Exercises: systems with three variables

The function graph has the equation $y=a{x}^2+bx+c$.

If a point lies on a graph, it means that the equation gives a true statement.

From the given three points, you can set up three equations, all of which must be satisfied.

Plug in point $\mathrm{R}$

$y=a{x}^2+bx+c$

Plug the point $\mathrm{R}(1|2)$ into the equation.

$2=a\cdot 1^2+b\cdot 1+c$

You get your first equation.

$\mathrm{I}2=a+b+c$

Plug in point $\mathrm{Q}$

$y=a{x}^2+bx+c$

Plug the point $\mathrm{Q}(-1|3)$ into the equation.

$3=a\cdot (-1{)}^2+b\cdot (-1)+c$

You get your second equation.

$\mathrm{II}3=a-b+c$

Plug in point $\mathrm{S}$

$y=a{x}^2+bx+c$

Plug the point $\mathrm{Q}(0|1)$ into the equation.

$1=a\cdot (0{)}^2+b\cdot 0+c$

You get your third equation.

$\mathrm{III}1=c$

So the system of equations is:

$\begin{array}{rrlllllll}\mathrm{I}& 2& =& a& +& b& +& c& \\ \mathrm{II}& 3& =& a& -& b& +& c& \\ \mathrm{III}& 1& =& & & & & c& \end{array}$

The following system of equations is given:

$\begin{array}{rrlllllll}\mathrm{I}& 2& =& a& +& b& +& c& \\ \mathrm{II}& 3& =& a& -& b& +& c& \\ \mathrm{III}& 1& =& & & & & c& \end{array}$

It follows directly from the third equation that:

$c=1$

Substitute $c=1$ into the other two equations:

$\begin{array}{rrlllllll}{\mathrm{I}}^{\prime }& 2& =& a& +& b& +& 1& \\ \mathrm{I}{\mathrm{I}}^{\prime }& 3& =& a& -& b& +& 1& \end{array}$

Careful: The information of the third equation is now contained in ${\mathrm{I}}^{\prime }$ and $\mathrm{I}{\mathrm{I}}^{\prime }$.

Now solve the new system of equations. Since the coefficients before the variable $b$ have different signs and the same absolute value, the addition method can be used.

Add the equation ${\mathrm{I}}^{\prime }$ and $\mathrm{I}{\mathrm{I}}^{\prime }$. You get a new equation ${\mathrm{I}}^{\prime \prime }$.

$\begin{array}{rlrllll}{\mathrm{I}}^{\prime }+\mathrm{I}{\mathrm{I}}^{\prime }\to {\mathrm{I}}^{\prime \prime }& |& 5& =& 2a& +& 2\end{array}$

Solve for a.

$a=\frac{3}{2}$

Substitute $a=\frac{3}{2}$ into equation ${\mathrm{I}}^{\prime }$ to determine the parameter $b$.

$\begin{array}{rrlllllll}{\mathrm{I}}^{\prime }& 2& =& \frac{3}{2}& +& b& +& 1& \end{array}$

Solve for $b$.

$b=-\frac{1}{2}$

Write down the solution set.

$𝕃=\{(a|b|c)\in {ℝ}^{\mathrm{𝟛}}|a=\frac{3}{2};b=-\frac{1}{2};c=1\}$

The function equation is $f(x)=\frac{3}{2}\cdot x^2-\frac{1}{2}\cdot x+1$.

You can solve the system of equations using the addition method.

$\begin{array}{rccccccc}\mathrm{I}& 4u& +& 3v& -& w& =& 2\\ \mathrm{II}& -3u& -& 4v& +& 5w& =& -5\\ \mathrm{III}& -2u& +& 2v& +& w& =& 6\end{array}$

Choose the variable $w$ and calculate the least common multiple of the coefficients of $w$:

Now multiply each of the equations so that each $w$ has a coefficient of $5$.

