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  1. Chapter 2 Class 8 Linear Equations in One Variable
  2. Serial order wise
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Ex 2.5, 7 Simplify and solve the following linear equations. 3 (t – 3) = 5 (2t + 1) 3 (t – 3) = 5 (2t + 1) 3t − 9 = 5 (2t + 1) 3t − 9 = 10t + 5 3t − 10t − 9 = 5 −7t − 9 = 5 −7t = 5 + 9 −7t = 14 t = 14/(−7) t = −2 Check:- L.H.S 3 (t − 3) = 3 (−2 − 3) = 3 (−5) = −15 R.H.S 5(2t + 1) = 5 (2(−2) + 1) = 5 (−4 + 1) = 5 (−3) = −15 ∴ LHS = RHS Hence Verified.

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Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 8 years. He provides courses for Maths and Science at Teachoo. You can check his NCERT Solutions from Class 6 to 12.