Question Number 68280 by mezihloic last updated on 08/Sep/19
![given that 432_n −413_n =11_(10) .find the value of n](https://www.tinkutara.com/question/Q68280.png)
$${given}\:{that}\:\mathrm{432}_{{n}} −\mathrm{413}_{{n}} =\mathrm{11}_{\mathrm{10}} .{find}\:{the}\:{value}\:{of}\:{n} \\ $$
Commented by TawaTawa last updated on 08/Sep/19
![4 × n^2 + 3 × n^1 + 2 × n^0 − (4 × n^2 + 1 × n^1 + 3 × n^0 ) = 11 4n^2 + 3n + 2 − (4n^2 + n + 3) = 11 4n^2 + 3n + 2 − 4n^2 − n − 3 = 11 3n + 2 − n − 3 = 11 2n − 1 = 11 2n = 11 + 1 2n = 12 n = ((12)/2) n = 6](https://www.tinkutara.com/question/Q68282.png)
$$\mathrm{4}\:×\:\mathrm{n}^{\mathrm{2}} \:+\:\mathrm{3}\:×\:\mathrm{n}^{\mathrm{1}} \:+\:\mathrm{2}\:×\:\mathrm{n}^{\mathrm{0}} \:−\:\left(\mathrm{4}\:×\:\mathrm{n}^{\mathrm{2}} \:+\:\mathrm{1}\:×\:\mathrm{n}^{\mathrm{1}} \:+\:\mathrm{3}\:×\:\mathrm{n}^{\mathrm{0}} \right)\:\:=\:\:\mathrm{11} \\ $$$$\mathrm{4n}^{\mathrm{2}} \:+\:\mathrm{3n}\:+\:\mathrm{2}\:−\:\left(\mathrm{4n}^{\mathrm{2}} \:+\:\mathrm{n}\:+\:\mathrm{3}\right)\:\:=\:\:\mathrm{11} \\ $$$$\mathrm{4n}^{\mathrm{2}} \:+\:\mathrm{3n}\:+\:\mathrm{2}\:−\:\mathrm{4n}^{\mathrm{2}} \:−\:\mathrm{n}\:−\:\mathrm{3}\:\:=\:\:\mathrm{11} \\ $$$$\mathrm{3n}\:+\:\mathrm{2}\:−\:\mathrm{n}\:−\:\mathrm{3}\:\:=\:\:\mathrm{11} \\ $$$$\mathrm{2n}\:−\:\mathrm{1}\:\:=\:\:\mathrm{11} \\ $$$$\mathrm{2n}\:\:=\:\:\mathrm{11}\:+\:\mathrm{1} \\ $$$$\mathrm{2n}\:\:=\:\:\mathrm{12} \\ $$$$\:\:\mathrm{n}\:\:=\:\:\frac{\mathrm{12}}{\mathrm{2}} \\ $$$$\:\:\mathrm{n}\:\:=\:\:\mathrm{6} \\ $$
Answered by mr W last updated on 08/Sep/19
![4n^2 +3n+2−(4n^2 +n+3)=11 2n−1=11 ⇒n=6](https://www.tinkutara.com/question/Q68283.png)
$$\mathrm{4}{n}^{\mathrm{2}} +\mathrm{3}{n}+\mathrm{2}−\left(\mathrm{4}{n}^{\mathrm{2}} +{n}+\mathrm{3}\right)=\mathrm{11} \\ $$$$\mathrm{2}{n}−\mathrm{1}=\mathrm{11} \\ $$$$\Rightarrow{n}=\mathrm{6} \\ $$