Previous in Relation and Functions Next in Relation and Functions | ||
Question Number 66755 by John Kaloki Musau last updated on 19/Aug/19 | ||
$${solve}\:{the}\:{simultaneous}\:{equations}: \\ $$$$\mathrm{3}{x}-{y}=\mathrm{9} \\ $$$${x}^{\mathrm{2}} -{xy}=\mathrm{4} \\ $$ | ||
Answered by $@ty@m123 last updated on 20/Aug/19 | ||
$${x}^{\mathrm{2}} −{x}\left(\mathrm{3}{x}−\mathrm{9}\right)=\mathrm{4} \\ $$$${x}^{\mathrm{2}} −\mathrm{3}{x}^{\mathrm{2}} +\mathrm{9}{x}=\mathrm{4} \\ $$$$\mathrm{2}{x}^{\mathrm{2}} −\mathrm{9}{x}+\mathrm{4}=\mathrm{0} \\ $$$${x}=\frac{\mathrm{9}\pm\sqrt{\mathrm{81}−\mathrm{32}}}{\mathrm{4}}=\frac{\mathrm{9}\pm\mathrm{7}}{\mathrm{4}}=\mathrm{4},\frac{\mathrm{1}}{\mathrm{2}} \\ $$$${y}=\mathrm{3},−\mathrm{7}\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$ \\ $$ | ||
Answered by John Kaloki Musau last updated on 21/Aug/19 | ||
$$\boldsymbol{{y}}=\mathrm{3}\boldsymbol{{x}}−\mathrm{9} \\ $$$$\boldsymbol{{x}}^{\mathrm{2}} −\mathrm{3}\boldsymbol{{x}}^{\mathrm{2}} +\mathrm{9}\boldsymbol{{x}}−\mathrm{4}=\mathrm{0} \\ $$$$\mathrm{2}\boldsymbol{{x}}^{\mathrm{2}} −\mathrm{8}\boldsymbol{{x}}−\boldsymbol{{x}}+\mathrm{4}=\mathrm{0} \\ $$$$\left(\mathrm{2}\boldsymbol{{x}}−\mathrm{1}\right)\left(\boldsymbol{{x}}−\mathrm{4}\right)=\mathrm{0} \\ $$$$\boldsymbol{{x}}=\frac{\mathrm{1}}{\mathrm{2}},\mathrm{4} \\ $$$$\boldsymbol{{x}}=\mathrm{3}+\frac{\mathrm{1}}{\mathrm{3}}\boldsymbol{{y}} \\ $$$$\mathrm{9}+\mathrm{2}\boldsymbol{{y}}+\frac{\mathrm{1}}{\mathrm{9}}\boldsymbol{{y}}^{\mathrm{2}} −\mathrm{3}\boldsymbol{{y}}−\frac{\mathrm{1}}{\mathrm{3}}\boldsymbol{{y}}^{\mathrm{2}} −\mathrm{4}=\mathrm{0} \\ $$$$\frac{\mathrm{2}}{\mathrm{9}}\boldsymbol{{y}}^{\mathrm{2}} +\boldsymbol{{y}}−\mathrm{5}=\mathrm{0} \\ $$$$\left(\boldsymbol{{y}}−\mathrm{3}\right)\left(\mathrm{2}\boldsymbol{{y}}+\mathrm{15}\right)=\mathrm{0} \\ $$$$\boldsymbol{{y}}=\mathrm{3},−\mathrm{7}\frac{\mathrm{1}}{\mathrm{2}} \\ $$ | ||
Answered by Kunal12588 last updated on 20/Aug/19 | ||
