Solve for x and y: 2x + 3y + 1 = 0\(\frac{7-4x}3\) = y

1.	Solve for x and y: 2x + 3y + 1 = 0\(\frac{7-4x}3\) = y
Answer» The given equations are: \(\frac{7-4x}3\) = y ⇒ 4x + 3y = 7 ……..(i) and 2x + 3y + 1 = 0 ⇒ 2x + 3y = -1 ……….(ii) On subtracting (ii) from (i), we get: 2x = 8 ⇒ x = 4 On substituting x = 4 in (i), we get: 16x + 3y = 7 ⇒ 3y = (7 – 16) = - 9 ⇒ y = - 3 Hence, the solution is x = 4 and y = - 3.

Answer»

The given equations are:

\(\frac{7-4x}3\) = y

⇒ 4x + 3y = 7 ……..(i)

and 2x + 3y + 1 = 0

⇒ 2x + 3y = -1 ……….(ii)

On subtracting (ii) from (i), we get:

2x = 8

⇒ x = 4

On substituting x = 4 in (i), we get:

16x + 3y = 7

⇒ 3y = (7 – 16) = - 9

⇒ y = - 3

Hence, the solution is x = 4 and y = - 3.

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