Solve:   fraction numerator x plus 3 over denominator 6 end fraction plus 1 space equals space fraction numerator 6 x minus 1 over denominator 3 end fraction


fraction numerator x plus 3 over denominator 6 end fraction plus 1 space equals space fraction numerator 6 x minus 1 over denominator 3 end fraction

Multiplying both sides by 6, we have             (∵   LCM of 3 and 6 is 6)

         6 cross times fraction numerator x plus 3 over denominator 6 end fraction plus 6 cross times 1 equals 6 cross times fraction numerator 6 x minus 1 over denominator 3 end fraction

or             x + 3 + 6 = 2(6x-1)

or                x + 9 = 12x - 2

    Transposing 9 to RHS and 12x to LHS, we have
                
                       x - 12x = -2 - 9
or                    -11x = -11

or                     x equals fraction numerator negative 11 over denominator negative 11 end fraction equals 1   (Dividing both sides by (-11))

∴                       x = 1

Check:         LHS = fraction numerator x plus 3 over denominator 6 end fraction plus 1 equals fraction numerator 1 plus 3 over denominator 6 end fraction plus 1

                               equals space 4 over 6 plus 1 equals space 2 over 3 plus 1
equals fraction numerator 2 plus 3 over denominator 3 end fraction equals 5 over 3

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                                          equals fraction numerator 6 minus 1 over denominator 3 end fraction equals 5 over 3

∴                          LHS = RHS






           


       

 

 




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If 1 is subtracted from both sides of x + 1 = 5, then we have:

  • x = 5 – 1

  • x + 2 4

  • x = 1 – 6


A.

x = 5 – 1

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If x – 1 = 6, then x = ?

  • x = 6 + 1

  • x = 6 - 1

  • x = 1 – 6


A.

x = 6 + 1

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If 1 is added to both sides of x – 1 = 18, then we have:

  • x – 1 + 1 = 18 – 1

  • x – 1 + 1 = 18 + 1

  • x – 1 + 1 = 18 – 2


B.

x – 1 + 1 = 18 + 1

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The digits of a two digit number differ by 5. If the digits are interchanged and the resulting number is added to the original number, then we get 99. Find the original number.


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 divided by22 = 2

∴                                              x = 2

i.e.                                Unit place digit = 2

∴                                 Tens place digit = 2 + 5 = 7

                            Thus the original number = 72.



                                               

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