Mathematics Chapter 2 Linear Equations In One Variable

3x-2=13
Transposing -2 to RHS, we have
Or                              3x=13+2
                                 3x=15
Or                               x=               (Dividing both sides by 3)
Or                               x=5

5x-9=8
Transposing 9 to RHS, we have
                                         5x=8+9=17
 ∴                                           x=                     (Dividing both sides by 5)

x-2=7
Transposing (-2) to RHS, we have
                                     x=7+2
∴                                    x=9

y + 3 = 10
Transposing 3 to RHS, we have
                                       y = 10 - 3
∴                                      y = 7

Given, x – 2 = 7
By adding 2 to both sides, we get
x – 2 + 2 = 7 + 2 ⇒ x = 7 + 2
⇒ x = 9

6 = z + 2
Transposing 2 to LHS, we have
                                     6 - 2 = z
or                                       4 = z
∴                                        z = 4

∴                             

Dividing both sides by 6, we have:
                                

∴                                

∴                                 

∴                              

∴                             

∴                            

7x-9=16
Transposing (-9) to RHS, we have
                                    7x=16+9=25
Dividing both sides by 7, we have
                                   

∴                                    

14y -8 = 13
Transposing 8 to RHS, we have
                                   14y = 13 + 8 = 21
Dividing both sides by 14, we have
                           
or                               

∴                               

17 + 6p = 9
  
   Transposing 17 to RHS, we have
                                        6p = 9 - 17 = -8
Dividing both sides by 6, we have
                             
                               
or                             

∴                              

∴                                

x = 5

x = -12

x = 20

x = 6

t = 6

Twice of
Let the required number to be subtracted be x.

∴                       
Transposing to RHS, we have
                           
or                           

∴                              
   Thus, the required number is

Let the length of the rectangle = 1 cm

∴          Since, width of the rectangle (b) = cm
and                                   perimeter = 17 cm
   We know that perimeter of a rectangle = 2(l+b) cm
Then,                             
or                                   
or                              
or                                  
or                                              (Transposing to RHS)
or                                           
   Dividing both sides by 2, we have
                                 
                                         

∴                     Length of the rectangle =

Let x be a multiple of 7.
∴       The next two consecutive multiples of 7 are x + 7 and x + 7 + 7 (i.e. x + 14).
∴        Three consecutive multiples of 7 are: x, x+7 and x+14
          According to the condition,
                           x + (x+7) + (x+14) = 63
  ∴                                          3x + 21 = 63
     Transposing 21 to RHS, we have
                                              3x = 63 - 21 = 42
     Dividing both sides by 3, we have
                                           
or                                             x = 14

∴                                          x + 7 = 14 + 7 = 21
and                                     x + 14 = 14 + 14 = 28
    Thus, the three consecutive multiples of 7 are: 14, 21 and 28.

Let the required number be x.
∴       According to the condition, we have
                              
or                                              (Multiply both sides by 2)
                                     
Adding both sides , we have
                                 
or                                          

∴                     The required number

Perimeter of the rectangular pool = 154 m
                           Let breadth = x metres

∴                                  Length = 2(breadth) + 2 metres.
                                             = (2x + 2) metres.
   Since,            perimeter of a rectangle = 2(length + breadth)
∴                       2[(2x+2) + x] = 154
or                           (2x + 2) + x =      (Dividing both sides by 2)
or                             2x + 2 + x = 77
or                                      3x + 2 = 77
   Transposing 2 to RHS,  we have
                                         3x = 77 - 2 = 75
   Dividing both sides by 3, we have
                                                                           
  

∴                                     Breadth = 25 m
                                       Length = 2(25) + 2 = 52 m
                                     

Base of the isosceles triangle =
Let the length of either equal sides = x cm.

∴                 Perimeter of the triangle =
                                                   =
    
But the perimeter of the triangle =

∴                               
  Transposing to RHS, we have
                                             2x =
or                                           
Dividing both sides by 2, we have
                                              
or                                            
     Thus, the required length =

Let the smaller number = x.

