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Question Number 6317 by sanusihammed last updated on 23/Jun/16

Solve the system of equation     x + y − z = 9   .......... equation (i)  x^2  + y^(2 ) + z^(2 ) = 99  ........... equation (ii)  y^2  = xz   ........... equation(iii)

$${Solve}\:{the}\:{system}\:{of}\:{equation}\: \\ $$$$ \\ $$$${x}\:+\:{y}\:−\:{z}\:=\:\mathrm{9}\:\:\:..........\:{equation}\:\left({i}\right) \\ $$$${x}^{\mathrm{2}} \:+\:{y}^{\mathrm{2}\:} +\:{z}^{\mathrm{2}\:} =\:\mathrm{99}\:\:...........\:{equation}\:\left({ii}\right) \\ $$$${y}^{\mathrm{2}} \:=\:{xz}\:\:\:...........\:{equation}\left({iii}\right) \\ $$

Commented by prakash jain last updated on 23/Jun/16

from equation (ii) and (iii)  (x−z)^2 =x^2 +z^2 −2xz=99−y^2 −2y^2 =99−3y^2   from (i)  (x−z)^2 =(9−y)^2   99−3y^2 =81−18y+y^2   4y^2 −18y−18=0  2y^2 −9y−9=0  y=((9±(√(81+72)))/4)=((9±(√(153)))/4)  y^2 =((81+153±18(√(153)))/(16))=((234±18(√(153)))/(16))=xz  x−z=9−((9±(√(153)))/4)=((27±(√(153)))/4)  x+z=(x−z)^2 +4xz

$${from}\:{equation}\:\left({ii}\right)\:{and}\:\left({iii}\right) \\ $$$$\left({x}−{z}\right)^{\mathrm{2}} ={x}^{\mathrm{2}} +{z}^{\mathrm{2}} −\mathrm{2}{xz}=\mathrm{99}−{y}^{\mathrm{2}} −\mathrm{2}{y}^{\mathrm{2}} =\mathrm{99}−\mathrm{3}{y}^{\mathrm{2}} \\ $$$${from}\:\left({i}\right) \\ $$$$\left({x}−{z}\right)^{\mathrm{2}} =\left(\mathrm{9}−{y}\right)^{\mathrm{2}} \\ $$$$\mathrm{99}−\mathrm{3}{y}^{\mathrm{2}} =\mathrm{81}−\mathrm{18}{y}+{y}^{\mathrm{2}} \\ $$$$\mathrm{4}{y}^{\mathrm{2}} −\mathrm{18}{y}−\mathrm{18}=\mathrm{0} \\ $$$$\mathrm{2}{y}^{\mathrm{2}} −\mathrm{9}{y}−\mathrm{9}=\mathrm{0} \\ $$$${y}=\frac{\mathrm{9}\pm\sqrt{\mathrm{81}+\mathrm{72}}}{\mathrm{4}}=\frac{\mathrm{9}\pm\sqrt{\mathrm{153}}}{\mathrm{4}} \\ $$$${y}^{\mathrm{2}} =\frac{\mathrm{81}+\mathrm{153}\pm\mathrm{18}\sqrt{\mathrm{153}}}{\mathrm{16}}=\frac{\mathrm{234}\pm\mathrm{18}\sqrt{\mathrm{153}}}{\mathrm{16}}={xz} \\ $$$${x}−{z}=\mathrm{9}−\frac{\mathrm{9}\pm\sqrt{\mathrm{153}}}{\mathrm{4}}=\frac{\mathrm{27}\pm\sqrt{\mathrm{153}}}{\mathrm{4}} \\ $$$${x}+{z}=\left({x}−{z}\right)^{\mathrm{2}} +\mathrm{4}{xz} \\ $$

