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Ex.3.7 (Optional)
Ex 3.7, 2 (Optional)
Ex 3.7, 3 (Optional)
Ex 3.7, 4 (Optional) Important
Ex 3.7, 5 (Optional)
Ex 3.7, 6 (Optional) Important
Ex 3.7, 7 (i) (Optional) You are here
Ex 3.7, 7 (ii) (Optional) Important
Ex 3.7, 7 (iii) (Optional)
Ex 3.7, 7 (iv) (Optional)
Ex 3.7, 7 (v) (Optional) Important
Ex 3.7, 8 (Optional)
Ex.3.7 (Optional)
Last updated at July 29, 2021 by Teachoo
Maths Crash Course - Live lectures + all videos + Real time Doubt solving!
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