Blog: `2x - 2y - 6z = -4, -3x + 2y + 6z = 1, x - y - 5z = -3` Solve the system of linear equations and check any solutions algebraically.

Wednesday, February 24, 2016

$\displaystyle{2}{x}-{2}{y}-{6}{z}=-{4},-{3}{x}+{2}{y}+{6}{z}={1},{x}-{y}-{5}{z}=-{3}$ Solve the system of linear equations and check any solutions algebraically.

EQ1: $\displaystyle{2}{x}-{2}{y}-{6}{z}=-{4}$

EQ2: $\displaystyle-{3}{x}+{2}{y}+{6}{z}={1}$

EQ3: $\displaystyle{x}-{y}-{5}{z}=-{3}$

To solve this system of equation, let's apply elimination method. In this method, a variable or variables should be removed in order to get the value of the other variable.

Let's eliminate y. To do so, add EQ1 and EQ2.

$\displaystyle{2}{x}-{2}{y}-{6}{z}=-{4}$

$\displaystyle+$ $\displaystyle-{3}{x}+{2}{y}+{6}{z}={1}$

$\displaystyle----------------$

$\displaystyle-{x}=-{3}$

Then, solve for x.

$\displaystyle\frac{{-{x}}}{{-{1}}}=\frac{{-{3}}}{{-{1}}}$

$\displaystyle{x}={3}$

Isolate y again.Consider EQ1 and EQ3.

$\displaystyle{2}{x}-{2}{y}-{6}{z}=-{4}$

$\displaystyle{x}-{y}-{5}{z}=-{3}$

To be able to eliminate y, multiply EQ3 by -2. Then, add the two equations.

$\displaystyle{2}{x}-{2}{y}-{6}{z}=-{4}$

$\displaystyle+$ $\displaystyle-{2}{x}+{2}{y}+{10}{z}={6}$

$\displaystyle---------------$

$\displaystyle{4}{z}={2}$

And, solve for z.

$\displaystyle\frac{{{4}{z}}}{{4}}=\frac{{2}}{{4}}$

$\displaystyle{z}=\frac{{1}}{{2}}$

Now that the values of the two variables are known, let's solve for the remaining variable. Let's plug-in x=3 and z=1/2 to EQ1.

$\displaystyle{2}{x}-{2}{y}-{6}{z}=-{4}$

$\displaystyle{2}{\left({3}\right)}-{2}{y}-{6}{\left(\frac{{1}}{{2}}\right)}=-{4}$

$\displaystyle{6}-{2}{y}-{3}=-{4}$

$\displaystyle{3}-{2}{y}=-{4}$

$\displaystyle{3}-{3}-{2}{y}=-{4}-{3}$

$\displaystyle-{2}{y}=-{7}$

(-2y)/(-2)=(-7)/(-2)

$\displaystyle{y}=\frac{{7}}{{2}}$

Therefore, the solution is $\displaystyle{\left({3},\frac{{7}}{{2}},\frac{{1}}{{2}}\right)}$ .

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