∴          The other number = x +15.
  According to the condition, we have
                x + [x+15] = 95
                     2x + 15 =95
or                              2x = 95 - 15 = 80
   Dividing both sides by 2, we have
                                       x = 80 + 2 = 40

∴

Let consecutive numbers be, x, x+1 and x+2.
According to the condition, we have
          x + [x+1] +[x+2] =51
or                x + x + 1 + x + 2 =51
or                              3x + 3 = 51
   Transposing 3 to RHS, we have
                                3x = 51 - 3 = 48
  Dividing both sides by 3, we have:
                                  or  x =16
Now                         x + 1 = 16 + 1 = 17
and                          x + 2 = 16 + 2 =18   
    Thus, the required three consecutive numbers are: 16, 17 and 18.

Let the three multiples of 8 are: x, x + 8 and x + 8 + 8 = x + 16.
According to the condition, we have
         [x] + [x + 8] + [x + 16] = 888
or            x + x + 8 + x + 16 =   888
or                           3x + 24 =   888
  
   Transposing 24 to RHS, we have
                               3x = 888 - 24 = 864
  
   Dividing both sides by 3, we have
                                   
                                          x = 864 3 = 288

∴                                    x + 8 = 288 + 8 =296
and                                 x + 16 = 288 + 16 = 304
     Thus, the required multiples of 8 are: 288, 296 and 304.

Let the integers be x, (x+1) and (x+2).

∴  According to the condition, we have
    
or                 2x + 3x +3 + 4x + 8 =74
or                            9x +11 = 74
or                                   9x = 74 - 11 = 63
    Dividing both sides by 9, we have
                                         x = 639 = 7

∴  The required integers are:
                                   7, (7+1) and (7+2) or 7, 8, and 9.

Since, ages of Rahul and Harron are in the ratio of 5:7.
Let their present ages are: 5x and 7x.

∴ 4 years later:     Age of Rahul = (5x + 4) years
                      Age of Haroon = (7x + 4 ) years
  According to the condition, we have
     (5x + 4) + (7x + 4) = 56
or        5x + 4 + 7x + 4 = 56
or              12x + 8 = 56
   Transposing 8 to RHS, we have
                                  12x = 56 - 8 = 48
  
Dividing both sides by 12, we have
                                    

∴       Present age of Rahul = 5x = 5 4 = 20 years
          Present age of Haroon = 7x = 7 4 = 28 years
 

Since, [number of boys]: [number of girls] = 7 : 5
  Let the number of boys = 7x
and     the number of girls = 5x
     According to the condition, we have
                                       7x = 5x +8
    Transposing 5x to LHS, we have
                                   
                                     7x - 5x =8
or                                           2x = 8
    Dividing both sides by 2,
                                           x = 8 2 = 4

∴                     Number of boys = 7x = 7 4 = 28
                      
                       Number of girls = 5x = 5 4 = 20

∴                   Total class strength = 28 + 20 = 48  students

Let the two numbers be 5x and 3x.
According to the condition, we have
                                  5x - 3x =18
or                                       2x = 18
  
         Dividing both sides by 2, we have
                                    
or                                    

∴                                      5x = 5 9 = 45
and                                    3x = 3 9 = 27
∴     The required numbers are 45 and 27.

Let the age of Baichung = x years

∴             His father's age = (x + 29) years
and      His grandfather's age = (x + 29 +26) years
                                       = (x + 55 ) Years
      According to the condition, we have
      [x] + [x+29] + [x+55] = 135 years
or             x + x + 29 + x + 55 = 135
or                         3x + 84 = 135
   Transposing 84 to RHS, we have
                            3x = 135 - 84
or                          3x = 51
   Dividing both sides by 3, we have
                                x = 51 3 = 17
Now,       Baichung's age = 17 years
           His father's age = 17 + 29 = 46 years
       His grandfather's age = 46 + 26 = 72 years

Let Ravi's present age = x years
    4 times the present age = 4 x = 4x years
15 years from now, Ravi's age = (x + 15) years
According to the condition, we have
                   x + 15 = 4x
Transposing 15 and 4x, we have
                                  x - 4x = -15
or                                    -3x = -15
  Dividing both sides by (-3), we have
                                      x = (-15) (-3) = 5

∴          Ravi's present age = 5 years.

