Example 7 - Chapter 3 Class 10 Pair of Linear Equations in Two Variables (Term 1)
Last updated at Dec. 18, 2020 by Teachoo
Examples
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Example 4 Important
Example 5
Example 6 Important
Example 7 You are here
Example 8
Example 9
Example 10 Important
Example 11 Important
Example 12
Example 13 Important
Example 14 Important
Example 15
Example 16 Important
Example 17
Example 18 Important
Example 19 Important
Transcript
Example 7 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
