How did this solution calculate the value of the x intercept of 2.1875 and the point 8.125. I think it may have to do with similar triangles and by writing the length of a to b split into a and a-3, using ratios. But I'm not sure?

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Question 3 (40 marks) For the beam shown below, calculate the support reactions and draw the shear force and bending moment diagrams indicating all the critical values. 10 kN/m 10 kN/m SFD Ax B A 3 m +/+ 5 m Ay k E21.8752.1875 = 23.926 KNM =-8.125x0.8125 = = = =-3.300 or = -3.301 KNM BMD 21.875KN + 2.1875m - -8.125 KN k 3 m 5 m +4 C 20KN D 2 m +4 = -8,1255 =-40.625 Kam = X202=20KNM * 42m 4 -20 kNm (= 0 ++ ) = Question 3 (40 marks) For the beam shown below, calculate the support reactions and draw the shear force and bending moment diagrams indicating all the critical values. 10 kN/m 10 kN/m SFD Ax B A 3 m +/+ 5 m Ay k E21.8752.1875 = 23.926 KNM =-8.125x0.8125 = = = =-3.300 or = -3.301 KNM BMD 21.875KN + 2.1875m - -8.125 KN k 3 m 5 m +4 C 20KN D 2 m +4 = -8,1255 =-40.625 Kam = X202=20KNM * 42m 4 -20 kNm (= 0 ++ ) =