Math, asked by roushankrraj12, 2 months ago

solve the following equations and till we're you know you solve till there only​

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Answered by vipashyana1
1

[tex]\mathfrak{\huge{Answer:-}} \\ v) \bold{\frac{2}{3} x = \frac{3}{8} x + \frac{7}{12}} \\ \frac{2x}{3} = \frac{3x}{8} + \frac{7}{12} \\ \frac{2x}{3} - \frac{3x}{8} = \frac{7}{12} \\ \frac{8(2x) - 3(3x)}{24} = \frac{7}{12} \\ \frac{16x - 9x}{24} = \frac{7}{12} \\ \frac{7x}{24} = \frac{7}{12} \\ cross \: multiply \\ 12(7x) = 7(24) \\ 84x = 168 \\ x = \frac{168}{84} \\ \boxed{\boxed{\bold{x=2}}} \\ vi) \bold{\frac{3x - 1}{5} - \frac{x}{7} = 3} \\ \frac{7(3x - 1) - 5(x)}{35} = 3 \\ \frac{21x - 7 - 5x}{35} = 3 \\ 21x - 5x - 7 = 3(35) \\ 16x - 7 = 105 \\ 16x = 105 + 7 \\ 16x = 112 \\ x = \frac{112}{16} \\ \boxed{\boxed{\bold{x = 7}}} \\ vii)\bold{\frac{y - 1}{3} - \frac{y - 2}{4} = 1} \\ \frac{4(y - 1) - 3(y - 2)}{12} = 1 \\ \frac{4y - 4 - 3y + 6}{12} = 1 \\ 4y - 3y - 4 + 6= 1 (12) \\ y + 2 = 12 \\ y = 12 - 2 \\ \boxed{\boxed{\bold{y = 10}}} \\ viii) \bold{\frac{x - 2}{4} + \frac{1}{3} = x - \frac{2x - 1}{3} } \\ \frac{3(x - 2) + 4(1)}{12} = \frac{3(x) - (2x - 1)}{3} \\ \frac{3x - 6 + 4}{12} = \frac{3x - 2x + 1}{3} \\ \frac{3x - 2}{12} = \frac{x + 1}{3} \\ cross \: multiply \\ 3(3x - 2) = 12(x + 1) \\ 9x - 6 = 12x + 12 \\ 12x - 9x = 6 - 12 \\ 3x = ( - 6) \\ x = \frac{( - 6)}{3} \\ \boxed{\boxed{\bold{x=(-2)}}} \\ ix) \bold{\frac{2x - 1}{3} - \frac{6x - 2}{5} = \frac{1}{3}} \\ \frac{5(2x - 1) - 3(6x - 2)}{15} = \frac{1}{3} \\ \frac{10x - 5 - 18x + 6}{15} = \frac{1}{3} \\ \frac{6 - 5 + 10x - 18x}{15} = \frac{1}{3} \\ \frac{1 - 8x}{15} = \frac{1}{3} \\ cross \: multiply \\ 3(1 - 18x) = 1(15) \\ 3 - 54x = 15 \\ 54 x= 3 - 15 \\ 54x = ( - 12) \\ \boxed{\boxed{\bold{x = \frac{( - 12)}{54} = \frac{( - 2)}{9} }}} \\ x) \bold{\frac{y + 7}{3} = 1 + \frac{3y - 2}{5}} \\ \frac{y + 7}{3} = \frac{5(1) + 3y - 2}{5} \\ \frac{y + 7}{3} = \frac{5 + 3y - 2}{5} \\ \frac{y + 7}{3} = \frac{5 - 2 + 3y}{5} \\ \frac{y + 7}{3} = \frac{3 + 3y}{5} \\ cross \: multiply \\ 5(y + 7) = 3(3 + 3y) \\ 5y + 35 = 9 + 9y \\ 9y - 5y = 35 - 9 \\ 4y = 26 \\\boxed{\boxed{\bold{y = \frac{26}{4} = \frac{13}{2} }}} \\ xi)\bold{\frac{2}{7} (x - 9) + \frac{x}{3} = 3} \\ \frac{2(x - 9)}{7} + \frac{x}{3} = 3 \\ \frac{2x - 18}{7} + \frac{x}{3} = 3 \\ \frac{3(2x - 18) + 7(x)}{21} = 3 \\ \frac{6x - 54 + 7x}{21} = 3 \\ 6x + 7x - 54 = 3(21) \\ 13x - 54 = 63 \\ 13x = 63 + 54 \\ 13x = 117 \\ x = \frac{117}{13} \\ \boxed{\boxed{\bold{x = 9}}} \\ xii)\bold{\frac{2x - 3}{5} + \frac{x + 3}{4} = \frac{4x + 1}{7}} \\ \frac{4(2x - 3) + 5(x + 3)}{20} = \frac{4x + 1}{7} \\ \frac{8x - 12 + 5x + 15}{20} = \frac{4x + 1}{7} \\ \frac{8x + 5x - 12 + 15}{20} = \frac{4x + 1}{7} \\ \frac{13x + 3}{20} = \frac{4x + 1}{7} \\ 20(4x + 1) = 7(13x + 3) \\ 80x + 20 = 91x + 21 \\ 91x - 80x = 20 - 21 \\ 11x = ( - 1) \\ \boxed{\boxed{\bold{x = \frac{( - 1)}{11} }}} \\ xiii)\bold{ \frac{3}{4} (7x - 1) - (2x - \frac{1 - x}{2} ) = x + \frac{3}{2} } \\ \frac{3(7x - 1)}{4} - 2x + \frac{1 - x}{2} = \frac{2x + 3}{2} \\ \frac{21x - 3}{4} - 2x + \frac{1 - x}{2} = \frac{2x + 3}{2} \\ \frac{21x - 3 - 4(2x) + 2(1 - x)}{4} = \frac{2x + 3}{2} \\ \frac{21x - 3 - 8x + 2 - 2x}{4} = \frac{2x + 3}{2} \\ \frac{21x - 8x - 2x - 3 + 2}{4} = \frac{2x + 3}{2} \\ \frac{11x - 1}{4} = \frac{2x + 3}{2} \\ cross \: multiply \\ 2(11x - 1) = 4(2x + 3) \\ 22x - 2 = 8x + 12 \\ 22x - 8x = 12 + 2 \\ 14x = 14 \\ x = \frac{14}{14} \\ \boxed{\boxed{\bold{x=1}}}[/tex]

Sorry but I am not sure about the 9th and 12th answer

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