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Question Number 66755 by John Kaloki Musau last updated on 19/Aug/19

solve the simultaneous equations:  3x-y=9  x^2 -xy=4

$${solve}\:{the}\:{simultaneous}\:{equations}: \\ $$$$\mathrm{3}{x}-{y}=\mathrm{9} \\ $$$${x}^{\mathrm{2}} -{xy}=\mathrm{4} \\ $$

Answered by $@ty@m123 last updated on 20/Aug/19

x^2 −x(3x−9)=4  x^2 −3x^2 +9x=4  2x^2 −9x+4=0  x=((9±(√(81−32)))/4)=((9±7)/4)=4,(1/2)  y=3,−7(1/2)

$${x}^{\mathrm{2}} −{x}\left(\mathrm{3}{x}−\mathrm{9}\right)=\mathrm{4} \\ $$$${x}^{\mathrm{2}} −\mathrm{3}{x}^{\mathrm{2}} +\mathrm{9}{x}=\mathrm{4} \\ $$$$\mathrm{2}{x}^{\mathrm{2}} −\mathrm{9}{x}+\mathrm{4}=\mathrm{0} \\ $$$${x}=\frac{\mathrm{9}\pm\sqrt{\mathrm{81}−\mathrm{32}}}{\mathrm{4}}=\frac{\mathrm{9}\pm\mathrm{7}}{\mathrm{4}}=\mathrm{4},\frac{\mathrm{1}}{\mathrm{2}} \\ $$$${y}=\mathrm{3},−\mathrm{7}\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$ \\ $$

Answered by John Kaloki Musau last updated on 21/Aug/19

y=3x−9  x^2 −3x^2 +9x−4=0  2x^2 −8x−x+4=0  (2x−1)(x−4)=0  x=(1/2),4  x=3+(1/3)y  9+2y+(1/9)y^2 −3y−(1/3)y^2 −4=0  (2/9)y^2 +y−5=0  (y−3)(2y+15)=0  y=3,−7(1/2)

$$\boldsymbol{{y}}=\mathrm{3}\boldsymbol{{x}}−\mathrm{9} \\ $$$$\boldsymbol{{x}}^{\mathrm{2}} −\mathrm{3}\boldsymbol{{x}}^{\mathrm{2}} +\mathrm{9}\boldsymbol{{x}}−\mathrm{4}=\mathrm{0} \\ $$$$\mathrm{2}\boldsymbol{{x}}^{\mathrm{2}} −\mathrm{8}\boldsymbol{{x}}−\boldsymbol{{x}}+\mathrm{4}=\mathrm{0} \\ $$$$\left(\mathrm{2}\boldsymbol{{x}}−\mathrm{1}\right)\left(\boldsymbol{{x}}−\mathrm{4}\right)=\mathrm{0} \\ $$$$\boldsymbol{{x}}=\frac{\mathrm{1}}{\mathrm{2}},\mathrm{4} \\ $$$$\boldsymbol{{x}}=\mathrm{3}+\frac{\mathrm{1}}{\mathrm{3}}\boldsymbol{{y}} \\ $$$$\mathrm{9}+\mathrm{2}\boldsymbol{{y}}+\frac{\mathrm{1}}{\mathrm{9}}\boldsymbol{{y}}^{\mathrm{2}} −\mathrm{3}\boldsymbol{{y}}−\frac{\mathrm{1}}{\mathrm{3}}\boldsymbol{{y}}^{\mathrm{2}} −\mathrm{4}=\mathrm{0} \\ $$$$\frac{\mathrm{2}}{\mathrm{9}}\boldsymbol{{y}}^{\mathrm{2}} +\boldsymbol{{y}}−\mathrm{5}=\mathrm{0} \\ $$$$\left(\boldsymbol{{y}}−\mathrm{3}\right)\left(\mathrm{2}\boldsymbol{{y}}+\mathrm{15}\right)=\mathrm{0} \\ $$$$\boldsymbol{{y}}=\mathrm{3},−\mathrm{7}\frac{\mathrm{1}}{\mathrm{2}} \\ $$

