Question

Chapter...linear equation in one variable....solve the following equation
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Answer 1

★ Solution (1) :-

$\sf \leadsto \dfrac{7x - 3}{2} = \dfrac{-2}{3}$

$\sf \leadsto 3(7x - 3) = -2(2)$

$\sf \leadsto 21x - 9 = -4$

$\sf \leadsto 21x = -4 + 9$

$\sf \leadsto 21x = 5$

$\sf \leadsto x = \dfrac{5}{21}$

★ Solution (2) :-

$\sf \leadsto \dfrac{2x - 5}{3} + \dfrac{3}{5} = 0$

$\sf \leadsto \dfrac{10x - 25 + 9}{15} = 0$

$\sf \leadsto \dfrac{10x - 16}{15} = 0$

$\sf \leadsto 10x - 16 = 0$

$\sf \leadsto 10x = 0 + 16$

$\sf \leadsto 10x = 16$

$\sf \leadsto x = \dfrac{16}{10}$

$\sf \leadsto x = \dfrac{8}{5}$

★ Solution (3) :-

$\sf \leadsto \dfrac{7x - 3}{5x - 2} = -2$

$\sf \leadsto 7x - 3 = -2(5x - 2)$

$\sf \leadsto 7x - 3 = -10x + 4$

$\sf \leadsto 7x + 10x = 4 + 3$

$\sf \leadsto 17x = 7$

$\sf \leadsto x = \dfrac{7}{17}$

★ Solution (4) :-

$\sf \leadsto \dfrac{3x + 1}{x - 2} = \dfrac{-6}{11}$

$\sf \leadsto 11(3x + 1) = -6(x - 2)$

$\sf \leadsto 33x + 11 = -6x + 12$

$\sf \leadsto 33x + 6x = 12 - 11$

$\sf \leadsto 39x = 1$

$\sf \leadsto x = \dfrac{1}{39}$

★ Solution (5) :-

$\sf \leadsto \dfrac{z + 9}{3(z + 5)} = -2$

$\sf \leadsto \dfrac{z + 9}{3z + 15} = -2$

$\sf \leadsto z + 9 = -2(3z + 15)$

$\sf \leadsto z + 9 = -6z - 30$

$\sf \leadsto z + 6z = -30 - 9$

$\sf \leadsto 7z = -39$

$\sf \leadsto z = \dfrac{-39}{7}$

★ Solution (6) :-

$\sf \leadsto \dfrac{5y - 2}{4(y + 2)} = \dfrac{1}{3}$

$\sf \leadsto \dfrac{5y - 2}{4y + 8} = \dfrac{1}{3}$

$\sf \leadsto 3(5y - 2) = 1(4y + 8)$

$\sf \leadsto 15y - 6 = 4y + 8$

$\sf \leadsto 15y - 4y = 8 + 6$

$\sf \leadsto 11y = 14$

$\sf \leadsto y = \dfrac{14}{11}$

★ Solution (7) :-

$\sf \leadsto m - \dfrac{m - 1}{2} = 1 - \dfrac{m - 2}{3}$

$\sf \leadsto \dfrac{2m - m - 1}{2} = \dfrac{3 - m - 2}{3}$

$\sf \leadsto \dfrac{m - 1}{2} = \dfrac{1 - m}{3}$

$\sf \leadsto 3(m - 1) = 2(1 - m)$

$\sf \leadsto 3m - 3 = 2 - 2m$

$\sf \leadsto 3m + 2m = 2 + 3$

$\sf \leadsto 5m = 5$

$\sf \leadsto m = \dfrac{5}{5}$

$\sf \leadsto m = 1$

★ Solution (8) :-

$\sf \leadsto \dfrac{3t - 2}{4} - \dfrac{2t + 3}{3} = \dfrac{2}{3} - t$

$\sf \leadsto \dfrac{9t - 6 - 8t + 12}{12} = \dfrac{2 - 3t}{3}$

$\sf \leadsto \dfrac{t + 6}{12} = \dfrac{2 - 3t}{3}$

$\sf \leadsto 3(t + 6) = 12(2 - 3t)$

$\sf \leadsto 3t + 18 = 24 - 36t$

$\sf \leadsto 3t + 36t = 24 - 18$

$\sf \leadsto 39t = 6$

$\sf \leadsto t = \dfrac{6}{39}$

★ Solution (9) :-

$\sf \leadsto \dfrac{2x + 1}{3} + 1 = \dfrac{x - 4}{6}$

$\sf \leadsto \dfrac{2x + 1 + 3}{3} = \dfrac{x - 4}{6}$

$\sf \leadsto \dfrac{2x + 4}{3} = \dfrac{x - 4}{6}$

$\sf \leadsto 6(2x + 4) = 3(x - 4)$

$\sf \leadsto 12x + 24 = 3x - 12$

$\sf \leadsto 12x - 3x = -12 - 24$

$\sf \leadsto 9x = -36$

$\sf \leadsto x = \dfrac{-36}{9}$

$\sf \leadsto x = -4$

$\:$

Hence, solved!!

Answer 2

Answer:

answer will 4 hope it will help you