Systems of Linear Equations and their Solutions
A system of linear equations has one or more equations to be solved simultaneously.

Example 1
a system of two equations with two unknowns
The solution to this system of linear equations is the set of ordered pairs (x,y) that must satisfy all equations in the system simultaneously.
The ordered pair (1,1) is a solution to the above system. You can check this by substituting x and y by 1 and 1 respectively in both equations included in the systems and see that both equations are satisfied.
Graphical Interpretation of the Solution to a 2 by 2 System of Equations
Below are shown the graphs of the two equations included in the system and the solution (1,1) is the point of intersection of the two lines whose equations are the two equations in the system.

Example 2
A system of three equations with three unknowns
\( \begin{equation}
\begin{array}{ccl}
x + y + z & = & 2 \\
x  y + z & = & 0 \\
x  y & = & 0
\end{array}
\end{equation}
\)
The ordered triple (1,1,0) is a solution to example 2 above because it satisfies all three equations simultaneously. You may check by substitution of the unknowns by their values in the ordered triple (1,1,0).
Graphical Interpretation of the Solution to a 3 by 3 System of Equations
Below are shown the graphs of the three equations included in the system which are represented by 3 planes in 3 dimensions. The point of intersection of the three planes is the point (1,1,0) as shown.
Both systems in examples 1 and 2 have solutions and are called consistent systems.
System with no solutions are called inconsistent

Example 3
The following system
\( \begin{equation}
\begin{array}{ccl}
2x + 2y & = & 1 \\
x + y & = & 1/2
\end{array}
\end{equation}
\)
has an infinite number of solutions. If you multiply all terms of the second equation by 2, you will obtain the first equation and therefore the two equations are equivalent to one equation only in two variable which has an infinite number of solutions. As an exercise, substitute and verify the the ordered pairs (0 , 1/2) , (1 , 3/2), (2 , 5/2)... are solutions to both equations in the system.
Solve Systems by the Method of Elimination
We first define equivalent systems of equations as systems with the same solution set. The method of elimination uses three elementary operations on systems listed below to rewrite a given system of equation to an equivalent one that is easier to solve.
1) If we multiply all terms of a given equation, in a given system, by a constant (not equal to zero), we obtain an equivalent system of equations.
2) If we add the two left sides and the two right sides of two equations, we obtain an equivalent system of equations.
3) If we interchange two equations, we obtain an equivalent system of equations.

Example 4
Use the method of elimination to solve the system of linear equations.
\( \begin{equation}
\begin{array}{ccl}
2x + 3y & = & 8 \\
3x  y & = &  5
\end{array}
\end{equation}
\)
Solution to example 4
multiply all terms in the second equation by 3 to obtain the equivalent system
\( \begin{equation}
\begin{array}{ccl}
2x + 3y & = & 8 \\
9x  3y & = &  15
\end{array}
\end{equation}
\)
Add the left sides and the right sides of the two equations to obtain an equivalent system given by
\( \begin{equation}
\begin{array}{ccl}
2x + 3y & = & 8 \\
7x & = &  7
\end{array}
\end{equation}
\)
Note: y has been eliminated from the second equation, hence the name: method of elimination
Solve the second equation for x.
\( x = 1 \)
Substitute x by 1 in the first equation.
\( 2(1) + 3y = 8 \)
solve the above equation for y
\( 2 + 3y = 8 \)
\( 3y = 6 \)
\( y = 2 \)
Write the solution to the system as an ordered pair.
\( (1,2) \)
Check that the solution obtained satisfies both equations in the given system.
first equation: Left Side:  2(1) + 3(2)= 2 + 6 = 8
Right Side: 8
second equation: Left Side: 3(1) (2)=  3  2 =  5
Right Side:  5
conclusion: The given system of equations is consistent and its solution is the ordered pair (1,2).