Answered by Kunal12588 last updated on 20/Aug/19

 determinant ((3,1),(y,x))=9,  determinant ((x,x),(y,x))=4  ⇒x determinant ((1,1),(y,x))=4  ⇒x determinant ((3,1),(y,x))−x determinant ((2,0),(y,x))=4  ⇒9x−2x^2 =4  ⇒2x^2 −9x+4=0  ⇒x=((9±(√(81−32)))/4)=((9±7)/4)=4,(1/2)  putting in one of the eq^n   y=12−9=3 and y=(3/2)−9=−((15)/2)=−7(1/2)   ((x),(y) )= ((4,(       (1/2))),(3,(−7(1/2))) )

$$\begin{vmatrix}{\mathrm{3}}&{\mathrm{1}}\\{{y}}&{{x}}\end{vmatrix}=\mathrm{9},\:\begin{vmatrix}{{x}}&{{x}}\\{{y}}&{{x}}\end{vmatrix}=\mathrm{4} \\ $$$$\Rightarrow{x}\begin{vmatrix}{\mathrm{1}}&{\mathrm{1}}\\{{y}}&{{x}}\end{vmatrix}=\mathrm{4} \\ $$$$\Rightarrow{x}\begin{vmatrix}{\mathrm{3}}&{\mathrm{1}}\\{{y}}&{{x}}\end{vmatrix}−{x}\begin{vmatrix}{\mathrm{2}}&{\mathrm{0}}\\{{y}}&{{x}}\end{vmatrix}=\mathrm{4} \\ $$$$\Rightarrow\mathrm{9}{x}−\mathrm{2}{x}^{\mathrm{2}} =\mathrm{4} \\ $$$$\Rightarrow\mathrm{2}{x}^{\mathrm{2}} −\mathrm{9}{x}+\mathrm{4}=\mathrm{0} \\ $$$$\Rightarrow{x}=\frac{\mathrm{9}\pm\sqrt{\mathrm{81}−\mathrm{32}}}{\mathrm{4}}=\frac{\mathrm{9}\pm\mathrm{7}}{\mathrm{4}}=\mathrm{4},\frac{\mathrm{1}}{\mathrm{2}} \\ $$$${putting}\:{in}\:{one}\:{of}\:{the}\:{eq}^{{n}} \\ $$$${y}=\mathrm{12}−\mathrm{9}=\mathrm{3}\:{and}\:{y}=\frac{\mathrm{3}}{\mathrm{2}}−\mathrm{9}=−\frac{\mathrm{15}}{\mathrm{2}}=−\mathrm{7}\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\begin{pmatrix}{{x}}\\{{y}}\end{pmatrix}=\begin{pmatrix}{\mathrm{4}}&{\:\:\:\:\:\:\:\frac{\mathrm{1}}{\mathrm{2}}}\\{\mathrm{3}}&{−\mathrm{7}\frac{\mathrm{1}}{\mathrm{2}}}\end{pmatrix} \\ $$

Answered by Mr Jor last updated on 24/Aug/19

y=3x−9  x^2 −3x^2 +9x=4  2x^2 −9x+4=0  (2x−1)(x−4)=0  x=(1/2),4  x=3+(1/3)  9+2y+(1/9)y^2 −3y−(1/3)y^2 −4=0  (2/9)y^2 +y−5=0  (y+3)(2y+15)=0  y=3,−7(1/2)

$$\boldsymbol{{y}}=\mathrm{3}\boldsymbol{{x}}−\mathrm{9} \\ $$$$\boldsymbol{{x}}^{\mathrm{2}} −\mathrm{3}\boldsymbol{{x}}^{\mathrm{2}} +\mathrm{9}\boldsymbol{{x}}=\mathrm{4} \\ $$$$\mathrm{2}\boldsymbol{{x}}^{\mathrm{2}} −\mathrm{9}\boldsymbol{{x}}+\mathrm{4}=\mathrm{0} \\ $$$$\left(\mathrm{2}\boldsymbol{{x}}−\mathrm{1}\right)\left(\boldsymbol{{x}}−\mathrm{4}\right)=\mathrm{0} \\ $$$$\boldsymbol{{x}}=\frac{\mathrm{1}}{\mathrm{2}},\mathrm{4} \\ $$$${x}=\mathrm{3}+\frac{\mathrm{1}}{\mathrm{3}} \\ $$$$\mathrm{9}+\mathrm{2}\boldsymbol{{y}}+\frac{\mathrm{1}}{\mathrm{9}}\boldsymbol{{y}}^{\mathrm{2}} −\mathrm{3}\boldsymbol{{y}}−\frac{\mathrm{1}}{\mathrm{3}}\boldsymbol{{y}}^{\mathrm{2}} −\mathrm{4}=\mathrm{0} \\ $$$$\frac{\mathrm{2}}{\mathrm{9}}\boldsymbol{{y}}^{\mathrm{2}} +\boldsymbol{{y}}−\mathrm{5}=\mathrm{0} \\ $$$$\left(\boldsymbol{{y}}+\mathrm{3}\right)\left(\mathrm{2}\boldsymbol{{y}}+\mathrm{15}\right)=\mathrm{0} \\ $$$$\boldsymbol{{y}}=\mathrm{3},−\mathrm{7}\frac{\mathrm{1}}{\mathrm{2}} \\ $$

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