Example 1

\( \left\{ \begin{array}{c} x + \large \frac{y}{3} \normalsize = 11 \\ \large \frac{x}{4} \normalsize + y = 11\; \end{array} \right. \qquad \qquad \) 2x2-system

Example 2

\(\left\{ \begin{array}{c} 300c + f = 4000\\500c + f = 6400 \end{array} \right. \qquad \qquad \) 2x2-system

Example 3

\(\left\{ \begin{array}{c}{x^2} + 2{y^2} = 33\\3x - 2y = 11 \end{array} \right. \qquad \qquad \) 2x2-system

Example 4

\( \left\{ \begin{array}{c}2u + 4v - 5w = 21\\4u + 3v + 4w = - 1\\u - 3v - 4w = - 1.5\end{array} \right. \qquad \qquad \) 3x3-system

In example 1:

The pair  \(x = 8\), \(y = 9 \;\) statisfies both equations and is therefore a solution of the system.
L={(8,9)}

In example 4:

The ordered triple   \(u = -0.5\), \(v = 3\), \(w=-2\; \) statisfies all three equations and is therefore a solution of the system.
L={(-0.5, 3, -2)}

Main Principle:

Solve one of the equations for either of the variables and replace this variable in the other equation by the expression just found. Thereby, the "substituted" variable gets eliminated.

In example 1:

\(\left\{ \begin{array}{l}\;x + \large \frac{y}{3} \normalsize = 11 \\ \large \frac{x}{4} \normalsize + y = 11\;\end{array} \right. \qquad \qquad \)

Solve the first equation for x :

\(x = 11 - \large \frac{y}{3}\)

Substitute x in the second equation by the expression \(11 - \large \frac{y}{3}\):

\( \large \frac{{11 - \frac{y}{3}}} {4} \normalsize + y = 11\)

After multiplying by 4 you get:

\( \begin{eqnarray} 11 - \frac{y}{3} + 4y &=& 44\quad &| \cdot 3 \\ 33 - y + 12y &=& 132\quad &| - 33 \\ 11y &=& 99\quad &|\;:11 \\ y &=& 9 \end{eqnarray} \)

Now plug \(y=9\) into the transformed equation \(x = 11 -\large \frac{y}{3}\):

\(x = 11 - \large \frac{9}{3} \normalsize = 8\)

The solution of the system is:  \( \quad x = 8\) , \(y = 9\)

Main Principle:

Solve both equations to either (but the same) variable and set them equal. Thereby, the "compared" variable gets eliminated.

In example 2:

\(\left\{ \begin{array}{l}300c + f = 4000\\500c + f = 6400 \end{array} \right.\)

Solve the first equation for f :\(\quad \; f = 4000 - 300c\)
Solve the second equation for f : \( \; f = 6400 - 500c\)

Set the right-hand parts equal:

\( \begin{eqnarray} 4000 - 300c &=& 6400 - 500c\quad &|\;+ 500c - 4000 \\ 200c &=& 2400\quad &|\;:200\\ c &=& 12 \end{eqnarray}\)

Now plug \(c=9\) into either of the equations with isolated f (e.g. \(f = 4000 - 300c\)):

\(f = 4000 - 3600 = 400\)

The solution of the system is: \(\quad c = 12\) , \(f = 400\)

Main Principle:

Multiply (if needed) one or both of the equations by suitable numbers, in order to get the same coefficients for one of the variables in both equations, once with positive, once with negative sign.
Then add both equations, thereby eliminating that variable.

In example 5

\( \left\{\begin{array}{c}5x + 2y = - 4\\ - 4x + y = 11\end{array} \right.\)

Multiplying equation 2 by (-2) and using it in place of the "older" one, we get the new (equivalent) system

\( \left\{\begin{array}{c}5x + 2y = - 4\\8x - 2y = -22 \end{array} \right. \)

Addition of the two equations leads to a new equation for the remaining variable x (whereas y has dropped out, which was the purpose of the previous multiplication).
It can easily be solved for x:

\( \begin{eqnarray} 13x &=& - 26\quad |\;:13\\x &=& -2 \end{eqnarray}\)

Finally, we use either of the original equations to get y (by inserting \(x=-2\)):

\(\begin{eqnarray} 5x + 2y &=& - 4\\ - 10 + 2y &=& - 4\quad |\; + 10\\y &=& 3 \end{eqnarray}\)

The solution of the system is:  \( x=- 2 \;,\;y = 3\)

Example 6

\( \left\{\;\begin{eqnarray}2x - y &=& - 2\\ - 4x + 2y &=& 4 \end{eqnarray} \right.\)

We multiply the first equation by 2, intending to eliminate y; the resulting system is

\( \left\{ \; \begin{eqnarray}4x - 2y &=& -4\\-4x + 2y &=& 4 \end{eqnarray} \right. \)

But now, the addition of the equations eliminates not only y, but also x; we get

\(0=0\)

This is an identity; the result means that both equations have the same solutions (which was obvious already after the multiplication, because the resulting equations were identical - except for the sign). Hence the solution set consists of all pairs (x,y) satisfying the equation \(2x-y=-2\).

