Solve the following system of equations: 3x - 2y + z = 1, x + y + z = 4 and 2x + 2y

Thursday, February 20, 2014

Solve the following system of equations: 3x - 2y + z = 1, x + y + z = 4 and 2x + 2y - 3z = 8

Since the
equation was solved for $\displaystyle{z}$ , replace all occurrences of $\displaystyle{z}$ in the other
equations with the solution $\displaystyle{\left(-{3}{x}+{2}{y}+{1}\right)}$ .

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

Remove the parentheses around the
expression $\displaystyle-{3}{x}+{2}{y}+{1}$ .

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

$\displaystyle{x}+{y}-{3}{x}+{2}{y}+{1}={4}$

$\displaystyle{2}{x}+{2}{y}-{3}{\left(-{3}{x}+{2}{y}+{1}\right)}={8}$

Combine all
similar terms in the polynomial
$\displaystyle{x}+{y}-{3}{x}+{2}{y}+{1}$ .

$\displaystyle{z}=-{3}{x}+{2}{y}+{1}$
$\displaystyle-{2}{x}+{3}{y}+{1}={4}$

$\displaystyle{2}{x}+{2}{y}-{3}{\left(-{3}{x}+{2}{y}+{1}\right)}={8}$

Multiply $\displaystyle-{3}$ by
each term inside the parentheses.

$\displaystyle{z}=-{3}{x}+{2}{y}+{1}$
$\displaystyle-{2}{x}+{3}{y}+{1}={4}$
$\displaystyle{2}{x}+{2}{y}+{9}{x}-{6}{y}-{3}={8}$

Combine all similar
terms in the polynomial $\displaystyle{2}{x}+{2}{y}+{9}{x}-{6}{y}-{3}$ .

$\displaystyle{z}=-{3}{x}+{2}{y}+{1}$
$\displaystyle-{2}{x}+{3}{y}+{1}={4}$
$\displaystyle{11}{x}-{4}{y}-{3}={8}$

Move all terms not containing $\displaystyle{x}$ to the
right-hand side of the equation.

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

$\displaystyle{11}{x}-{4}{y}-{3}={8}$

Divide each
term in the equation by $\displaystyle-{2}$ .

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

$\displaystyle{x}=\frac{{{3}{y}}}{{2}}-$
3/2

$\displaystyle{11}{x}-{4}{y}-{3}={8}$

Since the
equation was solved for $\displaystyle{x}$ , replace all occurrences of $\displaystyle{x}$ in the other
equations with the solution $\displaystyle{\left(\frac{{{3}{y}}}{{2}}-\frac{{3}}{{2}}\right)}.$

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

$\displaystyle{x}=\frac{{{3}{y}}}{{2}}-\frac{{3}}{{2}}$

$\displaystyle{11}{\left(\frac{{{3}{y}}}{{2}}-\frac{{3}}{{2}}\right)}-{4}{y}-{3}={8}$

Multiply $\displaystyle{11}$ by each
term inside the parentheses.

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

$\displaystyle{x}=\frac{{{3}{y}}}{{2}}-\frac{{3}}{{2}}$

$\displaystyle\frac{{{33}{y}}}{{2}}-\frac{{{33}}}{{2}}-{4}{y}-{3}={8}$

Combine $\displaystyle\frac{{{33}{y}}}{{2}}-{4}{y}$ into a single
expression by finding the least common denominator
(LCD). The LCD of $\displaystyle\frac{{{33}{y}}}{{2}}-{4}{y}$ is $\displaystyle{2}$ .

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

$\displaystyle{x}=\frac{{{3}{y}}}{{2}}-\frac{{3}}{{2}}$

$\displaystyle\frac{{{25}{y}}}{{2}}-\frac{{{33}}}{{2}}-{3}={8}$

Combine $\displaystyle-\frac{{{33}}}{{2}}-{3}$ into a single
expression by finding the least common denominator
(LCD). The LCD of $\displaystyle-\frac{{{33}}}{{2}}-{3}$ is $\displaystyle{2}$ .