Let the required rational number be x.

∴   According to the condition, we have
                     
or                            
  Transposing to RHS, we have
                  
Dividing both sides by , we have
                         
                               

∴          The required integer =

Let the number of:
                   Rs 100-notes = 2x
                     Rs 50-notes = 3x
                      Rs 10-notes = 5x

∴                 Value of Rs 100-notes = 2x Rs 100 = Rs 200x
                   Value of Rs 50-notes = 3x Rs 50 = Rs 150x
   Value of Rs 10 we have 10-notes = 5x Rs 10 = Rs 50x
 
According to the condition, we have
       Rs 200x + Rs 150 x + Rs 50x = Rs 4,00,000
or                           200x + 150x +50x = 400000
or                                        400x = 400000
    Dividing both sides by 400, we have
                                            x = 400000 400 = 1000

∴                                       2x = 2  1000 = 2000
                                         3x = 3 1000 = 3000
                                         5x = 5 1000 = 5000
Thus, the number of
                                  Rs 100-notes =2000
                                      Rs 50-notes = 3000
                                      Rs 10-notes = 5000
   

Let the number of Rs 5 coins = x

∴         The number of Rs 2 coins = 3x

∵                  Total number of coins = 160

∴                   Number of Re 1 coins = 160 - 3x - x
                                                 = (160 - 4x)
Now, value of:
                                Rs 5- coins = Rs 5 x = Rs 5x
                                Rs 2-coins = Rs 2 3x = Rs 6x
                                Rs 1-coins = Rs 1 (160-4x) = Rs (160-4x)
According to the condition, we have
                        5x + 6x + (160 - 4x) = 300
or                          5x + 6x + 160 - 4x = 300
or                                         7x +160 = 300
   
     Transposing 160 to RHS, we have
                                       
                                             7x =300 - 160 =140
     Dividing both sides by 7, we have
                                             

∴ Number of:
                                  Rs 5-coins = 20
                                  Rs 2-coins = 320 =60
                                  Rs 1-coins = 160 - (204) = 80

Let the number of winners =x

∴  Number of participants who are not winners = (63 - x)

∴  Prize money given to:                Winners = xRs 100 = Rs 100 x
                 Non-winner participants = Rs 25 (63-x)
                                                = Rs 25 63 - Rs 25x
                                                 = Rs 1575 - Rs 25x
According to the condition, we have
                                100x + 1575 - 25x = 3000
or                                          75x = 3000 - 1575       [Transposing 1575 to RHS]
or                                           75x = 1425
or                                              x = 1425 75 = 19
    Thus, the number of winners = 19

32 and 42

18, 19 and 20

110, 121 and 132

110 and 44

Rs 50 6
Rs 2010
Rs 109

We have,
           2x - 2 = x + 3
Transposing (-2) to RHS, we have
                            2x = x + 3 + 2
or                           2x = x +5
  Transposing x to LHS, we have
                             2x - x = 5

∴                                  x = 5 

We have
             
Transposing to RHS, we have

                   
Transpiring 5x to LHS, we have
               
or             
or               
   Dividing both sides by , we have
                          
                             

∴                             

3x = 2x + 18
Transposing 2x from RHS to LHS, we have
                        3x - 2x = 18
or                             x = 18

