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Rule 1: Same quantity can be added to both sides of the equation.
Rule 2: Same quantity can be subtracted from both sides of the equation.
Rule 3: Both sides of an equation can be multiplied by the same number.
Rule 4: Both sides of an equation can be divided by the same non zero number.
Transposition Method When any term or equation is shifted from one side to another side, then the sign of the term changes. This is known as transposition. The changes that takes place are: Addition Changes to subtractionExample : x+8 =11 x = 11-8 (+8 becomes – 8 when shifted) x = 3 Subtraction Changes to Addition.Example: y-9 = 11 y= 11+9 (-9 becomes +9 when shifted) y=20
Multiplication Changes to divisionExample: 5* = 20 5 * x = 20 (x is multiplied by 5 which on shifting 20 divides by 5) x= 20/5 x = 4 Division changes to MultiplicationExample: x/6 = 2 x= 2 X 6 (Division by 6 changes to multiplication by 6 on shifting) x = 12
Application of Linear Equation to word problems Follow the steps given here to solve word problems successfully.
Read the problem thoroughly.
Note what is given and what need to be find out
Denote the unknown quantity with any literal, say x,y,z etc.
Translate the statement of the given problem in to algebraic equation.
Solve the equation for the unknown
The solution of an equation becomes the value of unknown.
Examples Problem : If three less then a number is 10 find the number Solution : let the number be x so x-3 = 10 x = 10+3 x = 13
Problem : A is twice as old as B. Three years ago A’s age was three times as of B find the age of A. Solution : let B’s age be x years, So A’s age = 2x Three years ago B’s Age = x-3 A’s Age = 2x-3 According to given condition 2x-3 = 3(x-3) 2x-3 = 3x-9 2x – 3x = -9 + 3 -x = -6 x = 6 B’s age = 6 years A’s age = 2*6 = 12 Years
mixed word problems Problem : One Number is 6 time the other. Their sum is 140 find the two numbers Solution let the other number be x then first number = 6x According to question 6x +x =140 7x =140 x = 140/7 x =20 First number = 6 *20 = 120 Other number = 20 Thus 120 & 20 are required two numbers
Problem : Gauri has a piggy bank it is full of 1 rupee and 50 paisa coins it contains three times as many 50 paisa coins as 1 rupee coins. The total amount in the piggy bank is 35 rupees how many coins are there of each kind in the piggy bank Solution: let the number of 1 rupee be x then the number of 50 paisa coins = 3x rupees 35 = 35 *100 paisa = 3500 paisa Rs. 1 = 100 paisa and x coins make 100x paisa coins of 50 paisa are 3x X 50 = 150 x paisa Total 250x paisa According to Question 250x = 3500 x = 3500/ 250 x= 14 Number of 1 Rupee Coin = 14 Number of 50 paisa coins = 3x = 3 X 14 = 42.
Problem: The length of a rectangle is 6m less than three times its breadth. Find the length and the breadth of the rectangle if its perimeter is 148m. Solution: Let the breadth of given rectangle be x m . Then , Length =(3x-6)m According to the question, Perimeter of Rectangle = 148m 2(3x-6+x)=148 2(4x-6)=148 8x-12=148 8x=148+12 8x=160 x=160/2=80
Problem: A 100 litre solution of acid and water contains 20 litres of acid. How many water must be added to make the solution 16% acidic? Solution: Let x litres of water be added to make the solution 16 % acidic. Then, the volume of solution = 100+x litres 16% of this is acid i.e, 16/100(100+x) = 20 litres 16(100+x)=20*100 1600+16x=2000 16x=2000-1600=400 x=400/16=25 litres
Problem: A number consists of 2 digits whose sum is 8. if 18 is added to it its digits are reversed. Find the number. Solution : Let the digit of the units place be x. then, the number at tens place=8-x original number = 10(8-x) +x = 80-10x+x =80-9x The reversed number = 10x+8-x =9x+8 According to question, 80-9x+18 = 9x+8 9x+9x=80+18-8 18x=90 x=90/18=5 Digit at units place = x = 5 And, digit at tens place = 8-x =8-5 =3 The number = 10*3+5 =30+5 =35