• Theory and Examples
• Exercises

### Systems of Equations

In the chapters Equations - Basics and Equations - Additional Aspects we discussed how to solve an equation for a single variable.

Now we turn to systems of equations: that means a set of two or more equations with the same variables.
An mxn-System (say: m times n system) consists of m equations with n unknowns.

### Examples

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

### Solutions

A solution to a system of equation is a set of values for the variables that satisfy all the equations simultaneously.

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)}

### Linear and Non-linear Systems

Systems 1,2 and 4 of the examples above are linear, while system 3 is not.
"Linear" means that each term is either a constant or the product of a constant and (the first power of) a single variable.

The linearity of a system makes an important difference for the decision which method we choose to solve it.

In the next paragraphs we will present (for the special case of linear 2x2-systems) three important solution methods:

• substitution
• comparison

All three methods try to achieve a common goal: they want to eliminate one of the unknowns in order to come up to one single equation with one variable. This equation can be solved then with the methods presented in the chapters Equations - Basics and Equations - Additional Aspects to obtain the value of the first variable. By inserting it into one of the equations, we find the second part of the solution.

Many of the methods presented here can also be applied to non-linear systems and to systems with more than two variables. We will get back to these aspects later, in separate sections.

### Substitution Method

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:  $\quad x = 8$ , $y = 9$

### Comparison Method

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:  $x=- 2 \;,\;y = 3$

Remarks:

• Instead of multiplying by (-2) and adding there had been the possibility of multiplying by +2 and then subtracting. Indeed you could eliminate the variable f in example 2 by a simple subtraction (without a preceding multiplication).
• Sometimes it makes sense to multiply both of the equations before adding them up.
here for an additional example.
•

### Solvability of Linear Systems

In all of the linear 2x2-systems presented above there was a unique solution; this is the normal case.

But there are other possibilities:
- the system has no solution
- the system has infinitely many solutions.

An equation with two variables x and y can be understood as the defining equation of a function y=f(x) whose graph is a straight line in the xy-plane. (For more details see chapter Introduction to Functions).
Therefore, the solution of a linear 2x2-system can be interpreted as the intersection point of two lines. If the lines do intersect, we have a unique solution; in the special cases we mentioned above we have either parallel lines (no solution) or coinciding lines (infinitely many solutions).

When calculating, these special cases can be recognized by the fact that the elimination of the one variable leads to the simultaneous elimination of the second one, too.

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.

Note: In case of no solutions you will get an "impossible" equation as $\;0=2\,$, for example.

### Non-linear Systems

In non-linear systems, equations may contain powers or products of the variables. Therefore, it might be difficult to find a suitable additive operation to eliminate one of the variables.

The first choice for solving non-linear systems is the substitution method; it is less ambitious than the comparison method, because it is sufficient to isolate the variable in one equation (for applying the comparison method, this would have to be done in both equations).

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:

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

### Systems of Equations with more than two Variables

With more than two equations and variables the procedure remains basically the same. Again we try to get a simpler system by eliminating the variables step by step.

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

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

As you can see the whole procedure has become much more complicated. The more unknowns you have the more difficult it becomes to keep track of the elimination process.

The usual approach for a systematic elimination of the variables is named after the mathematician Carl Friedrich Gauss (1777-1855),, therefore known as Gauss-Elimination. It can be used for hand-calculation, but its main importance is based on the fact that the algorithm is particularly suitable for computers.

You will learn more about systems of linear equations (with three and more variables) in the course Mathematics for Management and Economics of the IM&E Programme.

1)

Solve the following systems of linear equations (try to choose a beneficial method):

a)

$\left\{ \; \begin{eqnarray}2x -3y &=& 5 \\ x + 2y &=& 6 \end{eqnarray} \right.$

b)

$\left\{ \; \begin{eqnarray}7x &=& -3y+1 \\ 7x &=& 2y-3 \end{eqnarray} \right.$

c)

$\left\{ \; \begin{eqnarray}0.3x + 0.4y &=& 0.5\\ 0.7x +1.1y &=& 1.6 \end{eqnarray} \right.$

d)

$\left\{ \; \begin{eqnarray}6x + 5y &=& - 6\\ y &=& 3 - 4x \end{eqnarray} \right.$

e)

$\left\{ \; \begin{eqnarray}8x -5y &=& 5.5 \\ - x + y &=& 1 \end{eqnarray} \right.$

f)

$\left\{ \; \begin{eqnarray}2x - 3y &=& 10\\ 7x + 2y &=& 2.5 \end{eqnarray} \right.$

2)

Solve the following non-linear systems:

a)

$\left\{ \; \begin{eqnarray} {x^2} - 2y &=& 0 \\ x + y &=& 4 \end{eqnarray} \right.$

b)

$\left\{ \; \begin{eqnarray} x + 2y &=& 2\\ xy - y &=& -3 \end{eqnarray} \right.$

3)

Solve the following systems of equations with three variables:

a)

$\left\{ \; \begin{eqnarray} x +4y-5z &=& 21 \\ 2x +3y+4z &=& -1 \\ x-6y-8z&=&-3 \end{eqnarray} \right.$

b)

$\left\{ \; \begin{eqnarray}3x+2y-8z &=& -5 \\ -x+2z &=& 3 \\ x+ y-5z &=& 7 \end{eqnarray} \right.$

4)

Solve the following systems of equations.
In a first step, before using the methods discussed in the theory section, you have to simplify the equations.

a)

$\left\{ \; \begin{eqnarray} \frac {x-y}{4}- \frac {3(y-2)}{7} &=& 2 \\ \frac {x - y}{x+y} &=& 2 \end{eqnarray} \right.$

b)

$\left\{ \; \begin{eqnarray}x(y+1) &=&(x+1)(y-2) \\ \frac {x + y}{2} &=& 2x+1 \end{eqnarray} \right.$

c)

$\left\{ \; \begin{eqnarray} \frac {x-1}{y} &=&2 \\{x^2}-y &=& 2 \end{eqnarray} \right.$

Theory with explanations and examples:

Video: Solving Linear Systems of Equations   (2 variables, addition method)

Video: Solving Linear Systems of Equations   (3 variables)

Exercises:

Math-Drills: Systems of Linear Equations  (many worksheets with exercises; different levels)

Math-Worksheets: Systems of Linear Equations  (worksheet with exercises for download; 2 and 3 variables)

Purplemath: Systems of Non-Linear Equations  (examples on pages 3, 4 and 5)