5t - 3 = 3t - 5
Transposing (-3) to RHS, we have
                            5t = 3t - 5 + 3
or                          5t = 3t - 2
Transposing 3t to LHS, we have
                              5t -3t = -2
or                                 2t = -2
Dividing both sides by 2, we have
                               
or                                 t = -1

5x + 9 = 5 + 3x
Transposing 9 to RHS, we have
                                    5x = 5 + 3x - 9
or                                  5x = -4 + 3x
Transposing 3x to LHS, we have
                           5x - 3x = -4
or                                 2x = -4
Dividing both sides by 2, we have
                              
or                              x = -2

4z + 3 = 6 + 2z
Transposing 3 to RHS, we have
                                 4z = 6 - 3 + 2z
or                               4z = 3 + 2z
Transposing 2z to LHS, we have
                                  4z - 2z =3
or                                      2z = 3
Dividing both sides by 2, we have
                                     

2x - 1 = 14 - x
Transposing (-1) to RHS, we have
                             2x = 14 -x +1
or                           2x = 15 -x
Transposing (-x) to LHS, we have
                              2x + x = 15
or                                    3x = 15
Dividing both sides by 3, we have
                                    

∴                                    x = 5

8x + 4 = 3 (x-1) +7
or                          8x + 4 = 3x - 3 + 7
or                          8x + 4 = 3x + 4
Transposing 4 to RHS, we have
                             8x = 3x + 4 - 4
Transposing 3x to LHS, we have
                              8x - 3x =0
or                                  5x = 0

∴                                    x = 0

∴                                      x = 40

∴                         

   ∴                            

∴                                    

x = 5

x = 5

Let the digit at unit place be 'x'.
∴                    The digit at the tens place = (x+5)

∴                                original number =10(x+5) + x
   
        With interchange of digits, the new number = 10x + (x + 5)
  
     Now, According to the condition, we have
             [original number] + [New number] = 99
or         [10(x + 5) + x] + [10x + (x + 5)] = 99
or         [10x + 50 +x] + [10x + x + 5] = 99
or                        11x + 50 + 11x + 5 = 99
or                                       22x + 55 = 99
    Transposing 55 to RHS, we have
                                                  22x = 99 - 55 = 44
    Dividing both sides by 22, we have
                                               x = 44 22 = 2

∴                                              x = 2
i.e.                                Unit place digit = 2

∴                                 Tens place digit = 2 + 5 = 7
                            Thus the original number = 72.

Let the number be 'x'.

∴       Subtracting we get
        
       According to the condition, we have
                       
                                         
or                                                8x - 20 = 3x
    Transposing 20 to RHS, we have
                                            
                                        8x = 3x +20
   
    Again transposing 3x to LHS, we have
                                              8x - 3x = 20
or                                                    5x = 20
or                                                 
                   Thus, required number = 4

                                                 
  

                                   Let positive number = 5x
                                   The other number = x

∴
      On adding 21 to both number:
      We get (5x+21) and (x+21)
      According to the condition, we have
                                          
or                                               2x + 42 = 5x +21
  Transposing 42 to RHS, we have
                                             2x = 5x + 21 - 42
  Transposing 5x to LHS, we have
                                           2x - 5x = -21
or                                             -3x = -21
   Dividing both sides by (-3), we have
                                            
or                                               x = 7
and                                             5x = 5  7 =35
     Thus, the required numbers are 7 and 35.    
                                        
     

    

                Let the unit's digit = x

∴                      The ten's digit = (9-x)

∴                The original number = 10(9-x) + x
                                            = 90 - 10x + x
                                            = 90- 9x
   On interchanging the digits, the new number = 10x + (9 - x)
                                                             = 10x + 9 - x
                                                             = 9x + 9
 According to the condition, we have
                         [New number] = [Original number] + 27
or                                 9x + 9 = 90 - 9x + 27
or                                 9x + 9 = 117 - 9x
or                                        9x = 117 - 9 -9x      [Transposing 9 to RHS]
or                                   9x + 9x = 108                [Transposing (-9x) to LHS]
or                                        18x = 108
   Dividing both sides by 18, we have
                                        

∴               The original number = 90 - (9 6)
                                           = 90 - 54 = 36