Example 5
Solve the system of linear equations using the method of elimination.
\( \begin{equation}
\begin{array}{ccl}
4x  y & = & 8 \\
8 x + 2 y & = &  5
\end{array}
\end{equation}
\)
Solution to Example 5
Multiply all terms in the first equation by 2.
\( \begin{equation}
\begin{array}{ccl}
8x  2y & = & 16 \\
8 x + 2 y & = &  5
\end{array}
\end{equation}
\)
Add the two equations to obtain the equation.
\( 0x + 0y = 11 \) or \( 0 = 11 \)
Conclusion: Because there are no values of x and y for which 0x + 0y = 11 is satisfied, the given system of equations has no solutions. This system is inconsistent.

Example 6
Use the method of elimination to solve the system of linear equations given by
\( \begin{equation}
\begin{array}{ccl}
2x  3y & = & 8 \\
4 x + 6 y & = &  16
\end{array}
\end{equation}
\)
Solution to Example 6
Multiply all terms in the first equation by 2 to obtain an equivalent system given by
\( \begin{equation}
\begin{array}{ccl}
4x  6y & = & 16 \\
4 x + 6 y & = &  16
\end{array}
\end{equation}
\)
add the two equations to obtain the system
\( \begin{equation}
\begin{array}{ccl}
4x  6y & = & 16 \\
0 x + 0 y & = & 0
\end{array}
\end{equation}
\)
Conclusion: Any value for x and y in the second equation is a solution. Hence it has an infinite number of solutions. We may select one of the two unknowns as being any real number t and solve for the second unknown. Let y = t , t being any real number. Substitute y by t in the top equation to obtain
\( 4x  6 t = 16 \)
Solve the above for x
\( x = (3/2) t + 4 \)
The solution set to the given system may be written as
\( ((3/2)t + 4 , t) \) with \( t \in \mathbb{R} \)
Solutions may be generated by substituting t by real numbers.
a) t = 0, solution (4 , 0) , b) t =  2, solution (1 , 2), c) t = 1/2, solution (19/4 , 1/2), ...and so on.
This system is consistent.

Example 7
Use the method of elimination to solve the system of linear equations.
\( \begin{equation}
\begin{array}{ccl}
x + y + 2 z & = & 3 \\
2x + 3y  2 z & = & 6 \\
4x  y + z & = & 5
\end{array}
\end{equation}
\)
Solution to example 7
multiply all terms in the first equation by 2 and add it to the second equation; this will eliminate x from the second equation.
\( \begin{equation}
\begin{array}{ccl}
x + y + 2 z & = & 3 \\
0x + 5y + 2z & = & 12 \\
4x  y + z & = & 5
\end{array}
\end{equation}
\)
multiply all terms in the first equation by  4 and add it to the third equation; this will eliminate x from the third equation.
\( \begin{equation}
\begin{array}{ccl}
x + y + 2 z & = & 3 \\
0x + 5y + 2z & = & 12 \\
0x  5y  7 z & = & 17
\end{array}
\end{equation}
\)
Add the second equation to the third equation; this will eliminate y from the third equation.
\( \begin{equation}
\begin{array}{ccl}
x + y + 2 z & = & 3 \\
0x + 5y + 2z & = & 12 \\
0x + 0y  5 z & = &  5
\end{array}
\end{equation}
\)
We now use back substitution to find the solution. The solution to the last equation \( 5 z =  5\) is
z = 1
We now substitute z by 1 in the second equation to obtain \( 5y + 2(1) = 12 \) and solve it for y to obtain
5 y = 10
y = 2
We finally substitute z by 1 and y by 2 in the first equation to obtain \( x + (2) + 2 (1) = 3 \) and solve for x to obtain
x = 1
We now put the solution in the order (x,y,z) to obtain the ordered triple (1,2,1) as the solution to the given system of linear equations.
conclusion: The given system of equations is consistent and its solution is the ordered triple (1,2,1).

Example 8
Use the method of elimination to solve the system of linear equations.
\( \begin{equation}
\begin{array}{ccl}
 x + y + 2 z & = & 1 \\
3x  3y  6 z & = &  5 \\
 4x + 4 y + 8z & = & 9
\end{array}
\end{equation}
\)
Solution to example 8
multiply all terms in the first equation by 3 and add it to the second equation.
\( \begin{equation}
\begin{array}{ccl}
 x + y + 2 z & = & 1 \\
0x + 0y + 0z & = &  2 \\
 4x + 4 y + 8z & = & 9
\end{array}
\end{equation}
\)
The second equation has no solution.
conclusion: The given system of equations has no solutions and is therefore inconsistent.

