Example 4 - Solve 5x - 2 (2x - 7) = 2 (3x - 1) + 7/2 - Chapter 2 - Examples

part 2 - Example 4 - Examples - Serial order wise - Chapter 2 Class 8 Linear Equations in One Variable
part 3 - Example 4 - Examples - Serial order wise - Chapter 2 Class 8 Linear Equations in One Variable

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Example 4 Solve 5x – 2 (2x – 7) = 2 (3x – 1) + 7/( 2)5x − 2 (2x − 7) = 2 (3x − 1) + 7/2 5x − 4x + 14 = 2 (3x − 1) + 7/2 5x − 4x + 14 = 6x − 2 + 7/2 x + 14 = 6x − 2 + 𝟕/𝟐 x + 14 = (12𝑥 − 4 + 7)/2 x + 14 = (𝟏𝟐𝒙 + 𝟑)/𝟐 2 (x + 14) = 12x + 3 2x + 28 = 12x + 3 12x + 3 = 2x + 28 12x − 2x + 3 = 28 10x + 3 = 28 10x = 28 − 3 10x = 25 x = 25/10 x = 𝟓/𝟐 L.H.S 5𝑥−2(2𝑥−7) = 5× 5/2−2(2×5/2−7) = 25/2−2(5−7) = 25/2−2 (−2) =25/2 + 4 =(25 + 8)/2 =𝟑𝟑/𝟐 R.H.S 2(3𝑥−1)+7/2 = 2(3× 5/2−1)+7/2 = 2(15/2−1)+7/2 =2((15 − 2)/2)+7/2 =2(13/2)+7/2 = 26/2+7/2 =𝟑𝟑/𝟐 ∴ LHS = RHS Hence Verified.

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Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science and Computer Science at Teachoo