$\begin{array}{rccccccc}5\cdot \mathrm{I}\to {\mathrm{I}}^{\prime }& 20u& +& 15v& -& 5w& =& 10\\ \mathrm{II}& -3u& -& 4v& +& 5w& =& -5\\ 5\cdot \mathrm{I}\mathrm{I}\mathrm{I}\to \mathrm{I}\mathrm{I}{\mathrm{I}}^{\prime }& -10u& +& 10v& +& 5w& =& 30\end{array}$

Then, add ${\mathrm{I}}^{\prime }$ to $\mathrm{II}$ and ${\mathrm{I}}^{\prime }$ to ${\mathrm{III}}^{\prime }$.

$\begin{array}{rccccccc}{\mathrm{I}}^{\prime }& 20u& +& 15v& -& 5w& =& 10\\ \mathrm{I}\mathrm{I}+{\mathrm{I}}^{\prime }\to \mathrm{I}{\mathrm{I}}^{\prime }& 17u& +& 11v& & & =& 5\\ \mathrm{I}\mathrm{I}{\mathrm{I}}^{\prime }+{\mathrm{I}}^{\prime }\to \mathrm{I}\mathrm{I}{\mathrm{I}}^{\prime \prime }& 10u& +& 25v& & & =& 40\end{array}$

You now solve the "small" system of equations consisting of $\mathrm{I}{\mathrm{I}}^{\prime }$ and $\mathrm{I}\mathrm{I}{\mathrm{I}}^{\prime \prime }$.

To do this, you select the variable $v$ and determine the lcm of its coefficients:

Now multiply the equations accordingly:

Subtract $\mathrm{I}{\mathrm{I}}^{\prime \prime }$ from $\mathrm{I}\mathrm{I}{\mathrm{I}}^{(4)}$ to eliminate $v$.

$\begin{array}{rccccc}\mathrm{I}\mathrm{I}{\mathrm{I}}^{(4)}-\mathrm{I}{\mathrm{I}}^{\prime \prime }\to \mathrm{I}\mathrm{I}{\mathrm{I}}^{(5)}& -315u& & & =& 315\end{array}$

Now you solve $\mathrm{I}\mathrm{I}{\mathrm{I}}^{(5)}$ for $u$ and put its value into ${\mathrm{II}}^{\prime \prime }$.

$\begin{array}{rccccc}\mathrm{I}\mathrm{I}{\mathrm{I}}^{(5)}& u& & & =& -1\\ \mathrm{I}{\mathrm{I}}^{\prime \prime }& 17u& +& 11v& =& 5\end{array}$

$\begin{array}{rccccc}u=-1\text{ in }\mathrm{I}{\mathrm{I}}^{\prime \prime }\to \mathrm{I}{\mathrm{I}}^{\prime \prime \prime }& 17\cdot (-1)& +& 5v& =& 11\\ \mathrm{I}\mathrm{I}{\mathrm{I}}^{(5)}& u& & & =& -1\end{array}$

$\begin{array}{rccccc}\mathrm{I}{\mathrm{I}}^{\prime \prime \prime }& & & v& =& 2\\ \mathrm{I}\mathrm{I}{\mathrm{I}}^{(5)}& u& & & =& -1\end{array}$

Now you put the two values into ${\mathrm{I}}^{\prime }$ and solve for $w$.

$\begin{array}{rrcl}u=-1\text{ and }v=2\text{ in }{\mathrm{I}}^{\prime }\to {\mathrm{I}}^{\prime \prime }& 20\cdot (-1)+15\cdot 2-5w& =& 10\\ & 10-5w& =& 10& |-10\\ & -5w& =& 0& |:(-5)\\ & w& =& 0\end{array}$

So in the end, you get the solution set

This system of equations can be solved using the addition method.

$\begin{array}{rrcrcrcr}\mathrm{I}& 2x& +& 10y& -& 5z& =& -1\\ \mathrm{II}& 10x& -& 30y& +& 3z& =& -1\\ \mathrm{III}& -4x& +& 15y& -& 2z& =& 1\end{array}$

Determine the least common multiple of the coefficients of $y$ (alternatively: of $x$ or $z$).

Then you multiply the equations such that all the coefficients of $y$ are $30$.

