(a) Solve the following equations, giving your answers as integers or as fractions in their simplest form. lla - 28c = d(1 -4c) (i) 12x + 8 =44 - 6x Divide By 1 - 40 assuming that (1 - 4+0 # 0) to Give (ii) 6 - (3 - x) = 21 - 52 dalla-28c 20x 1- 40 (iii) 5 4x - 1 (iv) 2a2 = 48a (i) Write out a correct rearrangement of the formula. [3] (b) Factorise the expression x2 + 2x - 35, and hence solve the equation (ii) Identify and explain, as if directly to the student, two of the [2] x2 + 2x - 35 = 0. mistakes they have made. (c) A student was asked to rearrange the formula 1la = - 2c(d + 7) to make d the subject (assuming that 1 - 4c / 0). The student's incorrect attempt is shown below. lla = " - 2cld + 7) Clear the fractions By multiplying By 2 lla = d- 4cld+1) Multiply out the Bracket lla = d- 4dc+28c Collect the d terms lla -28c =d-4dc Factorise