x + y = 5 … (1)

2x – 3y = 4 … (2)

Elimination method:

Multiplying equation (1) by 2, we GET equation (3)

2x + 2y = 10 … (3)

2x − 3y = 4 … (2)

Subtracting equation (2) from (3), we get

5y = 6⇒ y = 6/5

Putting value of y in (1), we get

x + = 5

⇒ x = 5 − 6/5 = 19/5

THEREFORE, x = 6/5 and y = 19/5

Substitution method:

x + y = 5 … (1)

2x − 3y = 4 … (2)

From equation (1), we get,

x = 5 − y

Putting this in equation (2), we get

2 (5 − y) − 3y = 4

⇒ 10 − 2y − 3y = 4

⇒ 5y = 6 ⇒ y = 6/5

Putting value of y in (1), we get

x = 5 −6/5 =

Therefore, x =6/5 and y = 19/5

3X + 4y = 10… (1)

2x – 2y = 2… (2)

Elimination method:

Multiplying equation (2) by 2, we get (3)

4x − 4y = 4 … (3)

3x + 4y = 10 … (1)

Adding (3) and (1), we get

7x = 14⇒ x = 2

Putting value of x in (1), we get

3 (2) + 4y = 10

⇒ 4y = 10 – 6 = 4

⇒ y = 1

Therefore, x = 2 and y = 1

Substitution method:

3x + 4y = 10… (1)

2x − 2y = 2… (2)

From equation (2), we get

2x = 2 + 2y

⇒ x = 1 + y … (3)

Putting this in equation (1), we get

3 (1 + y) + 4y = 10

⇒ 3 + 3y + 4y = 10

⇒ 7y = 7⇒ y = 1

Putting value of y in (3), we get x = 1 + 1 = 2

Therefore, x = 2 and y = 1

3x − 5y – 4 = 0 … (1)

9x = 2y + 7… (2)

Elimination method:

Multiplying (1) by 3, we get (3)

9x − 15y – 12 = 0… (3)

9x − 2y – 7 = 0… (2)

Subtracting (2) from (3), we get

−13y – 5 = 0

⇒ −13y = 5

⇒ y = -5/13

Putting value of y in (1), we get

3x – 5(-5/13) − 4 = 0

⇒ 3x = 4 −25/13 = 52 - 25 /13 =27/13

⇒ x = 27/13 × 3 = 9/ 13

Therefore, x = 9/13 and y = -5/13

Substitution Method:

3x − 5y – 4 = 0 … (1)

9x = 2y + 7… (2)

From equation (1), we can SAY that

3x = 4 + 5y⇒ x = 4+5y/ 3

Putting this in equation (2), we get

9 (4+5y/3) − 2y = 7

⇒ 12 + 15y − 2y = 7

⇒ 13y = −5 ⇒ y = -5/13

Putting value of y in (1), we get

3x – 5(-5/13) = 4

⇒ 3x = 4 − = 25/13 = 52-25/13 = 27/13

⇒ x = 9/13

Therefore, x = 9/13 and y = -5/13

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