Answered by Rasheed Soomro last updated on 24/Jun/16

x + y − z = 9   ............... (i)  x^2  + y^(2 ) + z^(2 ) = 99  ...........  (ii)  y^2  = xz   ........... .............(iii)  (i)^2  : (x + y − z)^2  = 9^2             x^2  + y^(2 ) + z^(2 ) +2xy−2yz−2zx=81            99+2xy−2yz−2(y^2 )=81  [ (ii)x^2  + y^(2 ) + z^(2 ) = 99 & (iii)xz=y^2 ]            2xy−2yz−2y^2 =−18           xy−yz−y^2 =−9  y(x−z^(−) −y)=−9  y(9−y−y)=−9   [From(i) x−z=9−y]  9y−2y^2 =−9  2y^2 −9y−9=0  y=((9±(√(81+72)))/4)=((9±(√(153)))/4)=    continue

$${x}\:+\:{y}\:−\:{z}\:=\:\mathrm{9}\:\:\:...............\:\left({i}\right) \\ $$$${x}^{\mathrm{2}} \:+\:{y}^{\mathrm{2}\:} +\:{z}^{\mathrm{2}\:} =\:\mathrm{99}\:\:...........\:\:\left({ii}\right) \\ $$$${y}^{\mathrm{2}} \:=\:{xz}\:\:\:...........\:.............\left({iii}\right) \\ $$$$\left({i}\right)^{\mathrm{2}} \::\:\left({x}\:+\:{y}\:−\:{z}\right)^{\mathrm{2}} \:=\:\mathrm{9}^{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:\:{x}^{\mathrm{2}} \:+\:{y}^{\mathrm{2}\:} +\:{z}^{\mathrm{2}\:} +\mathrm{2}{xy}−\mathrm{2}{yz}−\mathrm{2}{zx}=\mathrm{81} \\ $$$$\:\:\:\:\:\:\:\:\:\:\mathrm{99}+\mathrm{2}{xy}−\mathrm{2}{yz}−\mathrm{2}\left({y}^{\mathrm{2}} \right)=\mathrm{81}\:\:\left[\:\left({ii}\right){x}^{\mathrm{2}} \:+\:{y}^{\mathrm{2}\:} +\:{z}^{\mathrm{2}\:} =\:\mathrm{99}\:\&\:\left({iii}\right){xz}={y}^{\mathrm{2}} \right] \\ $$$$\:\:\:\:\:\:\:\:\:\:\mathrm{2}{xy}−\mathrm{2}{yz}−\mathrm{2}{y}^{\mathrm{2}} =−\mathrm{18} \\ $$$$\:\:\:\:\:\:\:\:\:{xy}−{yz}−{y}^{\mathrm{2}} =−\mathrm{9} \\ $$$${y}\left(\overline {{x}−{z}}−{y}\right)=−\mathrm{9} \\ $$$${y}\left(\mathrm{9}−{y}−{y}\right)=−\mathrm{9}\:\:\:\left[{From}\left({i}\right)\:{x}−{z}=\mathrm{9}−{y}\right] \\ $$$$\mathrm{9}{y}−\mathrm{2}{y}^{\mathrm{2}} =−\mathrm{9} \\ $$$$\mathrm{2}{y}^{\mathrm{2}} −\mathrm{9}{y}−\mathrm{9}=\mathrm{0} \\ $$$${y}=\frac{\mathrm{9}\pm\sqrt{\mathrm{81}+\mathrm{72}}}{\mathrm{4}}=\frac{\mathrm{9}\pm\sqrt{\mathrm{153}}}{\mathrm{4}}= \\ $$$$ \\ $$$${continue} \\ $$

Commented by Rasheed Soomro last updated on 24/Jun/16

A defferent  way  to determine value of y in   the question. Don′t claim to be better than  the way in comment by sir prakash above.

$${A}\:{defferent}\:\:{way}\:\:{to}\:{determine}\:{value}\:{of}\:{y}\:{in}\: \\ $$$${the}\:{question}.\:{Don}'{t}\:{claim}\:{to}\:{be}\:{better}\:{than} \\ $$$${the}\:{way}\:{in}\:{comment}\:{by}\:{sir}\:{prakash}\:{above}. \\ $$

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