Question

Solve the equations and check the answer (No need to solve the No.1☺️ So I erase it)​

Answer 1

Answer :-

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$\sf (ii) \leadsto3(x - 3) = 5(2x - 1) \\ \\ \sf\leadsto3x - 9 = 10x - 5 \\ \\ \sf \leadsto - 9 + 5 = 10x - 3x \\ \\ \sf \leadsto - 4 = 7x \\ \\ \boxed{\sf \leadsto \green{x = - \dfrac{ 4}{7}} }$

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$\sf (iii) \leadsto \dfrac{2y + 9}{3} = 3y + 10 \\ \\ \sf \leadsto2y + 9 = 3(3y + 10) \\ \\ \sf \leadsto2y + 9 = 9y + 30 \\ \\ \sf \leadsto2y - 9y = 30 - 9 \\ \\ \sf \leadsto - 7y = 21 \\ \\ \sf \leadsto y = \dfrac{21}{ - 7} \\ \\ \boxed{\sf \leadsto \pink{y = - 3 }}$

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$\sf (iv) \leadsto 2(x - 1) - 3(x - 2) = 4(x - 3) + 5(x - 4) \\ \\ \sf \leadsto 2x - 2 - 3x + 6 = 4x - 12 + 5x - 20 \\ \\ \sf \leadsto4 - x = 9x - 32 \\ \\ \sf \leadsto32 + 4 = 9x + x \\ \\ \sf \leadsto36 = 10x \\ \\ \sf \leadsto \dfrac{36}{10} = x \\ \\ \boxed{ \sf \leadsto \red{\dfrac{18}{5} = x}}$

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$\sf (v) \leadsto \dfrac{2x}{3} + \dfrac{5}{6} = \dfrac{13}{6} \\ \\ \sf \leadsto \dfrac{2x}{3} = \dfrac{13}{6} - \dfrac{5}{6} \\ \\ \sf \leadsto \dfrac{2x}{3} = \dfrac{13 - 5}{6} \\ \\ \sf \leadsto \dfrac{2x}{3} = \dfrac{8}{6} \\ \\ \sf \leadsto x = \dfrac{8}{6} \times \dfrac{3}{2} \\ \\ \boxed{\sf \leadsto \purple{x = 2}}$

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$\sf (vi) \leadsto \dfrac{x + 2}{3} + 5 = 17 \\ \\ \sf \leadsto \dfrac{x + 2}{3} = 17 - 5 \\ \\\sf \leadsto \dfrac{x + 2}{3} = 12 \\ \\ \sf \leadsto x + 2 = 12(3) \\ \\ \sf \leadsto x + 2 = 36 \\ \\ \sf \leadsto x = 36 - 2 \\ \\ \boxed{ \sf \leadsto \orange{x = 34}}$

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$\sf (vii) \leadsto 3y + \dfrac{5}{8} = \dfrac{11}{8} \\ \\ \sf \leadsto3y = \dfrac{11}{8} - \dfrac{5}{8} \\ \\ \sf \leadsto3y = \dfrac{11 - 5}{8} \\ \\ \sf \leadsto3y = \dfrac{6}{8} \\ \\\sf \leadsto y = \dfrac{6}{8 \times 3} \\ \\ \boxed{ \sf \leadsto \pink{y = \dfrac{1}{4} }}$

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$\sf (viii) \leadsto \dfrac{3m + 2}{3} = \dfrac{17}{3} \\ \\ \sf \leadsto 3m + 2 = \dfrac{17 \times 3}{3} \\ \\ \sf \leadsto3m + 2 = 17 \\ \\ \sf \leadsto3m = 17 - 2 \\ \\ \sf \leadsto3m = 15 \\ \\ \sf \leadsto m = \dfrac{15}{3} \\ \\ \boxed{\sf \leadsto \green{m = 5} }$

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$\sf (ix) \leadsto 2.5x + 3.5 = 6 \\ \\ \sf \leadsto2.5x = 6 - 3.5 \\ \\ \sf \leadsto2.5x = 2.5 \\ \\ \sf \leadsto x = \dfrac{2.5}{2.5} \\ \\ \boxed{\sf \leadsto \purple{x = 1}}$

Answer 2

Step-by-step explanation:

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