Solve (2x^2 - 3x - 14)/(x^2 + 6x + 8) = -6/(x+3).

LHS: Factorise numerator and denominator, which gives:((2x - 7)(x + 2))/((x + 2)(x + 4))Cancel out (x + 2), gives:(2x - 7)/(x + 4)Multiply both sides by (x + 4)(x + 3) to cancel the denominators, gives:(2x - 7)(x + 3) = -6(x + 4)Expanding brackets:2x^2 - x - 21 = -6x - 24Make RHS = 0:2x^2 + 5x + 3 = 0Solve with factorisation/complete the square/quadratic formula:(2x + 3)(x + 1) = 0x = -2/3, -1

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Answered by Joseph L. Maths tutor

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