Question and Answers Forum

All Questions      Topic List

Arithmetic Questions

Previous in All Question      Next in All Question      

Previous in Arithmetic      Next in Arithmetic      

Question Number 176774 by Ar Brandon last updated on 26/Sep/22

How many distinct y exist satisfying  ∣x^2 −8x+18∣+∣y−3∣=5

$$\mathrm{How}\:\mathrm{many}\:\mathrm{distinct}\:{y}\:\mathrm{exist}\:\mathrm{satisfying} \\ $$$$\mid{x}^{\mathrm{2}} −\mathrm{8}{x}+\mathrm{18}\mid+\mid{y}−\mathrm{3}\mid=\mathrm{5} \\ $$

Answered by mr W last updated on 26/Sep/22

x,y∈Z  ∣(x−4)^2 +2∣+∣y−3∣=5  2+∣y−3∣=5 ⇒y−3=±3 ⇒y=0, 6  3+∣y−3∣=5 ⇒y−3=±2 ⇒y=1, 5  4 dictinct values of y exist.

$${x},{y}\in\mathbb{Z} \\ $$$$\mid\left({x}−\mathrm{4}\right)^{\mathrm{2}} +\mathrm{2}\mid+\mid{y}−\mathrm{3}\mid=\mathrm{5} \\ $$$$\mathrm{2}+\mid{y}−\mathrm{3}\mid=\mathrm{5}\:\Rightarrow{y}−\mathrm{3}=\pm\mathrm{3}\:\Rightarrow{y}=\mathrm{0},\:\mathrm{6} \\ $$$$\mathrm{3}+\mid{y}−\mathrm{3}\mid=\mathrm{5}\:\Rightarrow{y}−\mathrm{3}=\pm\mathrm{2}\:\Rightarrow{y}=\mathrm{1},\:\mathrm{5} \\ $$$$\mathrm{4}\:{dictinct}\:{values}\:{of}\:{y}\:{exist}. \\ $$

Commented by Ar Brandon last updated on 27/Sep/22

Thank you Sir. I now get it.

Commented by mr W last updated on 27/Sep/22

∣(x−4)^2 +2∣+∣y−3∣=5  let t=x−4  ∣t^2 +2∣+∣y−3∣=5  t^2  can only be 0, 1, 4, 9, 16 etc.  ∣t^2 +2∣ can only be 2, 3, 6, 11, 18 etc.  with ∣t^2 +2∣=2 you get 2+∣y−3∣=5.  with ∣t^2 +2∣=3 you get 3+∣y−3∣=5.  with ∣t^2 +2∣=6 you get 6+∣y−3∣=5 ⇒no solution!

$$\mid\left({x}−\mathrm{4}\right)^{\mathrm{2}} +\mathrm{2}\mid+\mid{y}−\mathrm{3}\mid=\mathrm{5} \\ $$$${let}\:{t}={x}−\mathrm{4} \\ $$$$\mid{t}^{\mathrm{2}} +\mathrm{2}\mid+\mid{y}−\mathrm{3}\mid=\mathrm{5} \\ $$$${t}^{\mathrm{2}} \:{can}\:{only}\:{be}\:\mathrm{0},\:\mathrm{1},\:\mathrm{4},\:\mathrm{9},\:\mathrm{16}\:{etc}. \\ $$$$\mid{t}^{\mathrm{2}} +\mathrm{2}\mid\:{can}\:{only}\:{be}\:\mathrm{2},\:\mathrm{3},\:\mathrm{6},\:\mathrm{11},\:\mathrm{18}\:{etc}. \\ $$$${with}\:\mid{t}^{\mathrm{2}} +\mathrm{2}\mid=\mathrm{2}\:{you}\:{get}\:\mathrm{2}+\mid{y}−\mathrm{3}\mid=\mathrm{5}. \\ $$$${with}\:\mid{t}^{\mathrm{2}} +\mathrm{2}\mid=\mathrm{3}\:{you}\:{get}\:\mathrm{3}+\mid{y}−\mathrm{3}\mid=\mathrm{5}. \\ $$$${with}\:\mid{t}^{\mathrm{2}} +\mathrm{2}\mid=\mathrm{6}\:{you}\:{get}\:\mathrm{6}+\mid{y}−\mathrm{3}\mid=\mathrm{5}\:\Rightarrow{no}\:{solution}! \\ $$

Commented by Ar Brandon last updated on 26/Sep/22

Great! Thank you, Sir!

Commented by Ar Brandon last updated on 27/Sep/22

Sir why  2+∣y−3∣ and 3+∣y−3∣ ?

$$\mathrm{Sir}\:\mathrm{why} \\ $$$$\mathrm{2}+\mid{y}−\mathrm{3}\mid\:\mathrm{and}\:\mathrm{3}+\mid{y}−\mathrm{3}\mid\:? \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com