Question

y-y+1/3=1+y-1/2 linear equations​

Answer 1

Answer:

⇢ ${\boxed{\sf{y = \dfrac{-1}{6}}}}$

Step-by-step explanation:

⇢ $\sf{y - y + \dfrac{1}{3} = 1 + y - \dfrac{1}{2}}$

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⇢ $\sf{y + -y + \dfrac{1}{3} = 1 + y + \dfrac{-1}{2}}$

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⇢ $\sf{(y + -y) + \bigg(\dfrac{1}{3} \bigg) = y + \bigg(1 + \dfrac{-1}{2}\bigg)}$

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⇢ $\sf{\dfrac{1}{3} = y + \dfrac{1}{2}}$

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⇢ $\sf{y + \dfrac{1}{2} = \dfrac{1}{3}}$

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⇢ $\sf{y + \dfrac{1}{2} - \dfrac{1}{2} = \dfrac{1}{3} - \dfrac{1}{2}}$

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⇢ ${\boxed{\sf{y = \dfrac{-1}{6}}}}$

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Therefore this is the required answer.

Answer 2

y-y+1/3=1+y-1/2

=> 3y-y-1 / 3 = 2+y-1 / 2

=> 2y-1 / 3 = 1+y / 2

=> 2 ( 2y - 1) = 3 ( 1 + y)

=> 4y- 2 = 3 + 3y

=> 4y - 3y = 3 + 2

=> y = 5 [ answer]