linear equations in one variable

Linear Equations in one variable : Linear Equations in one variable Class - 8

Introduction : Introduction Algebraic Expressions: 3x , 2x-5 , 3a+b , 4xy+2 , abc+a+b+c , x2+1 , y+y2 Equations: 3x=9 , 4x-3=9 , 6z+10=-2 Linear expressions: 3x , 3x+1 , 5y-7 , 14-5p (the highest power of the variable appearing in the expression is 1) Thus , linear equations in one variable are like 2x-3 =7

In brief: : In brief: An algebraic equation an an equality involving variables. It has an equality sign. The expression on the left of the equality sign is the Left Hand Side (LHS). The expression on the right of the equality sign is the Right Hand Side (RHS). In an equation the values of the expressions on the LHS and RHS are equal. This happens to be true only for certain values of the variable. These values are the solutions of the equation.

Solving equations which have linear expressions on one side and numbers on the other side : Solving equations which have linear expressions on one side and numbers on the other side E.g 1 Find the solution of 5x – 9 =16 Solution: Step 1 – Adding 9 on both the sides 5x – 9 + 9 = 16 +9 5x = 25 Step 2 – Next divide both the sides by 5 5x/5 = 25/5 x = 5 (required solution)

Solve x/3 + 5/2 = -3/2 : Solve x/3 + 5/2 = -3/2 Solution: Step 1 – Transposing 5/2 to the RHS, we get x/3 = - 3/2 - 5/2 x/3 = -8/2 x/3 = -4 Step 2 – Multiply both the sides by 3 x/3 x 3 = -4 x 3 x = -12 (required solution)

Some applications( word problems) : Some applications( word problems)

Sum of two numbers is 95. If one exceeds the other by 15 , find the numbers. : Sum of two numbers is 95. If one exceeds the other by 15 , find the numbers. Solution: Two given conditions are One number exceeds the other by 15 Their sum is 95 If one number is taken to be y , the other number would be y + 15

Solution contd; : Solution contd; From the other given condition , the equation would be: y + ( y + 15 ) = 95 2y + 15 = 95 Transposing 15 to RHS, 2y = 95 – 15 2y = 80 Dividing both the sides by 2 y = 40 This is one number The other number is ( y + 15 ) = 40 + 15 = 55 The desired numbers are 40 and 55.

The ages of Aman and Atul are in the ratio 5:7. Four years later,the sum of their ages will be 56 years. What are their present ages? : The ages of Aman and Atul are in the ratio 5:7. Four years later,the sum of their ages will be 56 years. What are their present ages? Solution : It is given that the ages of Aman and Atul are in the ratio 5:7 Let the present age of Aman be 5x years Let the present age of Atul be 7x years 4 years later, Aman’s age = (5x + 4) years Atul’s age = (7x + 4) years From the given condition: 4yrs later,sum of their ages = 56 yrs

Solution contd; : Solution contd; Therefore , the equation formed would be: (5x + 4) + (7x + 4) = 56 5x + 4 + 7x + 4 = 56 12x + 8 = 56 Transposing 8 on RHS 12x = 56 – 8 12x = 48 dividing by 12 on both sides x = 4 Therefore, present age of Aman = 5x = 5x4 = 20yrs present age of Atul = 7x = 7x4 = 28yrs

Solving equations having variable on both the sides : Solving equations having variable on both the sides Solve 2x – 3 = x + 2 Solution: By transposing 3 to RHS 2x = x + 2 + 3 2x = x + 5 By transposing x to LHS 2x – x = 5 Therefore , x = 5 ( required solution)

Some more applications( word problems ) : Some more applications( word problems )

The digits of a two – digit number differ by 3 . If the digits are interchanged , and the resulting number is added to the original number , we get 143. What can be the original number? : The digits of a two – digit number differ by 3 . If the digits are interchanged , and the resulting number is added to the original number , we get 143. What can be the original number?

Solution: : Solution: Let the digit in the unit’s place be b. From the condition given, the digit in the ten’s place will be (b+3)

Solution contd; : Solution contd; Therefore , original number = 10 ( b+3 ) + b = 10b +30 + b = 11b + 30 After interchanging ,

Solution contd; : Solution contd; Resulting number = 10 (b) + (b+3) = 10b + b + 3 = 11b + 3 From the given condition: (11b+30) + (11b+3) = 143 11b + 30 + 11b + 3 = 143 22b + 33 = 143 By transposing 33 on RHS

Solution contd; : Solution contd; 22b = 143 – 33 22b = 110 by dividing both sides by 22 b = 5 Therefore , original number = 11b + 30 = 11(5) + 30 = 55 + 30 = 85 (required solution)

Reducing equations to simpler forms : Reducing equations to simpler forms E.g. Solve 15(y-4) – 2(y-9) + 5(y+6) = 0 By opening the brackets (15y-60) – (2y-18) + (5y+30) = 0 15y – 60 – 2y +18 +5y +30 = 0 Bringing like terms together 15y – 2y + 5y – 60 + 18 + 30 = 0 18y – 12 = 0 By transposing 12 on RHS 18y = 12 Dividing both the sides by 18 y = 12/18 y = 2/3 (required solution)

Solve 3(5a-7) – 2(9a-11) = 4(8a – 13) - 17 : Solve 3(5a-7) – 2(9a-11) = 4(8a – 13) - 17 By opening the brackets (15a -21) – (18a -22) = (32a – 52 ) – 17 15a – 21 – 18a + 22 = 32a – 52 – 17 by bringing the like terms together 15a – 18a – 21 + 22 = 32a – 69 - 3a + 1 = 32a – 69 by transposing - 3a -32a = -69 – 1 - 35a = -70 Dividing both the sides by (-35) a = 2 (required solution)

Slide 20 : Assignment questions

The perimeter of a rectangle is 18 cm and its breadth is 4 cm . Find the length of the rectangle. : The perimeter of a rectangle is 18 cm and its breadth is 4 cm . Find the length of the rectangle.

Solution: : Solution: Let the length of the rectangle be x cm. Perimeter = 2 (length + breadth) = 2 ( x + 4) The perimeter is given to be 18 cm Therefore , 2(x +4) = 18 2x + 8 = 18 by transposing 8 on RHS 2x = 18 – 8 2x = 10 By dividing both the sides by 2 x = 5 Therefore , length of the rectangle is 5 cm.

Solve 5 ( 3y + 4 ) = - 2 ( 2 - 6y ) : Solve 5 ( 3y + 4 ) = - 2 ( 2 - 6y )

Solution : Solution 5 (3y+4) = -2 (2-6y) 15y + 20 = -4 + 12y by transposing 15y – 12y = -4 – 20 3y = -24 dividing both the sides by 3 y = - 8 (required solution)

ANY DOUBTS? : ANY DOUBTS?

Slide 26 : THANK YOU STUDENTS!