Question

(a) 5x - 11 = 3x + 9 (b) 3y + 4 = 7 - 2y (c) 9 - 2(x - 5) = x + 10 (d) 5(y - 1) = 3(2y - 5) - (1 - 3y) (e) 2(x - 1) - 6x = 10 - 2(x - 4) (f) x/3 - (x - 2)/2 = 7/3

Answer 1

$\orange{1.)} \: \: \: \rm5x - 11 = 3x + 9 \\ \tt \to5 x- 3x = 9+ 11 \\ \tt \to2x = 20 \\ \tt \to x = \frac{20}{2} \: \: = \red{ 10}$

$\orange{2.)} \: \: \rm3y + 4 = 7 - 2y \\ \tt \to3y + 2y = 7 - 4 \\ \tt \to5y = 3 \\ \tt \to y = \red{\frac{3}{5} } \: \: \: or \: \: \: \red{0.6}$

$\orange{3.) } \: \: \: \rm9 - 2(x - 5) = x + 10 \\ \tt \to9 - 2x + 10 = x + 10 \\ \tt \to \: \: - 2x + 19 = x + 10 \\ \tt \to \: \: \: - 2x - x = 10 - 19 \\ \tt \to \: \: \cancel- 3x = \cancel - 9 \\ \tt \to \: 3x = 9 \\ \tt \to \:x = \frac{9}{3} \: \: \: \: = \red3$

$\small \orange{4.) }\: \: \rm \: 5(y - 1) = 3(2y - 5) - (1 - 3y) \\ \tt \to5y - 5 = 6y - 15 - 1 + 3y \\ \tt \to5y - 5 = (6 + 3)y - 16 \\ \tt \to5y - 5 = 9y - 16 \\ \tt \to5y - 9y = \: - 16 + 5 \\ \tt \to \: \: \cancel - 4y \: = \: \cancel - 11 \\ \tt \to \: \: 4y = 11 \\ \tt \to \: \: y = \red{\frac{11}{4} } \: \: \: \: or \: \: \: \: \red{2.75}$

$\small \orange{5.)} \: \: \: \rm2(x - 1) - 6x = 10 - 2(x - 4) \\ \tt \to2x - 2 - 6x = 10 - 2x + 8 \\ \tt \to(2 - 6)x - 2 = 18 - 2x \\ \tt \to - 4x - 2 = 18 - 2x \\ \tt \to - 4x + 2x = 18 + 2 \\ \to \tt - 2x = 20 \\ \tt \to \:x = \frac{20}{ - 2} \: \: = \red{ - 10}$

$\rm \orange{6.)} \: \: \: \frac{x}{3} - \frac{x - 2}{2} = \frac{7}{3} \\ \small \boxed{ \rm{}taking \: \: LCM \: \: of \: \: denominators \: \: in \: \:LHS} \\ \to \tt \frac{2(x) - 3(x - 2)}{6} = \frac{7}{3} \\ \\ \to \tt 2x - 3x + 6 = \frac{\cancel6 \times 7}{ \cancel3} \\ \\ \to \tt - x + 6 = 2 \times 7 \\ \to \tt - x + 6 = 14 \\ \to \tt - x = 14 - 6 \\ \to \tt - x = 8 \\ \small \boxed{ \rm{}Multiplying \: \: both \: \: sides \: \: by \: \: (- 1) } \\ \to \tt ( - x)( - 1) = 8( - 1) \\ \to \tt \: \: x \: = \red{ - 8}$