Blog: `xy - 1 = 0, 2x - 4y + 7 = 0` Solve the system graphically or algebraically. Explain your choice of method.

Saturday, February 14, 2015

$\displaystyle{x}{y}-{1}={0},{2}{x}-{4}{y}+{7}={0}$ Solve the system graphically or algebraically. Explain your choice of method.

Since the equations of the system are not transcendental equations, you may solve the system algebraically.

You need to use substitution method, hence, you may use the second equation to express x in terms of y, such that:

$\displaystyle{2}{x}={4}{y}-{7}\Rightarrow{x}=\frac{{{4}{y}-{7}}}{{2}}$

Replacing $\displaystyle\frac{{{4}{y}-{7}}}{{2}}$ for x in equation $\displaystyle{x}{y}-{1}={0}$ , yields:

$\displaystyle{\left(\frac{{{4}{y}-{7}}}{{2}}\right)}\cdot{y}-{1}={0}\Rightarrow{4}{{y}}^{{2}}-{7}{y}-{2}={0}$

You may use quadratic formula, such that:

$\displaystyle{y}_{{{1},{2}}}=\frac{{{7}\pm\sqrt{{{49}+{32}}}}}{{8}}$

$\displaystyle{y}_{{{1},{2}}}=\frac{{{7}\pm\sqrt{{{81}}}}}{{8}}$

$\displaystyle{y}_{{{1},{2}}}=\frac{{{7}\pm{9}}}{{8}}$

$\displaystyle{y}_{{1}}=\frac{{{7}+{9}}}{{8}}\Rightarrow{y}_{{1}}={2}\Rightarrow{x}_{{1}}=\frac{{1}}{{2}}$

$\displaystyle{y}_{{1}}=\frac{{{7}-{9}}}{{8}}\Rightarrow{y}_{{1}}=-\frac{{1}}{{4}}\Rightarrow{x}_{{2}}=-{4}$

Hence, evaluating the solutions to the given system, yields $\displaystyle{\left(\frac{{1}}{{2}},{2}\right)}$ and $\displaystyle{\left(-{4},-\frac{{1}}{{4}}\right)}.$

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Saturday, February 14, 2015

$\displaystyle{x}{y}-{1}={0},{2}{x}-{4}{y}+{7}={0}$ Solve the system graphically or algebraically. Explain your choice of method.

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