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Chapters

Exercise 1
Exercise 2
Exercise 3
Exercise 4
Exercise 5
Exercise 6
Exercise 7
Exercise 8
Exercise 9
Exercise 10
Exercise 11
Exercise 12
Exercise 13
Exercise 14
Exercise 15
Solution of exercise 1
Solution of exercise 2
Solution of exercise 3
Solution of exercise 4
Solution of exercise 5
Solution of exercise 6
Solution of exercise 7
Solution of exercise 8
Solution of exercise 9
Solution of exercise 10
Solution of exercise 11
Solution of exercise 12
Solution of exercise 13
Solution of exercise 14
Solution of exercise 15

Solve the following equations:

Exercise 1

$2x = 6$

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Exercise 2

$2x -3 = 6 + x$

Exercise 3

$2 (2x - 3) = 6 + x$

Exercise 4

$\frac{x - 1}{6} - \frac{x - 3}{2} = -1$

Exercise 5

$\frac{3}{4} (2x + 4) = x + 19$

Exercise 6

$4 (x - 10) = -6 (2 - x) -6x$

Exercise 7

$2 (x + 1) - 3(x - 2) = x + 6$

Exercise 8

$\frac{x - 1}{4} - \frac{x - 5}{36} = \frac{x + 5}{9}$

Exercise 9

$\frac{3x + 1}{7} - \frac{2 - 4x}{3} = \frac{-5x - 4}{14} + \frac{7x}{6}$

Exercise 10

$\frac{5}{x - 7} = \frac{3}{x - 2}$

Exercise 11

$\frac{4}{x - 3} = \frac{5}{x - 2}$

Exercise 12

$6 (\frac{x + 1}{8} - \frac{2x - 3}{16}) = 3(\frac{3}{4}x - \frac{1}{4}) - \frac{3}{8} (3x - 2)$

Exercise 13

$2 - [-2 \cdot (x + 1) - \frac{x - 3}{2}] = \frac{2x}{3} - \frac{5x - 3}{12} + 3x$

Exercise 14

$\frac{2}{3} [ x - (1 - \frac{x - 2}{3})] + 1 = x$

Exercise 15

$2 - [-2 \cdot (x + 1) - \frac{x - 3}{2}] = \frac{2x}{3} - \frac{5x - 3}{12} + 3x$