$\begin{array}{rrcrcrcr}3\cdot \mathrm{I}\to {\mathrm{I}}^{\prime }& 6x& +& 30y& -& 15z& =& -3\\ \mathrm{II}& 10x& -& 30y& +& 3z& =& -1\\ 2\cdot \mathrm{I}\mathrm{I}\mathrm{I}\to \mathrm{I}\mathrm{I}{\mathrm{I}}^{\prime }& -8x& +& 30y& -& 4z& =& 2\end{array}$

Add ${\mathrm{I}}^{\prime }$ and $\mathrm{II}$ and subtract ${\mathrm{I}}^{\prime }$ from $\mathrm{I}\mathrm{I}{\mathrm{I}}^{\prime }$ to eliminate the terms with $y$.

$\begin{array}{rrcrcrcr}{\mathrm{I}}^{\prime }& 6x& +& 30y& -& 15z& =& -3\\ \mathrm{I}\mathrm{I}+{\mathrm{I}}^{\prime }\to \mathrm{I}{\mathrm{I}}^{\prime }& 16x& & & -& 12z& =& -4\\ \mathrm{I}\mathrm{I}{\mathrm{I}}^{\prime }-{\mathrm{I}}^{\prime }\to \mathrm{I}\mathrm{I}{\mathrm{I}}^{\prime \prime }& -14x& & & +& 11z& =& 5\end{array}$

Now first solve the "small" system of equations consisting of $\mathrm{I}{\mathrm{I}}^{\prime }$ and $\mathrm{I}\mathrm{I}{\mathrm{I}}^{\prime \prime }$.

To do this, first determine the lcm of the coefficients of $z$, and then multiply the equations so that the lcm is in front of the $z$.

$\begin{array}{rccccr}11\cdot \mathrm{I}{\mathrm{I}}^{\prime }\to \mathrm{I}{\mathrm{I}}^{\prime \prime }& 176x& -& 132z& =& -44\\ 12\cdot \mathrm{I}\mathrm{I}{\mathrm{I}}^{\prime \prime }\to \mathrm{I}\mathrm{I}{\mathrm{I}}^{\prime \prime \prime }& -168x& +& 132z& =& 60\end{array}$

Then add $\mathrm{I}\mathrm{I}{\mathrm{I}}^{\prime \prime \prime }$ and $\mathrm{I}{\mathrm{I}}^{\prime \prime }$ to eliminate the term with $z$.

$\begin{array}{rccccr}\mathrm{I}{\mathrm{I}}^{\prime \prime }& 176x& -& 132z& =& -44\\ \mathrm{I}\mathrm{I}{\mathrm{I}}^{\prime \prime \prime }+\mathrm{I}{\mathrm{I}}^{\prime \prime }\to \mathrm{I}\mathrm{I}{\mathrm{I}}^{(4)}& 8x& & & =& 16\end{array}$

Now you solve $\mathrm{I}\mathrm{I}{\mathrm{I}}^{(4)}$ for $x$ and plug the value into $\mathrm{I}{\mathrm{I}}^{\prime \prime }$.

$\mathrm{I}\mathrm{I}{\mathrm{I}}^{(4)}x=2$

$\begin{array}{rrcr}x=2\text{ in }\mathrm{I}{\mathrm{I}}^{\prime \prime }\to \mathrm{I}{\mathrm{I}}^{\prime \prime \prime }& 176\cdot 2-132z& =& -44\\ & 352-132z& =& -44& |-352\\ & -132z& =& -396& |:(-132)\\ & z& =& 3\end{array}$

The values $x=2$ and $z=3$ can then be substituted into ${\mathrm{I}}^{\prime }$ to determine $y$:

$\begin{array}{rrcr}{\mathrm{I}}^{\prime }\to {\mathrm{I}}^{\prime \prime }& 6\cdot 2+30y-15\cdot 3& =& -3\\ & -33+30y& =& -3& |+33\\ & 30y& =& 30& |:30\\ & y& =& 1\end{array}$

Now, you can write down the solution set.

$L=\{(2;1;3)\}$