             Let the digit at unit place = x

∴                 The digit at tens place = 3x
                         The number = 10(3x) + x
                                         = 30x + x
                                         = 31x
   On interchanging the digits, the number = 10x + 3x
                                                       =13x
   According to the condition, we have
                                             31x + 13x = 88
or                                                    44x = 88
    Dividing both sides by 44, we have
                                     
or                                        x = 2

∴                               The number = 31 2
                                                 = 62

              Let Shobo's present age = x years
∴                 Mother's present age = 6x years

    After 5 years:           Shobo's age = (x + 5) years
                               Mother's age = (6x +5) years
    According to the condition, we have
                    (Mother's present age) = (Shobo's age after 5 years)
i.e.                                      
or                                         
or                                              2x = x + 5
or                                              2x - x =5            (Transposing x to LHS)
or                                                    x = 5
∴                    Shobo's present age = 5 years
                      Mother's present age = 6 5 = 30 years

Let the length = 11x metres
and      breadth = 4x metres

∴         Perimeter = 2(Length + Breadth)
                        = 2(11x + 4x)
                        = 2 15x
                         = 30x

∴          Cost of fencing = Rs 100 30x
                                = Rs 3000x
  But the cost of fencing is Rs 75000.

∴                              3000x = 75000.
or                                    

∴                                 Length = 11 25 = 275 metres
                                   Breadth = 4 25 = 100 metres

Let the length of cloth for trousers = 2x metres

∴    The length of cloth for shirts = 3x metres
              Cost of trouser's cloth = 2x Rs 90
                                           = Rs 180 x
              Cost of shirt's cloth = 3x Rs 50
                                        = Rs 150 x
S.P. of trouser's cloth at 10% profit = Rs
                                              
                                               = Rs 198 x
S.P. of shirt's cloth at 12% profit = Rs
                                          
                                            = Rs 168x

∴                               Total S.P. = Rs 198x + Rs 168x
                                              = Rs 366x
  But the total                      S.P. = Rs 36600

∴                                      366x = 36600
or                                             [Dividing both sides by 366]
                                                = 100

∴                                         2x = 2 100 = 200
   Thus, he bought cloth for trousers = 200 metres.

Let the number of deer = x

∴ Number of deer:     Grazing in the field =
                            Playing nearly = (Remaining no of deer)
                                               =
                                               =  
                                               =
                                 Drinking water = 9

∴                                   
  Transposing 9 to RHS, and x to LHS, we have
                             
or                           
or                                    
   Multiply by 8 on both sides we have
  
                              
or                               -x = -72
or                               x = 72
     Thus, the number of deer in herd = 72

Let the present age of granfather = x years

∴       Present age of granddaughter = years
  According to the condition, we have
      [Present age of granddaughter] + 54 = x
or                                      
   Transposing 54 to RHS and x to LHS, we have
                
                  
or                 
or                    
or                                [Multiplying both sides by 10]
or                                    [Dividing both sides by -9]
  
or                      
and                      

∴  Present age of grandfather = 60 years
   Present age of granddaughter = 6 years

Let present age of Son = x years

∴          Present age of Aman = 3x years
   Ten years ago           Son's age = (x-10) years
                            Aman's age = (3x-10) years
  According to the condition, we have
      [Son's age (10 years ago)]5 = [Aman's age (10 years ago)]
or                        (x -10) 5 = (3x -10)
or                          5x - 50 = 3x - 10
   Transposing (-50) to RHS and 3x to LHS, we have
                                     5x - 3x = -10 + 50
or                                         2x = 40
or                                             [Dividing both sides by 2]
and                                        3x = 3 20 = 60

∴               Son's present age = 20 years
 
              Aman's present age = 60 years

47 ( or 74)

85 (or 58)

93 (or 39)

10 years, 20 years

60m; 40 m

∴                       x = 1
Check:         LHS =
                              
                     RHS =
                                         

∴                          LHS = RHS

∴                                 
Check:                    LHS =
                                               
                                                   =
                                                  
                                                   =
                            
                            RHS =
                               
                                   =
                                   =
                            
 ∴                          LHS = RHS

∵  LCM of 2, 4 and 6 = 12

∴   Multiplying both sides by 12, we have
     
or                         6n -9n +10n = 252
or                                  7n = 252
or                                

∴                                      n = 36
Check:                            LHS =
                                            =
                                           
                                            =
                                           
                                            = 18 -27 + 30
                                            = 21 = LHS

∴                                        LHS = RHS

∵   LCM of 3, 6 and 2 is 6.