Solution of exercise 1

$2x = 6$

$x = \frac{6}{2}$

$x = 3$

Solution of exercise 2

$2x - 3 = 6 + x$

$2x -x = 6 + 3$

$x = 9$

Solution of exercise 3

$2 (2x - 3) = 6 + x$

$4x - 6 = 6 + x$

$4x - x = 6 + 6$

$3x = 12$

$x = \frac{12}{3}$

$x = 4$

Solution of exercise 4

$\frac{x - 1}{6} - \frac{x - 3}{2} = -1$

LCM $(6, 2) = 6$

$x - 1 - 3 (x - 3) = -6$

$x - 1 - 3x + 9 = -6$

$x - 1 - 3x + 9 = -6$

$x - 3x = -6 - 9 + 1$

$-2x = -14$

$x = \frac{14}{2}$

$x = 7$

Solution of exercise 5

$\frac{3}{4} (2x + 4) = x + 19$

$\frac{6}{4}x + \frac{12}{4} = x + 19$

$\frac{3}{2}x + 3 = x + 19$

$3x + 6 = 2x + 38$

$3x - 2x = 38 - 6$

$x = 32$

Solution of exercise 6

$4 (x - 10) = -6 (2 - x) -6x$

$4x - 40 = -12 + 6x - 6x$

$4x - 6x + 6x = -12 + 40$

$4x = 28$

$x = 7$

Solution of exercise 7

$2 (x + 1) - 3(x - 2) = x + 6$

$2x + 2 - 3x + 6 = x + 6$

$2x - 3x - x = 6 -2 - 6$

$-2x = -2$

$x = 1$

Solution of exercise 8

$\frac{x - 1}{4} - \frac{x - 5}{36} = \frac{x + 5}{9}$

LCM $(4, 36, 9) = 36$

$9 (x - 1) - (x - 5) = 4 (x + 5)$

$9x - 9 - x + 5 = 4x + 20$

$9x - x - 4x = 20 + 9 - 5$

$4x = 24$

$x = 6$

Solution of exercise 9

$\frac{3x + 1}{7} - \frac{2 - 4x}{3} = \frac{-5x - 4}{14} + \frac{7x}{6}$

LCM $(7, 3, 14, 6) = 42$

$6(3x + 1) - 14 (2 - 4x) = 3 (-5x - 4) + 49x$

$18x + 6 - 28 + 56x = -15x - 12 + 49x$

$18x + 56x + 15x - 49x = -12 - 6 + 28$

$40x = 10$

$4x = 1$

$x = \frac{1}{4}$

Solution of exercise 10

$\frac{5}{x - 7} = \frac{3}{x - 2}$

$5 (x - 2) = 3 (x - 7)$

$5x - 10 = 3x - 21$

$5x - 3x = -21 + 10$

$2x = -11$

$x = -\frac{11}{2}$

Solution of exercise 11

$\frac{4}{x - 3} = \frac{5}{x - 2}$

$4 (x - 2) = 5 (x - 3)$

$4x - 8 = 5x - 15$

$-8 + 15 = 5x - 4x$

$x = 7$

Solution of exercise 12

$6 (\frac{x + 1}{8} - \frac{2x - 3}{16}) = 3 (\frac{3}{4}x - \frac{1}{4}) - \frac{3}{8}(3x - 2)$

$\frac{6(x +1)}{8} - \frac{6 (2x - 3)}{16} = \frac{9}{4}x - \frac{3}{4} - \frac{9}{8}x + \frac{6}{8}$

$\frac{6x + 6}{8} - \frac{12x - 18}{16} = \frac{9}{4}x - \frac{3}{4} - \frac{9}{8}x + \frac{6}{8}$

LCM $(8, 16, 4) = 16$

$2(6x + 6) - (12x - 18) = 36x - 12 - 28x + 12$

$12+ 18 = 36x - 18x$

$18x = 30$

$3x = 5$

$x = \frac{5}{3}$

Solution of exercise 13

$2 - [-2 \cdot (x + 1) - \frac{x - 3}{2}] = \frac{2x}{3} - \frac{5x - 3}{12} + 3x$

$2 - (-2x - 2 - \frac{x - 3}{2}) = \frac{2x}{3} - \frac{5x - 3}{12} + 3x$

$2 + 2x + 2 + \frac{x - 3}{2} = \frac{2x}{3} - \frac{5x - 3}{12} + 3x$

$24 + 24x + 24 + 6 \cdot (x - 3) = 8x - (5x - 3) + 36x$

$24 + 24x + 24 + 6x - 18 = 8x - 5x + 3 + 36x$

$24x + 6x - 8x + 5x - 36x = 3 - 24 - 24 + 18$

$-9x = -27$

$x = 3$

Solution of exercise 14

$\frac{2}{3} [ x - (1 - \frac{x - 2}{3})] + 1 = x$

$\frac{2}{3} (x - 1 + \frac{x - 2}{3}) + 1 = x$

$\frac{2}{3}x - \frac{2}{3} + \frac{2x - 4}{9} + 1 = x$

$6x - 6 + 2x - 4 + 9 = 9x$

$-x = 1$

$x = -1$

Solution of exercise 15

$2 - [-2 \cdot (x + 1) - \frac{x - 3}{2}] = \frac{2x}{3} - \frac{5x - 3}{12} + 3x$

$2 - (-2x - 2 - \frac{x - 3}{2}) = \frac{2x}{3} - \frac{5x - 3}{12} + 3x$

$2 + 2x + 2 + \frac{x - 3}{2} = \frac{2x}{3} - \frac{5x - 3}{12} + 3x$

$24 + 24x + 24 + 6 \cdot (x - 3) = 8x - (5x - 3) + 36x$

$24 + 24 x + 24 + 6x - 18 = 8x - 5x + 3 + 36x$

$24x + 6x - 8x + 5x - 36x = 3 - 24 - 24 + 18$

$-9x = -27$

$x = 3$

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