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Mix questions - Equation given

Question 7
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Last updated at May 29, 2023 by Teachoo

Question 7 Solve the pair of equations: 2/π₯+ 3/π¦=13 5/π₯β4/π¦=β2 2/π₯+ 3/π¦=13 5/π₯β4/π¦=β2 So, our equations become 2u + 3v = 13 5u β 4v = β2 Hence, our equations are 2u + 3v = 13 β¦(3) 5u β 4v = β 2 β¦(4) From (3) 2u + 3v = 13 2u = 13 β 3V u = (13 β 3π£)/2 Putting value of u (4) 5u β 4v = - 2 5((13 β 3π£)/2)β4π£=β2 Multiplying 2 both sides 2 Γ 5((13 β 3π£)/2)β"2 Γ" 4π£="2 Γ"β2 5(13 β 3v) β 8v = β4 65 β 15v β 8v = β4 β 15v β 8v = β 4 β 65 β 23v = β 69 v = (β69)/(β23) v = 3 Putting v = 3 in (3) 2u + 3v = 13 2u + 3(3) = 13 2u + 9 = 13 2u = 13 β 9 2u = 4 u = 4/2 u = 2 Hence, u = 2, v = 3 is the solution But we have to find x & y u = π/π 2 = 1/π₯ x = π/π v = π/π 3 = 1/π¦ y = π/π Hence, x = 1/2 , y = 1/3 is the solution of the given equation