∴    Multiplying both sides by 6, we have
           6x+67-6 = 6-6
or               6x + 42 - 16x = 17 - 15x
or               (6-16)x + 42 = 17 - 15x
or                         -10x + 42 = 17 - 15x
Transposing 42 to RHS and -15x to LHS, we have
      
                          -10x + 15x = 17 - 42
or                               5x = -25
or                                   (Dividing both sides by 5)

∴                                x = -5
Check:                       LHS = x + 7 -
                                      = (-5) + 7 -
                                     = -5 +7 +
                                     = 2 + =
                              RHS =
                                    
                                    =
                                    =
                                     = 

∴                            LHS = RHS

∵  LCM of 3 and 5 is 15.

∴   Multiplying both sides by 15, we have
                   
or                     5(x-5) = 3(x-3)
or                      5x - 25 = 3x - 9
  Transposing (-25) to RHS and 3x to LHS, we have
                    
                       5x -3x = -9 + 25
or                            2x = 16
or                                       (Dividing both sides by 2)

∴                               x = 8
   Check:                    LHS =
                                     
                                      =
                                     
                                      =
                               
                                RHS =
                                      
                                       =

∴                                LHS = RHS

∵     LCM of 4 and 3 is 12.

∴        Multiplying both sides by 12, we have
         
or         3(3t-2) - 4(2t+3) = (4 2) - 12t
or            9t - 6 - 8t - 12 = 8 - 12t
or         (9-8)t - (6+12) = 8 - 12t
or                       t-18 = 8 - 12t
Transposing -18 to RHS and -12t to LHS, we have
                       t +12t = 8 + 18
or                        13t = 26
or                                             (Dividing both sides by 2)

∴                              t = 2
Check:                  LHS =
                                 =
                               
                                  =
                                
                                  =
                                 
                                   =
                         RHS =
                              
                             =
                             
                             =
                    
 

∴                       LHS = RHS

∴  Multiplying both sides by 6, we have
           6 m - 6
or                  6m -3(m-1) = 6 - 2(m-2)
or                      6m - 3m + 3 = 6 -2m + 4
or                     (6-3)m + 3 = (6+4) - 2m
or                            3m + 3 = 10 - 2m
Transposing 3 to RHS and -2m to LHS, we have
                           3m + 2m = 10 - 3
or                                  5m =7
or                                        (Dividing both sides by 5)
Check:                       LHS =
                                      =
                                      =
                                      =
                                RHS =
                                      =
                                     
                                      =
                                      =

∴                                LHS = RHS

3(t-3) = 5(2t+1)
or                              3t - 9 = 10t + 5
Transposing (-9) to RHS and 10t to LHS, we have
                                  3t -10t = 5 + 9
or                                     -7t = 14
or                                 t =   (Dividing both sides by -7)

∴                                 t = -2

15(y-4) -2(y-9)+5(y+6) = 0
Opening the brackets, we have
  15y -60 - 2y + 18 + 5y + 30 =0
Collecting the like terms,
 
(15-2+5)y + (-60 + 18 +30) =0
or                     18y + (-12) =0
Transposing (-12) to RHS, we have
                                18y = 12
or                                  (Dividing both sides by 18)

