`2x + 3y + 3z = 7, 4x + 18y + 15z = 44` Solve the system of linear equations and check any solutions algebraically.

Thursday, September 22, 2011

$\displaystyle{2}{x}+{3}{y}+{3}{z}={7},{4}{x}+{18}{y}+{15}{z}={44}$ Solve the system of linear equations and check any solutions algebraically.

You should notice that the system is indeterminate, since the number of variables is larger than the number of equations.

$\displaystyle{2}\cdot{\left({2}{x}+{3}{y}+{3}{z}\right)}={14}\Rightarrow{4}{x}+{6}{y}+{6}{z}={14}\Rightarrow{4}{x}={14}-{6}{y}-{6}{z}$

Replace $\displaystyle{14}-{6}{y}-{6}{z}$ for 4x in the second equation, such that:

$\displaystyle{4}{x}={44}-{18}{y}-{15}{z}\Rightarrow{14}-{6}{y}-{6}{z}={44}-{18}{y}-{15}{z}$

$\displaystyle{12}{y}+{9}{z}={30}\Rightarrow{4}{y}+{3}{z}={10}\Rightarrow{4}{y}={10}-{3}{z}\Rightarrow{y}=\frac{{5}}{{2}}-{\left(\frac{{3}}{{4}}\right)}{z}$

Replace back $\displaystyle\frac{{5}}{{2}}-{\left(\frac{{3}}{{4}}\right)}{z}$ for y in equation $\displaystyle{4}{x}={14}-{6}{y}-{6}{z}$ , such that:

$\displaystyle{4}{x}={14}-{6}{\left(\frac{{5}}{{2}}-{\left(\frac{{3}}{{4}}\right)}{z}\right)}-{6}{z}$

$\displaystyle{4}{x}={14}-{15}+{\left(\frac{{9}}{{2}}\right)}{z}-{6}{z}$

$\displaystyle{4}{x}=-{1}-{\left(\frac{{3}}{{2}}\right)}{z}\Rightarrow{x}=-\frac{{1}}{{4}}-{\left(\frac{{3}}{{8}}\right)}{z}$

Hence, evaluating the solutions to the system, yields $\displaystyle{x}=-\frac{{1}}{{4}}-{\left(\frac{{3}}{{8}}\right)}{z},{y}=\frac{{5}}{{2}}-{\left(\frac{{3}}{{4}}\right)}{z},{z}={z}.$

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Thursday, September 22, 2011

$\displaystyle{2}{x}+{3}{y}+{3}{z}={7},{4}{x}+{18}{y}+{15}{z}={44}$ Solve the system of linear equations and check any solutions algebraically.

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