Question

Choose the pair of equations which satisfy the point (1,-1) (a) 4x–y=3 , 4x+y=3 ( b) 4x+y=3 , 3x+2y=1 (c) 2x+3y=5 , 2x+3y=−1 (d) 2x+y=3 , 2x–y=1

Answer 1

Step-by-step explanation:

hope the above answer will help you

Answer 2

(b) 4x + y = 3, 3x + 2y = 1

Step-by-step explanation:

(a) 4x - y = 3, 4x + y = 3

The given equations are

4x - y = 3 ... ... (1)

4x + y = 3 ... ... (2)

• Putting x = 1, y = - 1 in LHS of (1), we get

LHS of (1)

= 4 (1) - (- 1) = 4 + 1 = 5 ≠ RHS of (1)

So, this pair does not satisfy (1, - 1).

(b) 4x + y = 3, 3x + 2y = 1

The given equations are

4x + y = 3 ... ... (1)

3x + 2y = 1 ... ... (2)

• Putting x = 1, y = - 1 in LHS of (1), we get

LHS of (1)

= 4 (1) + (- 1) = 4 - 1 = 3 = RHS of (1)

• Putting x = 1, y = - 1 in LHS of (2), we get

LHS of (2)

= 3 (1) + 2 (- 1) = 3 - 2 = 1 = RHS of (2)

So, this pair satisfies (1, - 1).

(c) 2x + 3y = 5, 2x + 3y = - 1

The given equations are

2x + 3y = 5 ... ... (1)

2x + 3y = - 1 ... ... (2)

• Putting x = 1, y = - 1 in LHS of (1), we get

LHS of (1)

= 2 (1) + 3 (- 1) = 2 - 3 = - 1 ≠ RHS of (1)

So, this pair does not satisfy (1, - 1).

(d) 2x + y = 3, 2x - y = 1

The given equations are

2x + y = 3 ... ... (1)

2x - y = 1 ... ... (2)

• Putting x = 1, y = - 1 in LHS of (1), we get

LHS of (1)

= 2 (1) + (- 1) = 2 - 1 = 1 ≠ RHS of (1)

So, this pair does not satisfy (1, - 1).

#SPJ3