Question and Answers Forum

All Questions      Topic List

Algebra Questions

Previous in All Question      Next in All Question      

Previous in Algebra      Next in Algebra      

Question Number 15051 by tawa tawa last updated on 07/Jun/17

Solve simultaneously x + y + z = 6 ............ equation (i) x^3 + y^3 + z^3 = 92 .......... equation (ii) x − y = z ........... equation (iii)

$$\mathrm{Solve}\:\mathrm{simultaneously} \\ $$$$ \\ $$$$\mathrm{x}\:+\:\mathrm{y}\:+\:\mathrm{z}\:=\:\mathrm{6}\:\:\:\:\:\:\:\:\:\:\:\:\:\:............\:\mathrm{equation}\:\left(\mathrm{i}\right) \\ $$$$\mathrm{x}^{\mathrm{3}} \:+\:\mathrm{y}^{\mathrm{3}} \:+\:\mathrm{z}^{\mathrm{3}} \:=\:\mathrm{92}\:\:\:\:\:\:\:\:\:..........\:\mathrm{equation}\:\left(\mathrm{ii}\right) \\ $$$$\mathrm{x}\:−\:\mathrm{y}\:=\:\mathrm{z}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:...........\:\mathrm{equation}\:\left(\mathrm{iii}\right) \\ $$

Answered by Tinkutara last updated on 07/Jun/17

y + z = x 2x = 6 ⇒ x = 3 y + z = 3 y^3 + z^3 = 92 − 27 = 65 (y + z)(y^2 + z^2 − yz) = 65 y^2 + z^2 − yz = ((65)/3) = (y + z)^2 − 3yz 9 − 3yz = ((65)/3) 4yz = ((−152)/9) (y − z)^2 = (y + z)^2 − 4yz = 9 + ((152)/9) y − z = ± ((√(233))/3) y + z = 3 y = ((9 ± (√(233)))/6) z = ((9 ∓ (√(233)))/6) x = 3

$${y}\:+\:{z}\:=\:{x} \\ $$$$\mathrm{2}{x}\:=\:\mathrm{6}\:\Rightarrow\:{x}\:=\:\mathrm{3} \\ $$$${y}\:+\:{z}\:=\:\mathrm{3} \\ $$$${y}^{\mathrm{3}} \:+\:{z}^{\mathrm{3}} \:=\:\mathrm{92}\:−\:\mathrm{27}\:=\:\mathrm{65} \\ $$$$\left({y}\:+\:{z}\right)\left({y}^{\mathrm{2}} \:+\:{z}^{\mathrm{2}} \:−\:{yz}\right)\:=\:\mathrm{65} \\ $$$${y}^{\mathrm{2}} \:+\:{z}^{\mathrm{2}} \:−\:{yz}\:=\:\frac{\mathrm{65}}{\mathrm{3}}\:=\:\left({y}\:+\:{z}\right)^{\mathrm{2}} \:−\:\mathrm{3}{yz} \\ $$$$\mathrm{9}\:−\:\mathrm{3}{yz}\:=\:\frac{\mathrm{65}}{\mathrm{3}} \\ $$$$\mathrm{4}{yz}\:=\:\frac{−\mathrm{152}}{\mathrm{9}} \\ $$$$\left({y}\:−\:{z}\right)^{\mathrm{2}} \:=\:\left({y}\:+\:{z}\right)^{\mathrm{2}} \:−\:\mathrm{4}{yz}\:=\:\mathrm{9}\:+\:\frac{\mathrm{152}}{\mathrm{9}} \\ $$$${y}\:−\:{z}\:=\:\pm\:\frac{\sqrt{\mathrm{233}}}{\mathrm{3}} \\ $$$${y}\:+\:{z}\:=\:\mathrm{3} \\ $$$${y}\:=\:\frac{\mathrm{9}\:\pm\:\sqrt{\mathrm{233}}}{\mathrm{6}} \\ $$$${z}\:=\:\frac{\mathrm{9}\:\mp\:\sqrt{\mathrm{233}}}{\mathrm{6}} \\ $$$${x}\:=\:\mathrm{3} \\ $$

Commented by tawa tawa last updated on 07/Jun/17

God bless you sir.

$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}. \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com