Given equations are: 3x – y – 3 = 0 ... (1) 2x + y – 12 = 0 ... (2) x – 2y – 1 = 0 ... (3) To Solve (1) and (2), (1) + (2) ⇒ 5x = 15 ⇒ x = 3 (2) ⇒ y = 12 - 6 = 6 Thus (1) and (2) intersect at C (3, 6). To solve (2) and (3), (2) - 2 (3) ⇒ 5y = 10 ⇒ y = 2 (2) ⇒ 2x = 12 - 2 = 10 ⇒ x = 5 Thus (2) and (3) intersect at B (5, 2). To solve (3) and (1), 2 (1) - (3) ⇒ 5x = 5 ⇒ x = 1 (3) ⇒ 1 - 2y = 1 ⇒ y = 0 Thus (3) and (1) intersect at A(1, 0). Area = $\int\limits_{1}^{3}$ (3x - 3) dx + $\int\limits_{3}^{5}$ (12 - 2x) dx - $\int\limits_{1}^{5} \frac{1}{2}$ (x - 1) dx = 3 $\left[ \frac{x^{2}}{2} - x \right]_{1}^{3}$ + $\left[ 12x - x^{2} \right]_{3}^{5}$ - $\frac{1}{2} \left[ \frac{x^{2}}{2} - x \right]_{1}^{5}$ = 3 $\left[ \left(\frac{9}{2} - 3\right) - \left(\frac{1}{2} - 1\right) \right]$ + [(60 - 25) - (36 - 9)] - $\frac{1}{2} \left[ \left(\frac{25}{2} - 5\right) - \left(\frac{1}{2} - 1\right) \right]$ = 3 $\left[ \frac{3}{2} + \frac{1}{2} \right]$ + [35 - 27] - $\frac{1}{2} \left[ \frac{15}{2} + \frac{1}{2} \right]$ = 6 + 8 - 4 = 10 sq. units