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Ex 3.7, 7 (Optional) - Solve the pair of linear equations - teachoo

Ex 3.7, 7 (Optional) - Chapter 3 Class 10 Pair of Linear Equations in Two Variables - Part 2
Ex 3.7, 7 (Optional) - Chapter 3 Class 10 Pair of Linear Equations in Two Variables - Part 3 Ex 3.7, 7 (Optional) - Chapter 3 Class 10 Pair of Linear Equations in Two Variables - Part 4

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Ex 3.7, 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

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Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.