



Examples
Example 2
Example 3
Example 4 Important
Example 5
Example 6 Important
Example 7
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 You are here
Example 19 Important
Last updated at Aug. 2, 2021 by Teachoo
Example 18 Solve the following pair of equations by reducing them to a pair of linear equations : 5/(𝑥 −1) + 1/(𝑦 −2) = 2 6/(𝑥 −1) – 3/(𝑦 −2) = 1 5/(𝑥 − 1) + 1/(𝑦 − 2) = 2 6/(𝑥 − 1) – 3/(𝑦 − 2) = 1 So, our equations become 5u + v = 2 6u – 3v = 1 Thus, our equations are 5u + v = 2 …(3) 6u – 3v = 1 …(4) From (3) 5u + v = 2 v = 2 – 5u Putting value of v in (4) 6u – 3v = 1 6u – 3(2 – 5u) = 1 6u – 6 + 15u = 1 6u + 15u = 1 + 6 21u = 7 u = 7/21 u = 1/3 Putting u = 1/3 in equation (3) 5u + v = 2 5(1/3) + v = 2 5/3 + v = 2 v = 2 – 5/3 v = (2(3) − 5)/3 v = (6 − 5)/3 v = 𝟏/𝟑 Hence, u = 1/3 & v = 1/3 But we need to find x & y u = 𝟏/(𝒙 − 𝟏) 1/3 = 1/(𝑥 − 1) x – 1 = 3 x = 3 + 1 x = 4 v = 𝟏/(𝒚 − 𝟐) 1/3 = 1/(𝑦 −2) y – 2 = 3 y = 3 + 2 y = 5 So, x = 4, y = 5 is the solution of our equations