Example 9
Use the method of elimination to solve the system of linear equations.
\( \begin{equation}
\begin{array}{ccl}
x  y + 6z & = &  5 \\
6 x 12 y  6 z & = & 24 \\
2 x + 4 y + 2 z & = & 8
\end{array}
\end{equation}
\)
Solution to example 9
multiply all terms in the first equation by 6 and add it to the second equation; this will eliminate x from the second equation.
\( \begin{equation}
\begin{array}{ccl}
x  y + 6z & = &  5 \\
0x  6 y  42 z & = & 6 \\
2 x + 4 y + 2 z & = & 8
\end{array}
\end{equation}
\)
Multiply all terms in the first equation by 2 and add it to the third equation; this will eliminate x from the third equation.
\( \begin{equation}
\begin{array}{ccl}
x  y + 6z & = &  5 \\
0 x  6 y  42 z & = & 6 \\
0 x + 2 y + 14 z & = & 2
\end{array}
\end{equation}
\)
Multiply all terms of the second equation by 1/3 and add it to the third equation.
\( \begin{equation}
\begin{array}{ccl}
x  y + 6z & = &  5 \\
0 x  6 y  42 z & = & 6 \\
0 x + 0 y + 0 z & = & 0
\end{array}
\end{equation}
\)
The last equations has an infinite number of solutions. We may select one of the unknowns as being a real number t and solve the other equations for the other unknowns. Let z = t where t is any real number. We now substitute z by t in the second equation and solve it for y.
\(  6 y = 6 + 42 t\)
\( y =  7t  1 \)
We now substitute z by t and y by  7t  1 in the first equation and solve for x.
\( x  (7t  1) + 6 t =  5 \)
Solve for x
\( x =  13 t  6 \)
conclusion: The given system of equations is consistent and its solution is the ordered triple \( ( 13 t  6 ,  7t  1 , t) \) where t is any real number.
The solutions to the above system may be generated by giving the parameter \( t \) any real number.
Examples: For \( t = 0 \) the solution is \( (6,1,0) \), for \( t = 1\) the solution is \((7,6,1)\), ... and so on.
Questions with Solution

Part 1
Use the method of elimination to solve the following systems of linear equations.
a)
\( \begin{equation}
\begin{array}{ccl}
x + 3y & = & 11 \\
4x  y & = &  11
\end{array}
\end{equation}
\)
b) \( \begin{equation}
\begin{array}{ccl}
2x + y & = & 8 \\
6x  3y & = & 10
\end{array}
\end{equation}
\)
c) \( \begin{equation}
\begin{array}{ccl}
x 2 y & = & 3 \\
3x + 6y & = & 9
\end{array}
\end{equation}
\)

Part 2
Use the method of elimination to solve the following systems of linear equations.
a)
\( \begin{equation}
\begin{array}{ccl}
2x + 3y  z & = &  1 \\
 x  y + 2z & = &  1 \\
3 x  2 y  z & = & 2
\end{array}
\end{equation}
\)
b) \( \begin{equation}
\begin{array}{ccl}
2x  y  5z & = & 1 \\
 x  y  z & = &  1 \\
3 x + 3 y + 3 z & = & 3
\end{array}
\end{equation}
\)

Part 3
Use the method of elimination to solve the following systems of linear equations.
\( \begin{equation}
\begin{array}{ccl}
x  2y + z + w & = &  6 \\
2 x  y  3z 2w & = & 5 \\
4 x + 2 y + 2z  w & = & 8 \\
x + 2 y + 3 z  w & = & 8
\end{array}
\end{equation}
\)

Part 4
Find all values of the parameter k so that the system of equation shown below has
a) One solution only b) No Solutions
\( \begin{equation}
\begin{array}{ccl}
x  2y & = &  5 \\
3 x  k y & = & 8
\end{array}
\end{equation}
\)

