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Linear Equations In One Variable

Question
CBSEENMA8002238
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Solve the following equation and check your results.

5t -3 = 3t -5

Solution

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
                                fraction numerator 2 t over denominator t end fraction equals fraction numerator negative 2 over denominator 2 end fraction
or                                 t = -1

Some More Questions From Linear Equations in One Variable Chapter

Solve the following equation:

2x -3 = 7

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