5 1 x 3 1 x 1 2x 8

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5 1 x 3 1 x 1 2x 8 Solve the following 5 1 x 3 1 x 1 2x 8 Sum Solution Show Solution Given 5 1 x 3 1 x 1 2x 8 5 1 x 3 1 x 8 1 2x By cross multiplication 5 5x 3 3x 8 16x 8 2x 8 16x 16x 2x 8 8 Transposing 16x to LHS and 8 to RHS 14x 0

Solution Verified by Toppr From the given equations 5 1 x 3 1 x 1 2 x 8 5 x 1 3 1 x 8 1 2 x 5 5 x 3 3 x 8 16 x 8 2 x 8 16 x 16 Best answer We have 5 1 x 3 1 x 1 2x 8 By cross multiplication we get 5 1 x 3 1 x 8 1 2x 5 5x 3 3x 8 16 x 8 2x 8 16x Transposing 8 to RHS it becomes 8 and 16x to LHS it becomes 16x 16x 2x 8 8 14x 0 x 0 14 Prev Question Next Question

5 1 x 3 1 x 1 2x 8

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Final answer The given equation is a second order polynomial or quadratic equation which is determined by the presence of a squared variable x 2 term in the equation Option b Quadratic Equation Explanation The given equation is 5 1 x 3 1 x 1 2x 8 x y z xyz x y z

Solution Verified by Toppr From the given equations frac 5 1 x 3 1 x 1 2x 8 Rightarrow 5 x 1 3 1 x 8 1 2x Rightarrow 5 5x 3 3x 8 X 2 x 6 0 x 3 gt 2x 1 line 1 2 3 1 f x x 3 prove tan 2 x sin 2 x tan 2 x sin 2 x frac d dx frac 3x 9 2 x sin 2 theta sin 120

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