Last updated at Dec. 16, 2024 by Teachoo
Question 7 Solve the following pair of linear equations: (iv) (𝑎−𝑏)𝑥+(𝑎+𝑏)𝑦=𝑎2 −2𝑎𝑏−𝑏2 (𝑎 + b) (x + y) = 𝑎2 + 𝑏2 (a − b) 𝑥 + (a + b) y = a2 − 2ab − b2 …(1) (a + b) (𝑥 + y) = a2 + b2 …(2) Solving equation (2) (a + b) (𝑥 + y) = a2 + b2 (a + b)𝒙 + (a + b)y = a2 + b2 (a + b) y = a2 + b2 − (a + b) 𝑥 y = (𝑎^2 + 𝑏^2 − (𝑎 + 𝑏)𝑥)/((𝑎 + 𝑏)) Putting y = (𝑎^2 + 𝑏^2 − (𝑎 + 𝑏)𝑥)/((𝑎 + 𝑏)) in equation (1) (a − b) 𝑥 + (a + b) y = a2 − 2ab − b2 (a − b) x + (a + b) [(𝑎^2 + 𝑏^2 − (𝑎 + 𝑏)𝑥)/((𝒂 + 𝒃))] = 𝑎2−2𝑎𝑏−𝑏2 (a − b)x + 𝒂𝟐 + 𝒃𝟐 − (a + b) x = 𝑎2−2𝑎𝑏−𝑏2 ax − bx + b2 − ax − bx = 2𝑎𝑏−𝑏2 − bx − bx + b2 = 2𝑎𝑏−𝑏2 –2bx + b2 = −2ab − b2 –2bx = −2ab − b2 − b2 –2bx = −2ab − 2b2 –2bx = −(2ab + 2b2) 2bx = 2ab + 2b2 2bx = 2b (a + b) x = (2𝑏(𝑎 + 𝑏))/2𝑏 x = a + b Put x = a + b in equation (3) y = (𝑎2+ 𝑏2− (𝑎 + 𝑏)𝑥)/(𝑎 + 𝑏) y = (𝑎2+ 𝑏2 − (𝑎 + 𝑏)(𝑎 + 𝑏))/(𝑎 + 𝑏) y = (𝑎2+ 𝑏2 − (𝑎 + 𝑏)^2)/(𝑎 + 𝑏) y = (𝑎2 + 𝑏2 − 𝑎2 − 𝑏2 − 2𝑎𝑏)/(𝑎 + 𝑏) y = (−𝟐𝒂𝒃)/(𝒂 + 𝒃) Therefore, x = a + b and y = (−𝟐𝒂𝒃)/(𝒂 + 𝒃).