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Learn in your speed, with individual attention - Teachoo Maths 1-on-1 Class


Transcript

Example 4 Solve the following pair of equations by substitution method: 7x – 15y = 2 x + 2y = 3 7x – 15y = 2 x + 2y = 3 From (1) 7x – 15y = 2 7x = 2 + 15y x = (𝟐 + 𝟏𝟓𝒚)/𝟕 Substituting the value of x in (2) x + 2y = 3 (2 + 15𝑦)/7 + 2𝑦=3 Multiplying both sides by 7 7 × ((2 + 15𝑦)/7) +7×2𝑦=7×3 (2 + 15y) + 14y = 21 15y + 14y = 21 – 2 29y = 21 – 2 29y = 19 y = 𝟏𝟗/𝟐𝟗 Putting value of y in equation (2) x + 2y = 3 x + 2(19/29) = 3 x + 38/29 = 3 x = 3 – 38/29 x = (3(29) − 38)/29 x = (97 − 38)/29 x = 𝟒𝟗/𝟐𝟗 Hence, x = 49/29,y=19/29 is the solution of the equation

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Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.