Menu Close

Solve-simultaneously-x-y-xy-3-i-x-1-y-1-4-ii-

Tinku Tara 0 Comments
Pin




Question Number 12023 by tawa last updated on 09/Apr/17
Solve simultaneously x + y − (√(xy)) = 3 .......... (i) (√(x + 1)) + (√(y + 1)) = 4 .......... (ii)
$$\mathrm{Solve}\:\mathrm{simultaneously} \\ $$$$\mathrm{x}\:+\:\mathrm{y}\:−\:\sqrt{\mathrm{xy}}\:=\:\mathrm{3}\:\:\:\:\:……….\:\left(\mathrm{i}\right) \\ $$$$\sqrt{\mathrm{x}\:+\:\mathrm{1}}\:+\:\sqrt{\mathrm{y}\:+\:\mathrm{1}}\:=\:\mathrm{4}\:\:\:\:\:……….\:\left(\mathrm{ii}\right) \\ $$
Answered by Mr Chheang Chantria last updated on 10/Apr/17
(i). x+y−(√(xy))≥x+y−((x+y)/2)=((x+y)/2) 3 ≥ ((x+y)/2) ⇔x+y≤6 . (√(x+1))+(√(y+1)) ≤(√(2(x+y+2))) ≤(√(2(6+2))) = 4 ⇒(√(x+1))+(√(y+1))≤4 by (ii) the sign equal when x=y=3
$$\left(\mathrm{i}\right).\:\boldsymbol{{x}}+\boldsymbol{{y}}−\sqrt{\boldsymbol{{xy}}}\geqslant\boldsymbol{{x}}+\boldsymbol{{y}}−\frac{\boldsymbol{{x}}+\boldsymbol{{y}}}{\mathrm{2}}=\frac{\boldsymbol{{x}}+\boldsymbol{{y}}}{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\mathrm{3}\:\:\:\:\:\:\:\:\:\:\:\:\:\geqslant\:\frac{\boldsymbol{{x}}+\boldsymbol{{y}}}{\mathrm{2}} \\ $$$$\Leftrightarrow\boldsymbol{{x}}+\boldsymbol{{y}}\leqslant\mathrm{6}\:\: \\ $$$$.\:\sqrt{\boldsymbol{{x}}+\mathrm{1}}+\sqrt{\boldsymbol{{y}}+\mathrm{1}}\:\leqslant\sqrt{\mathrm{2}\left(\boldsymbol{{x}}+\boldsymbol{{y}}+\mathrm{2}\right)} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\leqslant\sqrt{\mathrm{2}\left(\mathrm{6}+\mathrm{2}\right)}\:=\:\mathrm{4} \\ $$$$\Rightarrow\sqrt{\boldsymbol{{x}}+\mathrm{1}}+\sqrt{\boldsymbol{{y}}+\mathrm{1}}\leqslant\mathrm{4}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\: \\ $$$$\boldsymbol{{by}}\:\left(\boldsymbol{{ii}}\right)\:\boldsymbol{{the}}\:\boldsymbol{{sign}}\:\boldsymbol{{equal}}\:\boldsymbol{{when}}\:\boldsymbol{{x}}=\boldsymbol{{y}}=\mathrm{3} \\ $$
Commented by Mr Chheang Chantria last updated on 10/Apr/17
this is for student. we have ((√x)−(√y))^2 ≥0 for x,y≥0 x−2(√(xy))+y≥0 x+y≥2(√(xy)) or ((x+y)/2)≥(√(xy)) or −((x+y)/2)≤−(√(xy))
