Last updated at Dec. 16, 2024 by Teachoo
Question 7 Solve the following pair of linear equations: px + qy = p – q qx + py = p + q px + qy = p − q Multiplying both sides by q q(px + qy) = q(p – q) pqx + q2y = pq − q2 qx − py = p + q Multiplying both sides by p p(qx + py) = p(p + q) pqx + p2y = p2 + pq Hence, our equations are pqx + q2y = pq − q2 …(1) pqx + p2y = p2 + pq …(2) From equation (1), pqx + q2y = pq − q2 q2y = pq − q2 − pqx y = (𝑝𝑞 − 𝑞^2 − 𝑝𝑞𝑥)/𝑞^2 y = (𝑞(𝑝 − 𝑞 − 𝑝𝑥))/𝑞^2 y = (𝑝 − 𝑞 − 𝑝𝑥)/𝑞 Putting y = (𝑝 − 𝑞 − 𝑝𝑥)/𝑞 in equation (2) pqx − p2y = p2 + pq pqx − p2[(𝑝 − 𝑞 − 𝑝𝑥)/𝑞] = p2 + pq pqx − [(𝒑^𝟐 (𝒑 − 𝒒 − 𝒑𝒙))/𝑞] = p2 + pq pqx − [(𝑝^3 − 𝑝^2 𝑞 − 𝑝^3 𝑥)/𝑞]= p2 + pq Multiplying q both sides q(pqx) − q × [(𝑝^3 − 𝑝^2 𝑞 − 𝑝^3 𝑥)/𝑞]= q × p2 + q × pq pq2x − (𝑝^3−𝑝^2 𝑞−𝑝^3 𝑥) = 𝑝2𝑞 + pq2 pq2x − 𝑝^3+"p2q + " 𝑝^3 𝑥=𝒑𝟐𝒒 + pq2 pq2x − 𝑝^3 " + " 𝑝^3 𝑥=𝒑𝟐𝒒 – 𝒑𝟐𝒒 + pq2 pq2x − 𝑝^3 " + " 𝑝^3 𝑥=pq2 "pq2x + " 𝑝^3 𝑥="pq2 "+ 𝑝3 "(pq2 + " 𝑝^3 ")" 𝑥="pq2 "+ 𝑝3 𝑥 = (𝑝𝑞^2+〖 𝑝〗^3)/(𝑝𝑞^2+〖 𝑝〗^3 ) 𝒙 = 1 Put 𝑥 = 1 in equation (1) pqx + q2y = pq − q2 pq(1) + q2y = pq − q2 pq + q2y = − q2 + pq q2y = −q2 y = (−𝑞^2)/𝑞^2 y = −1 Hence, x = 1 and y = −1
About the Author
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