∴                                    

3(5z - 7) - 2(9z - 11) = 4(8z - 13) - 17
Opening the brackets, we have
             15z - 21 - 18z + 22 = 32z -52 - 17
Collecting the like terms on both sides,
          (15-18)z + (-21 + 22) = 32z + (-52 -17)
or                       -3z + 1 = 32z -69
Transposing 1 to RHS and 32z to LHS, we have
                         -3z - 32z = -69 - 1
or                            -35z = -70
or                                    (Dividing both sides by -35)

∴                             z = 2

0.25(4f-3) = 0.05(10f -9)
  Opening the brackets, we have
         0.25 4f - 3 0.25 = 0.05 10f - 0.05 9
or     
or                
Transposing to RHS and to LHS, we have
               
or                   
or                             (Dividing both sides by )
or                        

∴                                                

6

15

-4

2

1

32 years and 40 years

Substituting x = 4 in  we get
                   LHS = x + 7 -
                          = 4 + 7 -
                          = 11 -
                          =
                      
                      RHS =
                             =
                             =
                            
                             =
Since,                       LHS = RHS

∴      x = 4 is a solution of the given equation.

We have

LCM of 3, 4, 2 and 4 is 60.

∴ The given can be expressed as:

           
             
               
               
                       
                                  
     Thus, is the required solution.

We have
 
         
                 3x -4 + 44 - 4x -3 = 2x + 4
 
                    3x -4x -2x = 4 + 3 - 44 + 4
                           3x -6x = 11 - 44
                             -3x = -33
                              x = 11

We have
   Since, LCM of 3 and 5 is 15.

∴      The given equation is expressed as:
         
         
           35x + 70 - (51-9x) = 90x - (20x + 10) - 75
              35x + 70 -51 + 9x = 90x - 20x - 10 - 75
            35x + 9x - 90x + 20x = -10 -75 + 51 - 70
                       64x - 90x = -155 + 51
                             -26x = -104
                              
   Thus, x = 4 is the solution of the given equation.
  

we have
      By cross multiplication, we have:
                             3[x - (7 - 8x)] = 2[9x - (3 + 4x)]
                          3x - 3(7 - 8x) = 2 9x -2(3 + 4x)
                        3x - 21 + 24x = 18x - 6 - 8x
                    3x + 24x - 18x + 8x = -6 + 21
                             35x - 18x = 15
                                   17x = 15
                                    


 

We have 
By cross multiplication, we get:
                    (2 + x)(7 - x) = (5 - x)(4 + x)
             2(7 - x) + x(7 - x) = 5(4 + x) - x(4 + x)
               14 - 2x + 7x - x²=20 + 5x - 4x -x²
        -x² + x² - 2x + 7x - 5x + 4x = 20 - 14
                 -7x + 7x + 4x = 6
                           4x = 6
                            x = or
  Thus, the solution of the given equation is .

Let the number be x.
       Since, we have [The number] - 65 = 9] - [The number]
                         x - 65 = 91 -x
                         x + x = 91 + 65
                             2x = 156
                               
                                      
                                        = 78
    Thus, the required number is 78.

Let the denominator of the fraction be x.

∴            Numerator = x - 2

∴           The fraction =
    Since, it becomes when 1 is added to its denominator
i.e.            
   By cross multiplication, we have
    
                     2(x - 2) = x + 1
                   2x - 4 = x + 1
                  2x - x = 1 + 4
                           x = 5

∴             Fraction

Let my present age be x years.

∴  After 24 years, my age will be (x + 24) years.
   4 years ago, my age was (x - 4) years.
   According to the given condition, we have
                          (x + 24) = 3(x - 4)
                        x + 24 = 3x - 12
                        x - 3x = - 12 - 24
                            -2x = -36
                             
   Thus, my present age is 18 years.

Let one of the numbers be x.

∴ The other number = (30 - x)
   According to the condition, we have
                            [∴ The ratio number is 2:3]
                     3x = 2(30 - x)         [By cross multiplication]
                      3x = 60 - 2x
                 3x + 2x = 60
                     5x = 60
                      

∴                        30 - x = 30 - 12 = 18
   Thus, the required numbers are 12 and 18.