$$\boldsymbol{{this}}\:\boldsymbol{{is}}\:\boldsymbol{{for}}\:\boldsymbol{{student}}. \\ $$$$\:\boldsymbol{{we}}\:\boldsymbol{{have}}\:\left(\sqrt{\boldsymbol{{x}}}−\sqrt{\boldsymbol{{y}}}\right)^{\mathrm{2}} \geqslant\mathrm{0}\:\:\:\mathrm{for}\:\boldsymbol{{x}},\boldsymbol{{y}}\geqslant\mathrm{0} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\boldsymbol{{x}}−\mathrm{2}\sqrt{\boldsymbol{{xy}}}+\boldsymbol{{y}}\geqslant\mathrm{0} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\boldsymbol{{x}}+\boldsymbol{{y}}\geqslant\mathrm{2}\sqrt{\boldsymbol{{xy}}} \\ $$$$\:\:\:\:\:\:\:\boldsymbol{{or}}\:\frac{\boldsymbol{{x}}+\boldsymbol{{y}}}{\mathrm{2}}\geqslant\sqrt{\boldsymbol{{xy}}} \\ $$$$\:\:\:\:\:\:\:\boldsymbol{{or}}\:−\frac{\boldsymbol{{x}}+\boldsymbol{{y}}}{\mathrm{2}}\leqslant−\sqrt{\boldsymbol{{xy}}} \\ $$$$ \\ $$
Commented by tawa last updated on 10/Apr/17
God bless you sir.
$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}. \\ $$
Answered by ajfour last updated on 09/Apr/17
squaring ...(ii) x+y+2+2(√((x+2)(y+1))) =16 3+(√(xy))+2+2(√((x+1)(y+1))) =16 2(√((x+1)(y+1))) =11−(√(xy)) squaring : 4(xy+1+x+y)=121+xy−22(√(xy)) 4(xy+1+3+(√(xy)) =121+xy−22(√(xy)) 3xy+26(√(xy))−105=0 if (√(xy)) =t 3t^2 +26t−105=0 t=(√(xy))=((−26+(√(676+1260)))/6) (√(xy)) =((−26+44)/6)=3 ....(iii) xy=9 from ....(i) x+y=3+(√(xy)) ⇒ x+y=6 ....(iv) (x−y)^2 =(x+y)^2 −4xy so, (x−y)^2 =36−4(9)=0 ⇒ x=y and as x+y=6 x=y=3 .
$${squaring}\:…\left({ii}\right) \\ $$$${x}+{y}+\mathrm{2}+\mathrm{2}\sqrt{\left({x}+\mathrm{2}\right)\left({y}+\mathrm{1}\right)}\:=\mathrm{16} \\ $$$$\mathrm{3}+\sqrt{{xy}}+\mathrm{2}+\mathrm{2}\sqrt{\left({x}+\mathrm{1}\right)\left({y}+\mathrm{1}\right)}\:=\mathrm{16} \\ $$$$\mathrm{2}\sqrt{\left({x}+\mathrm{1}\right)\left({y}+\mathrm{1}\right)}\:=\mathrm{11}−\sqrt{{xy}} \\ $$$${squaring}\:: \\ $$$$\mathrm{4}\left({xy}+\mathrm{1}+{x}+{y}\right)=\mathrm{121}+{xy}−\mathrm{22}\sqrt{{xy}} \\ $$$$\mathrm{4}\left({xy}+\mathrm{1}+\mathrm{3}+\sqrt{{xy}}\:=\mathrm{121}+{xy}−\mathrm{22}\sqrt{{xy}}\right. \\ $$$$\mathrm{3}{xy}+\mathrm{26}\sqrt{{xy}}−\mathrm{105}=\mathrm{0} \\ $$$${if}\:\sqrt{{xy}}\:={t} \\ $$$$\mathrm{3}{t}^{\mathrm{2}} +\mathrm{26}{t}−\mathrm{105}=\mathrm{0} \\ $$$${t}=\sqrt{{xy}}=\frac{−\mathrm{26}+\sqrt{\mathrm{676}+\mathrm{1260}}}{\mathrm{6}} \\ $$$$\sqrt{{xy}}\:=\frac{−\mathrm{26}+\mathrm{44}}{\mathrm{6}}=\mathrm{3}\:\:\:\:\:….\left({iii}\right) \\ $$$${xy}=\mathrm{9} \\ $$$${from}\:….