An equation of the form ax + by + c = 0 where a, b, c ∈ R, a ≠ 0 and b ≠ 0 is a linear equation in two variables. While considering the system of linear equations, we can find the number of solutions by comparing the coefficients of the equations. Also, we can find whether the system of equations has no solution or infinitely many solutions by graphical method. In this article, we will learn how to find if a system of equations has no solution or infinitely many solutions. Show
Let us consider the pair of linear equations in two variables x and y. a1x + b1y + c1 = 0 a2x + b2y + c2 = 0 Here a1, b1, c1, a2, b2, c2 are all real numbers. Note that a12 + b12 ≠ 0, a22 + b22 ≠ 0. Case 1. If (a1/a2) = (b1/b2) = (c1/c2), then there will be infinitely many solutions. This type of equation is called a dependent pair of linear equations in two variables. If we plot the graph of this equation, the lines will coincide. Case 2. If (a1/a2) = (b1/b2) ≠ (c1/c2), then there will be no solution. This type of equation is called an inconsistent pair of linear equations. If we plot the graph, the lines will be parallel. The graph is shown below. Solved ExamplesExample 1: How many solutions does the following system have? y = -3x + 9 y = -3x – 7 (A) One solution Solution: Given equations are y = -3x + 9 y = -3x – 7 Here (a1/a2) = (b1/b2) ≠ (c1/c2). So this system of equations has no solution. Another method: Without graphing them, we can see that both have the same slope -3 which means lines are parallel. Hence the system of equations has no solution. So option (B) is the answer. Example 2: Determine whether the following system of equations have no solution, infinitely many solution or unique solutions. x+2y = 3, 2x+4y = 15 Solution: Given equations are x+2y = 3 2x+4y = 15 a1 = 1, b1 = 2, c1 = -3 a2 = 2, b2 = 4, c2 = -15 a1/a2 = ½ b1/b2 = ½ c1/c2 = 1/5 a1/a2 = b1/b2≠c1/c2 So, the system of equations has no solution. Related Links:
Frequently Asked QuestionsA Linear equation is an equation that has one or more variables having degree one. An example of a Linear equation in two variables: 2x + 3y + 4 = 0. If a1/a2 = (b1/b2) ≠
(c1/c2), then there will be no solution. If (a1/a2) = (b1/b2) = (c1/c2), then there will be
infinitely many solutions. Learning Objectives¶By the end of this section you should be able to:
Unique Solution¶The example shown previously in this module had a unique solution. The structure of the row reduced matrix was \[\begin{split}\begin{vmatrix} 1 & 1 & -1 & | & 5 \\ 0 & 1 & -5 & | & 8 \\ 0 & 0 & 1 & | & -1 \end{vmatrix}\end{split}\] and the solution was \[x = 1\] \[y = 3\] \[z = -1\] As you can see, each variable in the matrix can have only one possible value, and this is how you know that this matrix has one unique solution No solution¶Let’s suppose you have a system of linear equations that consist of: \[x + y + z = 2\] \[y - 3z = 1\] \[2x + y + 5z = 0\] The augmented matrix is \[\begin{split}\begin{vmatrix} 1 & 1 & 1 & | & 2 \\ 0 & 1 & -3 & | & 1 \\ 2 & 1 & 5 & | & 0 \end{vmatrix}\end{split}\] and the row reduced matrix is \[\begin{split}\begin{vmatrix} 1 & 0 & 4 & | & 1 \\ 0 & 1 & -3 & | & 1 \\ 0 & 0 & 0 & | & -3 \end{vmatrix}\end{split}\] As you can see, the final row states that \[0x + 0y + 0z = -3\] which impossible, 0 cannot equal -3. Therefore this system of linear equations has no solution. Let’s use python and see what answer we get. import numpy as py from scipy.linalg import solve A = [[1, 1, 1], [0, 1, -3], [2, 1, 5]] b = [[2], [1], [0]] x = solve(A,b) x --------------------------------------------------------------------------- LinAlgError Traceback (most recent call last) <ipython-input-1-afc47691740d> in <module>() 5 b = [[2], [1], [0]] 6 ----> 7 x = solve(A,b) 8 x C:\Users\Said Zaid-Alkailani\Anaconda3\lib\site-packages\scipy\linalg\basic.py in solve(a, b, sym_pos, lower, overwrite_a, overwrite_b, debug, check_finite, assume_a, transposed) 217 return x 218 elif 0 < info <= n: --> 219 raise LinAlgError('Matrix is singular.') 220 elif info > n: 221 warnings.warn('scipy.linalg.solve\nIll-conditioned matrix detected.' LinAlgError: Matrix is singular. As you can see the code gives us an error suggesting there is no solution to the matrix. Infinite Solutions¶Let’s suppose you have a system of linear equations that consist of: \[-3x - 5y + 36z = 10\] \[-x + 7z = 5\] \[x + y - 10z = -4\] The augmented matrix is \[\begin{split}\begin{vmatrix} -3 & -5 & 36 & | & 10 \\ -1 & 0 & 7 & | & 5 \\ 1 & 1 & -10 & | & -4 \end{vmatrix}\end{split}\] and the row reduced matrix is \[\begin{split}\begin{vmatrix} 1 & 0 & -7 & | & -5 \\ 0 & 2 & -3 & | & 1 \\ 0 & 0 & 0 & | & 0 \end{vmatrix}\end{split}\] As you can see, the final row of the row reduced matrix consists of 0. This means that for any value of Z, there will be a unique solution of x and y, therefore this system of linear equations has infinite solutions. Let’s use python and see what answer we get. import numpy as py from scipy.linalg import solve A = [[-3, -5, 36], [-1, 0, 7], [1, 1, -10]] b = [[10], [5], [-4]] x = solve(A,b) x C:\Users\Said Zaid-Alkailani\Anaconda3\lib\site-packages\scipy\linalg\basic.py:223: RuntimeWarning: scipy.linalg.solve
Ill-conditioned matrix detected. Result is not guaranteed to be accurate.
Reciprocal condition number: 3.808655316038273e-19
' condition number: {}'.format(rcond), RuntimeWarning)
array([[-12.],
[ -2.],
[ -1.]])
As you can see we get a different type of error from this code. It states that the matrix is ill-conditioned and that there is a RuntimeWarning. This means that the computer took to long to find a unique solution so it spat out a random answer. When RuntimeWarings occur, the matrix is likely to have infinite solutions. How do you determine if there is one solution infinitely many solutions or no solution?Some equations have infinitely many solutions. In these equations, any value for the variable makes the equation true. You can tell that an equation has infinitely many solutions if you try to solve the equation and get a variable or a number equal to itself.
What is the solution for infinitely many solutions?An equation can have infinitely many solutions when it should satisfy some conditions. The system of an equation has infinitely many solutions when the lines are coincident, and they have the same y-intercept. If the two lines have the same y-intercept and the slope, they are actually in the same exact line.
What is the difference between no solution and infinitely many solutions?If the equation is untrue then the system has no solution. If the equation is always true then there are infinitely many solutions.
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Infinitely many solutions no solution one solution
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