Question

22. In the figure, DE || BC, find x

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Answer 1

Answer:

$x = 1$

Step-by-step explanation:

$In \: ΔABC \\ DE||BC \\ \frac{AD}{BD} = \frac{AE}{EC} \: (BPT \: thereom) \\ \frac{4x - 3}{3x - 1} = \frac{8x - 7}{5x - 3} \\ Cross \: multiply \\ (4x - 3)(5x - 3) = (8x - 7)(3x - 1) \\ 4x(5x - 3) - 3(5x - 3) = 8x(3x - 1) - 7(3x - 1) \\ 20 {x}^{2} - 12x - 15x + 9 = 24 {x}^{2} - 8x - 21x + 7 \\ 20 {x}^{2} - 27x + 9 = 24 {x}^{2} - 29x + 7 \\ 20 {x}^{2} - 24 {x}^{2} - 27x + 29x + 9 - 7 = 0 \\ ( - 4 {x}^{2} ) + 2x + 2 = 0 \\ 4 {x}^{2} - 2x - 2 = 0 \\ 2 {x}^{2} - x - 1 = 0 \\ 2 {x}^{2} - 2x + x - 1 = 0 \\ (2 {x}^{2} - 2x) + (x - 1) = 0 \\ 2x(x - 1) + 1(x - 1) = 0 \\ (x - 1)(2x + 1) = 0 \\ (x - 1 = 0)(2x + 1 = 0) \\ x = 1 \: and \: x = \frac{( - 1)}{2} \\ Therefore, \: x = 1$