Solve these simultaneous equations: 3xy = 1, and y = 12x + 3
From first equation: 3xy = 1 => x = 1/(3y)Substitute expression for x into second equation: y = 12x + 3 => y = 12(1/3y) + 3 = 4/y +3Multiply through by y: y2 = 4 + 3y => y2 - 3y - 4 = 0Factorise: (y-4)(y+1) = 0 => y = 4, y = -1 are solutionsx = 1/3y = 1/12, -1/3
Solutions are (1/12,4) and (-1/3, -1)
HM
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