Solve:[(y – (4 – 3y))/ (2y – (3 + 4y))] = 1/5

1.	Solve:[(y – (4 – 3y))/ (2y – (3 + 4y))] = 1/5
Answer» We have, [(y – (4 – 3y))/ (2y – (3 + 4y))] = 1/5 (y – 4 + 3y)/ (2y – 3 – 4y) = 1/5 (-4y – 4)/ (2y – 3) = 1/5 By cross multiplication, we get 5 × (-4y – 4) = 1 × (2y – 3) 20y – 20 = 2y – 3 Transposing – 20 to RHS it becomes 20 and 6y to LHS it becomes -6y. 20y – 2y = 20 – 3 22 y = 17 y = 17/22

Answer»

We have,

[(y – (4 – 3y))/ (2y – (3 + 4y))] = 1/5

(y – 4 + 3y)/ (2y – 3 – 4y) = 1/5

(-4y – 4)/ (2y – 3) = 1/5

By cross multiplication, we get

5 × (-4y – 4) = 1 × (2y – 3)

20y – 20 = 2y – 3

Transposing – 20 to RHS it becomes 20 and 6y to LHS it becomes -6y.

20y – 2y = 20 – 3

22 y = 17

y = 17/22

Discussion