The Elimination Method in Systems - Questions with Solutions
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Use the method of elimination to solve systems of linear equations. Examples and questions with detailed solutions and explanations are presented. Also the graphical interpretation of the solution of 2 by 2 and 3 by 3 system of equations are presented.
The method of solving systems of equations using Cramer's Rule is also included in this website.
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.
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
\( ( \dfrac{4t+2}{3} , -\dfrac{7t-1}{3} , t) \) , \( t \in \mathbb{R} \)
- 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
\( y = -\dfrac{7}{k-6} \)
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