 3x=14-8 (transposing 8) => 3x=6 => x=6/3 (transposing 3) =>x=2 Slide 8: 2a-3=5 Solution: 2a=5+3 (transposing 3) => 2a=8 => a = 8/2 (transposing2) => a = 4 Slide 9: (3) 5n/8 =20 Solution: 5n=20×8 => n =204×8/51 => n=4×8=32 Slide 10: SOLVING AN EQUATION WITH VARIABLE ON BOTH THE SIDES Transpose the terms containing the variable, to one side and the constants to the other side . E.g.: (1) Solve 10y-3=7y+9 Solution: 10y-7y = 9+3 (transposing 7y to the left & 3 to the right) => 3y = 12 => y = 12/3 => y = 4 Slide 11: (2) Solve 2(x-5) + 3(x-2) = 8+7(x-4) Solution: 2x-10+3x-6=8+7x-28 (removing the brackets) => 5x-16 = 7x-20 => 5x-7x = -20+16 => -2x = -4 => x = -4/-2 => x = 2 Slide 12: SOLVING WORD PROBLEMS A number increased by 8 equal 15. Find the number? Solution: Let the number be ‘x’ Given, the number increased by 8 equal 15. => x+8 = 15 => x = 15-8 => x = 7 Slide 13: A number is decreased by 15 and the new number so obtained is multiplied by 3; the result is 81.Find the number? Solution: Let the number be ‘x’ The number decreased by 15 = x-15 The new number (x-15) multiplied by 3 = 3(x-15) Given 3(x-15) = 81 => 3x-45 = 81 => 3x = 81 + 45 => 3x = 126 => x = => x = 42 Slide 14: 3) A man is 26 years older than his son. After 10 years, he will be three times as old as his son. Find their present ages . Solution: let son’s present age= x years Then father’s age = x+26 After ten years, Son’s age = x+10 Father’s age = x+26+10 =x+36 Given, x+36 = 3(x+10) => x+36 =3x+30 => x-3x =30-36 => -2x =-6 => x = =>x=3 Son’s age = 3 years Father’s age = 3 + 26=29years You do not have the permission to view this presentation. In order to view it, please contact the author of the presentation.