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

$\displaystyle{x}=\frac{{{3}{y}}}{{2}}-\frac{{3}}{{2}}$

$\displaystyle\frac{{{25}{y}}}{{2}}-\frac{{{39}}}{{2}}={8}$

Move all
terms not containing $\displaystyle{y}$ to the right-hand side of the
equation.

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

$\displaystyle{x}=\frac{{{3}{y}}}{{2}}-$
3/2

$\displaystyle\frac{{{25}{y}}}{{2}}=\frac{{{55}}}{{2}}$

Multiply $\displaystyle{55}\cdot{2}$ to get $\displaystyle{110}$ in the
numerator.

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

$\displaystyle{x}=\frac{{{3}{y}}}{{2}}-$
3/2

$\displaystyle{25}{y}=\frac{{{110}}}{{2}}$

Reduce the
expression by canceling out all common factors from the
numerator and denominator.

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

$\displaystyle{x}=\frac{{{3}{y}}}{{2}}-\frac{{3}}{{2}}$

$\displaystyle{25}{y}={55}$

Since the equation
was solved for $\displaystyle{y}$ , replace all occurrences of $\displaystyle{y}$ in the other equations
with the solution $\displaystyle{\left(\frac{{11}}{{5}}\right)}.$

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

$\displaystyle{x}=\frac{{{3}{y}}}{{2}}-\frac{{3}}{{2}}$

$\displaystyle{y}=\frac{{11}}{{5}}$

Since the equation was solved for $\displaystyle{y}$
, replace all occurrences of $\displaystyle{y}$ in the other equations with the
solution $\displaystyle{\left(\frac{{11}}{{5}}\right)}$

$\displaystyle{z}=-{3}{x}+{2}{\left(\frac{{11}}{{5}}\right)}+{1}$

$\displaystyle{x}=\frac{{{3}{\left(\frac{{11}}{{5}}\right)}}}{{2}}-\frac{{3}}{{2}}$

$\displaystyle{y}=\frac{{11}}{{5}}$

Multiply $\displaystyle{2}$ by each
term inside the parentheses.

$\displaystyle{z}=-{3}{x}+\frac{{22}}{{5}}+{1}$

$\displaystyle{x}=\frac{{{3}{\left(\frac{{11}}{{5}}\right)}}}{{2}}-\frac{{3}}{{2}}$

$\displaystyle{y}=\frac{{11}}{{5}}$

Combine $\displaystyle\frac{{22}}{{5}}+$
1 $\displaystyle\int{o}{a}{\sin{{g{\le}}}}$ expression by finding the least common
denominator (LCD). The LCD of $\displaystyle\frac{{22}}{{5}}+$
1 $\displaystyle{i}{s}$ 5 $\displaystyle.$

$\displaystyle{z}=-{3}{x}+\frac{{27}}{{5}}$

$\displaystyle{x}=\frac{{{3}{\left(\frac{{11}}{{5}}\right)}}}{{2}}$
- 3/2

$\displaystyle{y}=\frac{{11}}{{5}}$

Remove the
single term factors from the
expression.

$\displaystyle{z}=\frac{{27}}{{5}}-{3}{x}$

$\displaystyle{x}=\frac{{33}}{{10}}-\frac{{3}}{{2}}$

$\displaystyle{y}=\frac{{11}}{{5}}$

Remove the parentheses from the
numerator.

$\displaystyle{z}=\frac{{27}}{{5}}-{3}{x}$

$\displaystyle{x}=\frac{{18}}{{10}}$

$\displaystyle{y}=\frac{{11}}{{5}}$

Combine $\displaystyle\frac{{33}}{{10}}-\frac{{3}}{{2}}$ into a single
expression by finding the least common denominator
(LCD). The LCD of $\displaystyle\frac{{33}}{{10}}-\frac{{3}}{{2}}$ is $\displaystyle{10}$ .

$\displaystyle{z}=\frac{{27}}{{5}}-{3}{x}$

$\displaystyle{x}=\frac{{9}}{{5}}$

$\displaystyle{y}=\frac{{11}}{{5}}$

Since the
equation was solved for $\displaystyle{x}$ , replace all occurrences of $\displaystyle{x}$ in the other
equations with the solution $\displaystyle{\left(\frac{{9}}{{5}}\right)}$

$\displaystyle{z}$
= 27/5 - 3(9/5)