1. Find the value:
(i) | 15 − 2 |
Solution:

 | 15 − 2 |

= | 13 |

= 13

(ii) | 4 − 9 |
Solution:

 | 4 − 9 |

= | − 5 |

= | 5 |

(iii) | 7 | × | − 4 |
Solution:

 | 7 | × | − 4 |

= 7 × 4

= 28



2. Solve:
(i) | 3x − 5 | = 1
Solution:

 | 3x − 5 | = 1

∴ 3x − 5 = + 1 OR 3x − 5 = − 1

∴ 3x = 1 + 5 OR 3x = − 1 + 5

∴ 3x = 6 OR 3x = 4

∴ \(\displaystyle x = \frac{6}{3}\) OR \(\displaystyle x = \frac{4}{3}\)

∴ \(\displaystyle x = 2\) OR \(\displaystyle x = \frac{4}{3}\)

(ii) | 7 − 2x | = 5
Solution:

 | 7 − 2x | = 5

∴ 7 − 2x = + 5 OR 7 − 2x = − 5

∴ 7 − 5 = 2x OR 7 + 5 = 2x

∴ 2 = 2x OR 12 = 2x

i.e. 2x = 2 OR 2x = 12

∴ \(\displaystyle x = \frac{2}{2}\) OR \(\displaystyle x = \frac{12}{2}\)

∴ \(\displaystyle x = 1\) OR \(\displaystyle x = 6\)

(iii) \(\displaystyle \left |\dfrac {8 - x}{2}\right | = 5\)
Solution:

 \(\displaystyle \left |\dfrac {8 - x}{2}\right | = 5\)

∴ \(\displaystyle 8 - x = 5 \times 2\) OR \(\displaystyle 8 - x = - 5 \times 2\)

∴ \(\displaystyle 8 - x = 10\) OR \(\displaystyle 8 - x = - 10\)

∴ \(\displaystyle 8 - 10 = x\) OR \(\displaystyle 8 + 10 = x\)

∴ \(\displaystyle - 2 = x\) OR \(\displaystyle 18 = x\)

∴ \(\displaystyle x = - 2\) OR \(\displaystyle x = 18\)

(iv) \(\displaystyle \left |5 + \dfrac {x}{4}\right | = 5\)
Solution:

 \(\displaystyle \left |5 + \dfrac {x}{4}\right | = 5\)

∴ \(\displaystyle 5 + \frac {x}{4} = + 5\) OR \(\displaystyle 5 + \frac {x}{4} = - 5\)

∴ \(\displaystyle \frac {x}{4} = 5 - 5\) OR \(\displaystyle \frac {x}{4} = - 5 - 5\)

∴ \(\displaystyle \frac {x}{4} = 0\) OR \(\displaystyle \frac {x}{4} = - 10\)

∴ \(\displaystyle x = 0 \times 4\) OR \(\displaystyle x = - 10 \times 4\)

∴ \(\displaystyle x = 0\) OR \(\displaystyle x = - 40\)





This page was last modified on
22 April 2026 at 23:07

© 1976 - 2026 Patwardhan Class, All Rights Reserved
  Patwardhan Class ... First Class!!!