1
Question
Simplify: 3y−[2y−{4−(y−2)}−5]
Open in App
Solution
3y−[2y−{4−(y−2)}−5] = 3y−[2y−{4−y+2}−5]= 3y−[2y−{6−y}−5]= 3y−[2y− 6+ y}−5]= 3y−[3y− 6}−5]= 3y−[3y− 6−5]= 3y−[3y− 11]= 3y− 3y+ 11= 11
3y−[2y−{4−(y−2)}−5]
= 3y−[2y−{4−y+2}−5]
= 3y−[2y−{6−y}−5]
= 3y−[2y− 6+ y}−5]
= 3y−[3y− 6}−5]
= 3y−[3y− 6−5]
= 3y−[3y− 11]
= 3y− 3y+ 11
= 11
Suggest Corrections
0
Similar questions
View More
Join BYJU'S Learning Program
Explore more
Join BYJU'S Learning Program
