Question

If `(2x + y + 2)/5 = (3x - y + 1)/3 = (3x + 2y + 1)/6` then

Options

• x = 1, y = 1

• x = –1, y = –1

• x = 1, y = 2

• x = 2, y = 1

Answer 1

x = 1, y = 1

Explanation:

Let the common value be t.

Then 2x + y + 2 = 5t, 3x – y + 1 = 3t, 3x + 2y + 1 = 6t.

Subtract the second from the third:

(3x + 2y + 1) – (3x – y + 1) = 6t – 3t 

⇒ 3y = 3t

⇒ y = t

Substitute y = t into 3x – y + 1 = 3t 

⇒ 3x – y + 1 = 3y 

⇒ 3x + 1 = 4y 

⇒ `x = (4y - 1)/3` 

Use 2x + y + 2 = 5y 

⇒ 2x + 2 = 4y 

⇒ Substitute x

⇒ `2((4y - 1)/3) + 2 = 4y` 

Solve: `(8y - 2)/3 + 2 = 4y` 

⇒ `(8y - 2 + 6)/3 = 4y` 

⇒ `(8y + 4)/3 = 4y` 

⇒ 8y + 4 = 12y 

⇒ 4 = 4y

⇒ y = 1 

Then `x = (4 xx 1 - 1)/3 = 1` 

Hence x = 1, y = 1.