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Question

(iv) x3+5y=13; 2x+y2=19

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Solution

x3+5y=13x+15y=39 ...(i) 2x+y2=19 4x+y=38 ...(ii) Multiplying eq (i) by 4, we get:4x+60y=156 ...(iii) Subtracting eq (i) from eq (iii), we get: 4x+60y=156 4x+ y= 38 - - - 59y = 118 y=2Substituting y=2 in (i), we get: x+15×2=39x=9Hence, x=9, y=2 is the solution of the given equations.

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