\left({i}\right) \\ $$$${x}+{y}=\mathrm{3}+\sqrt{{xy}} \\ $$$$\Rightarrow\:{x}+{y}=\mathrm{6}\:\:\:….\left({iv}\right) \\ $$$$\left({x}−{y}\right)^{\mathrm{2}} =\left({x}+{y}\right)^{\mathrm{2}} −\mathrm{4}{xy} \\ $$$${so},\:\left({x}−{y}\right)^{\mathrm{2}} =\mathrm{36}−\mathrm{4}\left(\mathrm{9}\right)=\mathrm{0} \\ $$$$\Rightarrow\:\:{x}={y} \\ $$$${and}\:{as}\:\:{x}+{y}=\mathrm{6} \\ $$$$\:\:\:{x}={y}=\mathrm{3}\:\:. \\ $$
Commented by tawa last updated on 09/Apr/17
wow. God bless you sir.
$$\mathrm{wow}.\:\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}. \\ $$
Answered by b.e.h.i.8.3.4.1.7@gmail.com last updated on 09/Apr/17
x+1+y+1+2(√(x+1))(√(y+1))=16 (√(xy))+5+2(√(x+1))(√(y+1))=16 2(√(x+1))(√(y+1))=11−(√(xy)) 4(xy+x+y+1)=121−22(√(xy))+xy x+y=t 4[(t−3)^2 +t+1]=121−22(t−3)+(t−3)^2 3(t−3)^2 +26(t−3)−105=0 t−3=((−26±(√(26^2 +4×3×105)))/(2×3))=((−26±44)/6)=3,−11.66 t=x+y=6⇒(√(xy))=3⇒xy=9 ⇒x=y=3 ■
$${x}+\mathrm{1}+{y}+\mathrm{1}+\mathrm{2}\sqrt{{x}+\mathrm{1}}\sqrt{{y}+\mathrm{1}}=\mathrm{16} \\ $$$$\sqrt{{xy}}+\mathrm{5}+\mathrm{2}\sqrt{{x}+\mathrm{1}}\sqrt{{y}+\mathrm{1}}=\mathrm{16} \\ $$$$\mathrm{2}\sqrt{{x}+\mathrm{1}}\sqrt{{y}+\mathrm{1}}=\mathrm{11}−\sqrt{{xy}} \\ $$$$\mathrm{4}\left({xy}+{x}+{y}+\mathrm{1}\right)=\mathrm{121}−\mathrm{22}\sqrt{{xy}}+{xy} \\ $$$${x}+{y}={t} \\ $$$$\mathrm{4}\left[\left({t}−\mathrm{3}\right)^{\mathrm{2}} +{t}+\mathrm{1}\right]=\mathrm{121}−\mathrm{22}\left({t}−\mathrm{3}\right)+\left({t}−\mathrm{3}\right)^{\mathrm{2}} \\ $$$$\mathrm{3}\left({t}−\mathrm{3}\right)^{\mathrm{2}} +\mathrm{26}\left({t}−\mathrm{3}\right)−\mathrm{105}=\mathrm{0} \\ $$$${t}−\mathrm{3}=\frac{−\mathrm{26}\pm\sqrt{\mathrm{26}^{\mathrm{2}} +\mathrm{4}×\mathrm{3}×\mathrm{105}}}{\mathrm{2}×\mathrm{3}}=\frac{−\mathrm{26}\pm\mathrm{44}}{\mathrm{6}}=\mathrm{3},−\mathrm{11}.\mathrm{66} \\ $$$${t}={x}+{y}=\mathrm{6}\Rightarrow\sqrt{{xy}}=\mathrm{3}\Rightarrow{xy}=\mathrm{9} \\ $$$$\Rightarrow{x}={y}=\mathrm{3}\:\:\:\blacksquare \\ $$$$ \\ $$$$ \\ $$
Commented by tawa last updated on 09/Apr/17
God bless you sir.
$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}. \\ $$

Pin

Leave a Reply

Your email address will not be published. Required fields are marked *