`x + y + z = 5, x - 2y + 4z = 13, 3y + 4z = 13` Solve the system of linear equations and check any solutions algebraically.

Saturday, June 18, 2016

You may use the substitution method to solve the system, hence, you need to use the first equation to write x in terms of y and z, such that:

$\displaystyle{x}+{y}+{z}={5}\Rightarrow{x}={5}-{y}-{z}$

You may now replace $\displaystyle{5}-{y}-{z}$ for x in equation $\displaystyle{x}-{2}{y}+{4}{z}={13}$ , such that:

$\displaystyle{5}-{y}-{z}-{2}{y}+{4}{z}={13}\Rightarrow-{3}{y}+{3}{z}={8}$

You may use the third equation, $\displaystyle{3}{y}+{4}{z}={13}$ , along with $\displaystyle-{3}{y}+{3}{z}={8}$ equation, such that:

$\displaystyle-{3}{y}+{3}{z}+{3}{y}+{4}{z}={8}+{13}\Rightarrow{7}{z}={21}\Rightarrow{z}={3}$

You may replace 3 for z in equation $\displaystyle{3}{y}+{4}{z}={13}:$

$\displaystyle{3}{y}+{12}={13}\Rightarrow{3}{y}={1}\Rightarrow{y}=\frac{{1}}{{3}}$

You may replace 3 for z and $\displaystyle\frac{{1}}{{3}}$ for y in equation $\displaystyle{x}={5}-{y}-{z}:$

$\displaystyle{x}={5}-\frac{{1}}{{3}}-{3}\Rightarrow{x}={2}-\frac{{1}}{{3}}\Rightarrow{x}=\frac{{5}}{{3}}$

Hence, evaluating the solution to the given system, yields $\displaystyle{x}=\frac{{5}}{{3}},{y}=\frac{{1}}{{3}},{z}={3}.$

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