The system has infinitely many solutions.

In example 3:

\( \;\left\{ \begin{array}{rl} \;{x^2} + 2{y^2} = 33\\3x -2y = 11 \end{array} \right. \)

The second equation (which is linear) can be solved for y easily:

\( \begin{eqnarray} 3x - 2y = 11\quad &|\, - 3x \\- 2y = 11 - 3x\quad &|\,:( - 2) \\ y = - 5.5 + 1.5x \end{eqnarray}\)

Now replace y with this term in the first equation:

\( {x^2} + 2 \cdot {( - 5.5 + 1.5x)^2} = 33 \)

Make use of the Binomial Formulas to simplify the left-hand expression:

\(\begin{eqnarray}{x^2} + 2 \cdot (30.25 - 16.5x + 2.25{x^2}) &=& 33\\{x^2} + 60.5 - 33x + 4.5{x^2} &=& 33 \\(5.5{x^2} - 33x + 27.5 &=& 0 \end{eqnarray}\)

This is a quadratic equation for x that can be solved by applying the quadratic formula (see Chapter Equations - Basics):

\({x_{1\,,2}} = \large \frac{{33 \pm \sqrt {{{( - 33)}^2} - 4 \cdot 5.5 \cdot 27.5} }}{{2 \cdot 5.5}} = \frac{{33 \pm \sqrt {484} }}{{11}} = \frac{{33 \pm 22}}{{11}}\)

So we have two possible solutions for x:

\({x_1} = 5 \quad {x_2} = 1\)

To each of these values you get a "partner" y by inserting it into the equation \(y = - 5.5 + 1.5x\) above:

\({y_1} = - 5.5 + 1.5 \cdot 5 = 2\)

\({y_2} = - 5.5 + 1.5 \cdot 1 = - 4\)

Therefore the non-linear system has two solutions:

\(L = \left\{ {(5,2);(1\,, - 4)} \right\}\)

Example 7

A linear 3x3-system:

\(\left\{ \begin{array}{c} 2x + y + 4z = - 2\\ - x - 2y + z = - 5\\3x - 8y - 7z = 23 \end{array} \right. \qquad\) (*)

Multiply, for example, the first equation by 2 and add the result to the second equation:

\( \left\{ {\begin{array}{*{20}{c}}{4x + 2y + 8z = - 4}\\ {- x - 2y + z = - 5}\end{array}} \right. \;\begin{array}{*{20}{c}} + \\ + \end{array} \quad \rightarrow \quad 3x + 9z = - 9\)

Of course you will need a second equation, because you can't determine two variables (x and z) from one single condition. It seems obvious that you have to make use of the information contained in equation 3; but above all it is important to eliminate the same variable y as before (even if it were easier to eliminate another one).

For example, multiply the second equation by (-4) and add the result to the third equation:

\(\left\{ {\begin{array}{*{20}{c}}{4x + 8y - 4z = 20}\\{3x - 8y -7z = 23}\end{array}} \right. \;\begin{array}{*{20}{c}} + \\ + \end{array} \quad \rightarrow \quad 7x - 11z = 43\)

With the two new equations we have a linear 2x2-System for x and z:

\( \left\{ \begin{array}{c} 3x + 9z = - 9\\7x - 11z = 43 \end{array} \right. \qquad\) (**)

In this new system we have to eliminate the next variable, say z.
This can be done by multiplying the first equation by 11 and the second one by 9, then by adding the resulting equations:

\( \left\{ {\begin{array}{*{20}{c}}{33x + 99z = - 99}\\{63x - 99z = 387}\end{array}} \right. \;\begin{array}{*{20}{c}} + \\ + \end{array} \quad \rightarrow \quad 96x = 288 \rightarrow \quad x = 3\)

That value \(x=3\) can be inserted into one of the equations of system (**), which gives us z:

For example, in equation 1: \(9 + 9z = - 9 \quad \rightarrow \quad z = - 2\)

And finally both values \(x=3\) and \(z=-2\) are inserted in equation 1 of the original system (*):

\(6 + y - 8 = - 2\quad \rightarrow \quad y = 0\)

(This last step-by-step procedure for finding the values of the variables is often referred to as backward substitution.)

The system has the solution

\(L = \left\{ {(3,0, - 2)} \right\}\)