Part 5
Find all values of the parameter k so that the system of equation shown below has an infinite number of solutions.
\( \begin{equation}
\begin{array}{ccl}
x  5y & = & 5 \\
4 x  2 k y & = & 20
\end{array}
\end{equation}
\)
Solutions to the Above Questions

Part 1
a)
Multiply all terms of the first equation by 4
\( \begin{equation}
\begin{array}{ccl}
4x + 12y & = & 44 \\
4x  y & = &  11
\end{array}
\end{equation}
\)
Add the first equation to the second equation
\( \begin{equation}
\begin{array}{ccl}
 4x + 12y & = & 44 \\
0x + 11 y & = & 33
\end{array}
\end{equation}
\)
Solve by back substitution to obtain
y = 3 and x = 2
The above system is consistent and has the solution (2,3).
b) Multiply the first equation by 3
\( \begin{equation}
\begin{array}{ccl}
6x + 3y & = & 24 \\
6x  3y & = & 10
\end{array}
\end{equation}
\)
Add the top equation to the second equation
\( \begin{equation}
\begin{array}{ccl}
6x + 3y & = & 24 \\
0x + 0y & = & 34
\end{array}
\end{equation}
\)
There are no values of x and y that satisfies the second equation. The system has no solutions and is therefore inconsistent.
c) Multiply the first equation by 3
\( \begin{equation}
\begin{array}{ccl}
3 x  6 y & = & 9 \\
3x + 6y & = & 9
\end{array}
\end{equation}
\)
Add the top equation to the second equation
\( \begin{equation}
\begin{array}{ccl}
3 x  6 y & = & 9 \\
0x + 0y & = & 0
\end{array}
\end{equation}
\)
The last equation has an infinite number of solutions. Let y = t, t being a real number.
Substitute y by z in the first equation and solve it for x.
3x  6(t) = 9
x = 2 t + 3
The system has an infinite number of solutions that may be written as (2 t + 3 , t) where t is any real number.

Part 2
a) Change the order of the equations and rewrite the given system as follows
\( \begin{equation}
\begin{array}{ccl}
 x  y + 2z & = &  1 \\
2x + 3y  z & = &  1 \\
3 x  2 y  z & = & 2
\end{array}
\end{equation}
\)
Multiply the top equation by 2 and add it to the second equation
\( \begin{equation}
\begin{array}{ccl}
 x  y + 2z & = &  1 \\
0x + y + 3z & = &  3 \\
3 x  2 y  z & = & 2
\end{array}
\end{equation}
\)
Multiply the top equation by 3 and add it to the third equation
\( \begin{equation}
\begin{array}{ccl}
 x  y + 2z & = &  1 \\
0x + y + 3z & = &  3 \\
0x  5 y + 5z & = & 1
\end{array}
\end{equation}
\)
Multiply the second equation by 5 and add it to the third equation
\( \begin{equation}
\begin{array}{ccl}
 x  y + 2z & = &  1 \\
0x + y + 3z & = &  3 \\
0x + 0y + 20 z & = & 16
\end{array}
\end{equation}
\)
Solve the system by back substitution
\( z =  4/5 \)
\( y =  3  3 z =  3 + 12/5 = 3/5 \)
\( x =  y + 2 z + 1 = 3/5 + 2(4/5) + 1 = 0\)
Solution is \( (0 , 3/5 , 4/5) \)
b) Multiply all terms of the second equation by 3 and add it to the third equation
\( \begin{equation}
\begin{array}{ccl}
2x  y  5z & = & 1 \\
 x  y  z & = &  1 \\
0x + 0y + 0 z & = & 0
\end{array}
\end{equation}
\)
Change the orders of equations 1 and 2
\( \begin{equation}
\begin{array}{ccl}
 x  y  z & = &  1 \\
2x  y  5z & = & 1 \\
0x + 0y + 0 z & = & 0
\end{array}
\end{equation}
\)
Multiply the first equation by 2 and add it to the second equation
\( \begin{equation}
\begin{array}{ccl}
 x  y  z & = &  1 \\
0x  3 y  7 z & = & 1 \\
0x + 0y + 0z & = & 0
\end{array}
\end{equation}
\)
The last equation has an infinite number of solutions.
Let z = t where t is any real number and solve by back substitution for y and x. The second equation gives
\( 3 y = 7 z  1 \) and \( y = (1/3)(7t 1) \)
The first equation gives
\( x = (1/3)(4t + 2) \)
The solution set may be written as
\( ( \dfrac{4t+2}{3} , \dfrac{7t1}{3} , t) \) , \( t \in \mathbb{R} \)

