Linear equation part 3

Some example of Linear equation:

Example 1: The sum of three consecutive multiples of 5 is 555. Find the numbers.

Solution: Alternative Method

We can also assume the three consecutive multiples of 5 to be 5x, 5x + 5 and 5x + 10

Now, we can frame the equation as

            5x + (5x + 5) + (5x + 10) = 555

Or,                               15x + 15 = 555

Or,                                          15x = 555 – 15 = 540

Or,                                              x = 540/15 

Or,                                           x = 36

             

             Therefore, the three required multiples of 5 are 5x = 5 ´ 36 = 180

                                                5x + 5 = 180 + 5 = 185

                                                5x + 10 = 180 + 10 = 190

Example 2: The length of a rectangle is 8 cm more that its breadth. If the perimeter of the rectangle is 108 cm, find the length and breadth of the rectangle.

Solution: Perimeter of the rectangle = 108 cm                                                          (i)

Let the breadth (b) of the rectangle = x cm  

Then the length (l) of the rectangle = (x + 8) cm

Perimeter of the rectangle = 2(l + b)

                                              = 2 (x + x + 8)

= 2(2x + 8)

= 4x + 16                                                                     (ii)

From (i) and (ii), we get

                               4x + 16 = 108

Or,                                  4x = 108 – 16

                                            = 92

 Or ,                                  x = 92/4

                                            = 23

Therefore, the breadth of the rectangle = x =23 cm and the length of the rectangle

= x + 8 = 23 + 8 = 31 cm.

Example 3: Two-third of a number is greater than three-fifths of the number by 6. Find the number.

Solution: Let the number be x

Two-thirds of the number = 2x/3

Three-fifths of the number = 3x/5

According to the given condition,

2x/3 – 3x/5 =6

Multiplying all the terms by 15, which is the LCM of 5 and 3, we get

                         10x – 9x = 90

Or                                  x = 90

Therefore, the required number is 90.

              Check 2/3 of 90 = 60

                           3/5 of 90= 54

And                      60 – 54 = 6

 Hence, the solution is correct.

Example 4: the sum of the digits of a two-digit number is 12. The number formed by interchanging the digits is greater than the original number by 54. Find the original number.

Solution: Let the units digit of the original number be x

Then the tens digit of the original number = 12 – x

The original number in expanded notation is

                              = 10(12 – x) + 1 ´ x

                              = 120 – 10x + x

                              = 120 – 9x

On interchanging the digits the unit digit is now

                               = 12 – x

And the tens digit is now

                                 = x

The number obtained by interchanging the digits in expanded notation

                                 = 10x + 1 ´ (12 – x)

                           = 10x + 12 – x

                           = 9x + 12

Applying the given condition,

The number obtained on interchanging the digits – The original number = 54

ð         9x + 12 – ( 120 – 9x) = 54

ð         9x + 12 – 120 + 9x =54

ð         18x – 108 = 54

ð         18x = 54 + 108 = 162

ð             x = 162/18 = 9

The units digit of the required number

                        = x

                        = 9

The tens digit

                       = 12 – x

                       = 12 – 9 = 3

So, the original number = 39

Check: The original number = 39

Number obtained by interchanging digits = 93

Their difference = 93 – 39 = 54

Hence, the solution is correct.