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Linear Equations In One Variable

Question
CBSEENMA8002313
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Solve the following linear equation:

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Solution
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∵  LCM of 3 and 5 is 15.

∴   Multiplying both sides by 15, we have
                    15 cross times fraction numerator x minus 5 over denominator 3 end fraction equals 15 cross times fraction numerator x minus 3 over denominator 5 end fraction
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                              x equals 16 over 2         (Dividing both sides by 2)

∴                               x = 8
   Check:                    LHS = fraction numerator x minus 5 over denominator 3 end fraction
                                     
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                                      = 3 over 3 equals 1
                               
                                RHS = space fraction numerator space straight x minus 3 over denominator 5 end fraction
                                      
                                       = fraction numerator 8 minus 3 over denominator 5 end fraction equals 5 over 5 equals 1

∴                                LHS = RHS




Some More Questions From Linear Equations in One Variable Chapter

Solve the following equation:

y + 3 = 10

Solve the following equation:

6 = z + 2

Solve the following equations:

x - 2 = 7

Solve the following equation:

6 = z + 2

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