Question

Subtract the second expression from the first expression:

2l² – 3lm + 5m², 3l² – 4lm + 6m²

Answer 1

Let A = 2l² – 3lm + 5m² and 

B = 3l² – 4lm + 6m² 

A – B = A + (- B) 

Additive inverse of B is 

– B = – (3t² – 4lm + 6m² ) 

= – 3l² + 4lm – 6m² 

∴ A – B = A + (- B) 

= (2l² – 3lm + 5m² ) + (- 3l² + 4lm – 6m² )

= 2l² – 3lm + 5m² – 3l² + 4lm – 6m² 

= 2l² – 3l² – 3lm + 4lm + 5m² – 6m²

= (2 – 3)l² + (- 3 + 4)lm + (5 – 6)m² 

= (- 1) l² + 1 lm + (- 1)m² 

∴ A – B = – l² + lm – m²