CameraIcon
CameraIcon
SearchIcon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

Subtract:

(i) − 5y2 from y2

(ii) 6xy from − 12xy

(iii) (ab) from (a + b)

(iv) a (b − 5) from b (5 − a)

(v) − m2 + 5mn from 4m2− 3mn + 8

(vi) − x2 + 10x − 5 from 5x− 10

(vii) 5a2 − 7ab + 5b2from 3ab − 2a2 −2b2

(viii) 4pq − 5q2 − 3p2from 5p2 + 3q2 pq

Open in App
Solution

(i) y2− (−5y2) = y2 + 5y2 = 6y2

(ii) −12xy − (6xy) = −18xy

(iii) (a+ b) − (a − b) = a + b −a + b = 2b

(iv) b(5 − a) − a (b − 5) = 5b− ab − ab + 5a

= 5a + 5b − 2ab

(v) (4m2− 3mn + 8) − (− m2 + 5mn)= 4m2 − 3mn + 8 + m2− 5 mn

= 4m2 + m2 − 3mn −5 mn + 8

= 5m2 − 8mn + 8

(vi) (5x− 10) − (− x2 + 10x −5) =5x − 10 + x2 − 10x+ 5

= x2 + 5x − 10x − 10 + 5

= x2 − 5x − 5

(vii) (3ab− 2a2 − 2b2) −(5a2− 7ab + 5b2)

= 3ab − 2a2 − 2b2− 5a2 + 7ab − 5 b2

= 3ab + 7ab − 2a2 − 5a2− 2b2 − 5 b2

= 10ab − 7a2 − 7b2

(viii) 4pq− 5q2 − 3p2 from 5p2+ 3q2 − pq

(5p2 + 3q2 − pq) −(4pq − 5q2− 3p2)

= 5p2 + 3q2 − pq −4pq + 5q2 + 3p2

= 5p2 + 3p2 + 3q2+ 5q2 − pq − 4pq

= 8p2 + 8q2 − 5pq


flag
Suggest Corrections
thumbs-up
0
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
MATHEMATICS
Watch in App
Join BYJU'S Learning Program
CrossIcon