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Question Number 139331 by 7770 last updated on 26/Apr/21
(√x)+(√y)=3 and (√(x+5))+(√(y+3))=5 find x and y
$$\sqrt{\boldsymbol{{x}}}+\sqrt{\boldsymbol{{y}}}=\mathrm{3}\:\:\boldsymbol{{and}} \\ $$$$\sqrt{\boldsymbol{{x}}+\mathrm{5}}+\sqrt{\boldsymbol{{y}}+\mathrm{3}}=\mathrm{5} \\ $$$$\boldsymbol{{find}}\:\boldsymbol{{x}}\:\boldsymbol{{and}}\:\boldsymbol{{y}} \\ $$
Answered by bemath last updated on 26/Apr/21
(1)x=9+y−6(√y) (2)x+5=25+y+3−10(√(y+3)) ⇒14+y−6(√y) = 28+y−10(√(y+3)) ⇒14−6(√y) = 28−10(√(y+3)) ⇒6(√y) = 10(√(y+3))−14 ⇒3(√y) = 5(√(y+3))−7 ⇒9y+42(√y)+49 = 25(y+3) 42(√y) = 16y+26 ⇒21(√y) = 8y+13 , (√y)=u ⇒8u^2 −21u+13=0 u = ((21+(√(441−32.13)))/(16))=((21+5)/(16)) u=((26)/(16)) ⇒y=(((26)/(16)))^2 =((676)/(256))
$$\left(\mathrm{1}\right)\mathrm{x}=\mathrm{9}+\mathrm{y}−\mathrm{6}\sqrt{\mathrm{y}}\: \\ $$$$\left(\mathrm{2}\right)\mathrm{x}+\mathrm{5}=\mathrm{25}+\mathrm{y}+\mathrm{3}−\mathrm{10}\sqrt{\mathrm{y}+\mathrm{3}} \\ $$$$\Rightarrow\mathrm{14}+\mathrm{y}−\mathrm{6}\sqrt{\mathrm{y}}\:=\:\mathrm{28}+\mathrm{y}−\mathrm{10}\sqrt{\mathrm{y}+\mathrm{3}} \\ $$$$\Rightarrow\mathrm{14}−\mathrm{6}\sqrt{\mathrm{y}}\:=\:\mathrm{28}−\mathrm{10}\sqrt{\mathrm{y}+\mathrm{3}} \\ $$$$\Rightarrow\mathrm{6}\sqrt{\mathrm{y}}\:=\:\mathrm{10}\sqrt{\mathrm{y}+\mathrm{3}}−\mathrm{14} \\ $$$$\Rightarrow\mathrm{3}\sqrt{\mathrm{y}}\:=\:\mathrm{5}\sqrt{\mathrm{y}+\mathrm{3}}−\mathrm{7} \\ $$$$\Rightarrow\mathrm{9y}+\mathrm{42}\sqrt{\mathrm{y}}+\mathrm{49}\:=\:\mathrm{25}\left(\mathrm{y}+\mathrm{3}\right) \\ $$$$\mathrm{42}\sqrt{\mathrm{y}}\:=\:\mathrm{16y}+\mathrm{26} \\ $$$$\Rightarrow\mathrm{21}\sqrt{\mathrm{y}}\:=\:\mathrm{8y}+\mathrm{13}\:,\:\sqrt{\mathrm{y}}=\mathrm{u} \\ $$$$\Rightarrow\mathrm{8u}^{\mathrm{2}} −\mathrm{21u}+\mathrm{13}=\mathrm{0} \\ $$$$\mathrm{u}\:=\:\frac{\mathrm{21}+\sqrt{\mathrm{441}−\mathrm{32}.\mathrm{13}}}{\mathrm{16}}=\frac{\mathrm{21}+\mathrm{5}}{\mathrm{16}} \\ $$$$\mathrm{u}=\frac{\mathrm{26}}{\mathrm{16}}\:\Rightarrow\mathrm{y}=\left(\frac{\mathrm{26}}{\mathrm{16}}\right)^{\mathrm{2}} =\frac{\mathrm{676}}{\mathrm{256}} \\ $$$$ \\ $$
Answered by MJS_new last updated on 26/Apr/21
x≥0∧y≥0 (1) y=(3−(√x))^2 (2) y=(5−(√(x+5)))^2 −3 ⇒ (3−(√x))^2 −(5−(√(x+5)))^2 +3=0 5(√(x+5))−3(√x)=9 squaring and transforming 17x+22=15(√x)(√(x+5)) squaring and transforming x^2 −((377)/(64))x+((121)/(16))=0 x_1 =4∧x_2 =((121)/(64)) ⇒ y_1 =1∧y_2 =((169)/(64))
$${x}\geqslant\mathrm{0}\wedge{y}\geqslant\mathrm{0} \\ $$$$\left(\mathrm{1}\right)\:{y}=\left(\mathrm{3}−\sqrt{{x}}\right)^{\mathrm{2}} \\ $$$$\left(\mathrm{2}\right)\:{y}=\left(\mathrm{5}−\sqrt{{x}+\mathrm{5}}\right)^{\mathrm{2}} −\mathrm{3} \\ $$$$\Rightarrow \\ $$$$\left(\mathrm{3}−\sqrt{{x}}\right)^{\mathrm{2}} −\left(\mathrm{5}−\sqrt{{x}+\mathrm{5}}\right)^{\mathrm{2}} +\mathrm{3}=\mathrm{0} \\ $$$$\mathrm{5}\sqrt{{x}+\mathrm{5}}−\mathrm{3}\sqrt{{x}}=\mathrm{9} \\ $$$$\mathrm{squaring}\:\mathrm{and}\:\mathrm{transforming} \\ $$$$\mathrm{17}{x}+\mathrm{22}=\mathrm{15}\sqrt{{x}}\sqrt{{x}+\mathrm{5}} \\ $$$$\mathrm{squaring}\:\mathrm{and}\:\mathrm{transforming} \\ $$$${x}^{\mathrm{2}} −\frac{\mathrm{377}}{\mathrm{64}}{x}+\frac{\mathrm{121}}{\mathrm{16}}=\mathrm{0} \\ $$$${x}_{\mathrm{1}} =\mathrm{4}\wedge{x}_{\mathrm{2}} =\frac{\mathrm{121}}{\mathrm{64}}\:\Rightarrow\:{y}_{\mathrm{1}} =\mathrm{1}\wedge{y}_{\mathrm{2}} =\frac{\mathrm{169}}{\mathrm{64}} \\ $$

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