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Question Number 204330 by universe last updated on 13/Feb/24

Answered by deleteduser1 last updated on 13/Feb/24

3a+2b+15=0 ∧ 6a+2b≤0⇒b≤−3a ⇒b=((−15−3a)/2)≤−3a⇒−15−3a≤−6a⇒3a≤15 ⇒a≤5⇒max(a∣a∈Z)=5

3a+2b+15=0∧6a+2b⩽0⇒b⩽−3a⇒b=−15−3a2⩽−3a⇒−15−3a⩽−6a⇒3a⩽15⇒a⩽5⇒max(a∣a∈Z)=5

Commented by universe last updated on 13/Feb/24

thank you sir

thankyousir

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