Example 4 - Chapter 2 Class 8 Linear Equations in One Variable

Last updated at April 16, 2024 by

Slide6.JPG

Slide7.JPG
Slide8.JPG

Go Ad-free

Transcript

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.

Davneet Singh's photo - Co-founder, Teachoo

Made by

Davneet Singh

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