$$\begin{vmatrix}{\mathrm{3}}&{\mathrm{1}}\\{{y}}&{{x}}\end{vmatrix}=\mathrm{9},\:\begin{vmatrix}{{x}}&{{x}}\\{{y}}&{{x}}\end{vmatrix}=\mathrm{4} \\ $$$$\Rightarrow{x}\begin{vmatrix}{\mathrm{1}}&{\mathrm{1}}\\{{y}}&{{x}}\end{vmatrix}=\mathrm{4} \\ $$$$\Rightarrow{x}\begin{vmatrix}{\mathrm{3}}&{\mathrm{1}}\\{{y}}&{{x}}\end{vmatrix}−{x}\begin{vmatrix}{\mathrm{2}}&{\mathrm{0}}\\{{y}}&{{x}}\end{vmatrix}=\mathrm{4} \\ $$$$\Rightarrow\mathrm{9}{x}−\mathrm{2}{x}^{\mathrm{2}} =\mathrm{4} \\ $$$$\Rightarrow\mathrm{2}{x}^{\mathrm{2}} −\mathrm{9}{x}+\mathrm{4}=\mathrm{0} \\ $$$$\Rightarrow{x}=\frac{\mathrm{9}\pm\sqrt{\mathrm{81}−\mathrm{32}}}{\mathrm{4}}=\frac{\mathrm{9}\pm\mathrm{7}}{\mathrm{4}}=\mathrm{4},\frac{\mathrm{1}}{\mathrm{2}} \\ $$$${putting}\:{in}\:{one}\:{of}\:{the}\:{eq}^{{n}} \\ $$$${y}=\mathrm{12}−\mathrm{9}=\mathrm{3}\:{and}\:{y}=\frac{\mathrm{3}}{\mathrm{2}}−\mathrm{9}=−\frac{\mathrm{15}}{\mathrm{2}}=−\mathrm{7}\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\begin{pmatrix}{{x}}\\{{y}}\end{pmatrix}=\begin{pmatrix}{\mathrm{4}}&{\:\:\:\:\:\:\:\frac{\mathrm{1}}{\mathrm{2}}}\\{\mathrm{3}}&{−\mathrm{7}\frac{\mathrm{1}}{\mathrm{2}}}\end{pmatrix} \\ $$ | ||
Answered by Mr Jor last updated on 24/Aug/19 | ||
$$\boldsymbol{{y}}=\mathrm{3}\boldsymbol{{x}}−\mathrm{9} \\ $$$$\boldsymbol{{x}}^{\mathrm{2}} −\mathrm{3}\boldsymbol{{x}}^{\mathrm{2}} +\mathrm{9}\boldsymbol{{x}}=\mathrm{4} \\ $$$$\mathrm{2}\boldsymbol{{x}}^{\mathrm{2}} −\mathrm{9}\boldsymbol{{x}}+\mathrm{4}=\mathrm{0} \\ $$$$\left(\mathrm{2}\boldsymbol{{x}}−\mathrm{1}\right)\left(\boldsymbol{{x}}−\mathrm{4}\right)=\mathrm{0} \\ $$$$\boldsymbol{{x}}=\frac{\mathrm{1}}{\mathrm{2}},\mathrm{4} \\ $$$${x}=\mathrm{3}+\frac{\mathrm{1}}{\mathrm{3}} \\ $$$$\mathrm{9}+\mathrm{2}\boldsymbol{{y}}+\frac{\mathrm{1}}{\mathrm{9}}\boldsymbol{{y}}^{\mathrm{2}} −\mathrm{3}\boldsymbol{{y}}−\frac{\mathrm{1}}{\mathrm{3}}\boldsymbol{{y}}^{\mathrm{2}} −\mathrm{4}=\mathrm{0} \\ $$$$\frac{\mathrm{2}}{\mathrm{9}}\boldsymbol{{y}}^{\mathrm{2}} +\boldsymbol{{y}}−\mathrm{5}=\mathrm{0} \\ $$$$\left(\boldsymbol{{y}}+\mathrm{3}\right)\left(\mathrm{2}\boldsymbol{{y}}+\mathrm{15}\right)=\mathrm{0} \\ $$$$\boldsymbol{{y}}=\mathrm{3},−\mathrm{7}\frac{\mathrm{1}}{\mathrm{2}} \\ $$ | ||