Part 3
Eliminate the terms in x from the second, third and fourth equations as follows:
Add 2 times the first equation to the second equation; add  4 times the first equation to the third and subtract the first equation from the fourth equation to obtain
\( \begin{equation}
\begin{array}{ccl}
x  2y + z + w & = &  6 \\
0x  5y  z + 0 w & = & 7 \\
0x + 10 y  2z  5w & = & 16\\
0x + 4 y + 2 z  2w & = &  2
\end{array}
\end{equation}
\)
Add 2 times the second equation to the third equation to eliminate y from the third equation.
\( \begin{equation}
\begin{array}{ccl}
x  2y + z + w & = &  6 \\
0x  5y  z + 0 w & = & 7 \\
0x + 0y  4z  5w & = & 2\\
0x + 4 y + 2 z  2w & = &  2
\end{array}
\end{equation}
\)
Add 4 times the second equation to 5 times the fourth equation to eliminate y from the fourth equation.
\( \begin{equation}
\begin{array}{ccl}
x  2y + z + w & = &  6 \\
0x  5y  z + 0 w & = & 7 \\
0x + 0y  4z  5w & = & 2\\
0x + 0y + 6 z  10w & = &  38
\end{array}
\end{equation}
\)
Add 3 times the third equation to 2 times the fourth equation to eliminate z from the fourth equation.
\( \begin{equation}
\begin{array}{ccl}
x  2y + z + w & = &  6 \\
0x  5y  z + 0 w & = & 7 \\
0x + 0y  4z  5w & = & 2\\
0x + 0y + 0z  35 w & = & 70
\end{array}
\end{equation}
\)
Use back substitution to find w, z, y and x.
Using the fourth equation, we obtain w = 2
Substitute w by 2 in the third equation and solve for z
\( 4 z  5(2) = 2\) gives z =  3
Substitute w by 2 and z by 3 in the second equation and solve for y
\(  5y  (3) + 0 (2) = 7 \) gives \( y = 2 \)
Substitute w by 2, z by 3 and y by 3 in the first equation and solve for x.
\( x  2(2) + (3) + 2 =  6 \) gives \( x = 1 \)
The given system is consistent and has the solution \((1,2,3,2) \)

Part 4
Find all values of the parameter k so that the system of equation shown below has
a) One solution only b) No Solutions
a)
\( \begin{equation}
\begin{array}{ccl}
x  2y & = &  5 \\
3 x  k y & = & 8
\end{array}
\end{equation}
\)
Add 3 times the first equation to the second equation
\( \begin{equation}
\begin{array}{ccl}
 x  2y & = &  5 \\
0x  (k  6) y & = & 7
\end{array}
\end{equation}
\)
Solve the second equation for y
\( y = \dfrac{7}{k6} \)
There is one solution for y if the denominator is equal to zero. Hence the system has one solution for any value of k not equal to 6 and no solution if k = 6.

Part 5
Find all values of the parameter k so that the system of equation shown below has an infinite number of solutions.
a)
\( \begin{equation}
\begin{array}{ccl}
x  5y & = & 5 \\
4 x  2 k y & = & 20
\end{array}
\end{equation}
\)
Subtract 4 times the first equation from the second equation and factor out y from the terms with y in the second equation.
\( \begin{equation}
\begin{array}{ccl}
x  5y & = & 5 \\
0 x  (2 k  20) y& = & 0
\end{array}
\end{equation}
\)
The last equation will have an infinite number of solutions if the coefficient of y is equal zero. Hence
\( 2 k  20 = 0\)
For \( k = 10 \) the given system has an infinite number of solutions.
References and Links on Systems of Equations
