Question and Answers Forum

All Questions      Topic List

None Questions

Previous in All Question      Next in All Question      

Previous in None      Next in None      

Question Number 111781 by mohammad17 last updated on 04/Sep/20

Commented by Dwaipayan Shikari last updated on 04/Sep/20

((2n+14)/(n+7))−((15)/(n+7))=2−((15)/(n+7))  n(−2,−4,,−6,−8,−10,−12,−22,8)

$$\frac{\mathrm{2}{n}+\mathrm{14}}{{n}+\mathrm{7}}−\frac{\mathrm{15}}{{n}+\mathrm{7}}=\mathrm{2}−\frac{\mathrm{15}}{{n}+\mathrm{7}} \\ $$$${n}\left(−\mathrm{2},−\mathrm{4},,−\mathrm{6},−\mathrm{8},−\mathrm{10},−\mathrm{12},−\mathrm{22},\mathrm{8}\right) \\ $$

Commented by mohammad17 last updated on 04/Sep/20

thank you sir

$${thank}\:{you}\:{sir} \\ $$

Commented by Aziztisffola last updated on 04/Sep/20

((2n−1)/(n+7))=2−((15)/(n+7))=k∈Z  ⇒((15)/(n+7))=2−k∈Z ⇒n+7∣15  n+7=+_− 15  ⇒n=8 ;n=−22  n+7=+_− 5⇒n=−2 ;n=−12  n+7=+_− 1⇒n=−6; n=−8  n+7=+_− 3⇒n=−4 ;n=−10

$$\frac{\mathrm{2n}−\mathrm{1}}{\mathrm{n}+\mathrm{7}}=\mathrm{2}−\frac{\mathrm{15}}{\mathrm{n}+\mathrm{7}}=\mathrm{k}\in\mathbb{Z} \\ $$$$\Rightarrow\frac{\mathrm{15}}{\mathrm{n}+\mathrm{7}}=\mathrm{2}−\mathrm{k}\in\mathbb{Z}\:\Rightarrow\mathrm{n}+\mathrm{7}\mid\mathrm{15} \\ $$$$\mathrm{n}+\mathrm{7}=\underset{−} {+}\mathrm{15}\:\:\Rightarrow\mathrm{n}=\mathrm{8}\:;\mathrm{n}=−\mathrm{22} \\ $$$$\mathrm{n}+\mathrm{7}=\underset{−} {+}\mathrm{5}\Rightarrow\mathrm{n}=−\mathrm{2}\:;\mathrm{n}=−\mathrm{12} \\ $$$$\mathrm{n}+\mathrm{7}=\underset{−} {+}\mathrm{1}\Rightarrow\mathrm{n}=−\mathrm{6};\:\mathrm{n}=−\mathrm{8} \\ $$$$\mathrm{n}+\mathrm{7}=\underset{−} {+}\mathrm{3}\Rightarrow\mathrm{n}=−\mathrm{4}\:;\mathrm{n}=−\mathrm{10} \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com