How to tell if a system of equations has no solution or infinitely many

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.

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Solved Examples
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Frequently Asked Questions
Define a Linear equation.
Give an example of a Linear equation in two variables.
Give the condition for a system of linear equations that has no solution.
Give the condition for a system of linear equations that has infinitely many solutions.
How do you tell if a system has no solution or infinitely many?
How do you tell if a system of equations has infinitely many?

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 Examples

Example 1: How many solutions does the following system have?

y = -3x + 9

y = -3x – 7

(A) One solution
(B) No solution
(C) Infinitely many solutions
(D) None of these

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.

Linear equations
System of linear equations using determinants
Techniques To Solve System Of Equations
Solving Linear Equations using Matrix

Frequently Asked Questions

Define a Linear equation.

A Linear equation is an equation that has one or more variables having degree one.

Give an example of a Linear equation in two variables.

An example of a Linear equation in two variables: 2x + 3y + 4 = 0.

Give the condition for a system of linear equations that has no solution.

If a1/a2 = (b1/b2) ≠ (c1/c2), then there will be no solution.

Give the condition for a system of linear equations that has infinitely many solutions.

If (a1/a2) = (b1/b2) = (c1/c2), then there will be infinitely many solutions.

Probably the most straightforward method (to fully distinguish between the various possibilities) that I've seen is transforming the corresponding augmented matrix into row-reduced echelon form. In this case, you would start with: $$\left[\begin{array}{ccc|c}1 & 3 & -1 & -4\\4 & -1 & 2 & 3\\2 & -1 & -3 & 1\end{array}\right]$$ Subtracting $4$ times the first row from the second, and $2$ times the first row from the third, we have: $$\left[\begin{array}{ccc|c}1 & 3 & -1 & -4\\0 & -13 & 6 & 19\\0 & -7 & -1 & 9\end{array}\right]$$ Subtracting $2$ times the third row from the second, we have: $$\left[\begin{array}{ccc|c}1 & 3 & -1 & -4\\0 & 1 & 8 & 1\\0 & -7 & -1 & 9\end{array}\right]$$ Adding $7$ times the second row to the third, we have: $$\left[\begin{array}{ccc|c}1 & 3 & -1 & -4\\0 & 1 & 12 & 1\\0 & 0 & 55 & 16\end{array}\right]$$

At this point, we have only zeroes below the main diagonal, but no zeroes on the diagonal, so a unique solution exists. Continuing to reduce until the $3\times 3$ portion of the augmented matrix is just the $3\times 3$ identity matrix, we have $$\left[\begin{array}{ccc|c}1 & 0 & 0 & 3/11\\0 & 1 & 0 & -73/55\\0 & 0 & 1 & 16/55\end{array}\right]$$ This tells us that $x=3/11,$ $y=-73/55,$ $z=16/55$ is the unique solution to the system.

Let's consider another system: $$\begin{cases}x+3y-z=4\\4x-y+2z=8\\2x-7y+4z=-3,\end{cases}$$ which has corresponding matrix $$\left[\begin{array}{ccc|c}1 & 3 & -1 & 4\\4 & -1 & 2 & 8\\2 & -7 & 4 & -3\end{array}\right].$$ Starting out the same way gets us $$\left[\begin{array}{ccc|c}1 & 3 & -1 & 4\\0 & -13 & 6 & -8\\0 & -13 & 6 & -11\end{array}\right],$$ and subtracting the second row from the third gives us $$\left[\begin{array}{ccc|c}1 & 3 & -1 & 4\\0 & -13 & 6 & -8\\0 & 0 & 0 & -3\end{array}\right].$$ Now we have only zeroes below the main diagonal, but we have a zero on the main diagonal, too. This tells us that either there are no solutions or there are infinitely-many. Translated back into terms of $x,y,z$ this is the equivalent system $$\begin{cases}x+3y+-z=4\\0x-13y+6z=-8\\0x+0y+0z=-3,\end{cases}$$ or alternatively $$\begin{cases}x=-\frac5{13}z+\frac{28}{13}\\y=\frac6{13}z+\frac8{13}\\0=3,\end{cases}$$ but there is no solution to the last equation, so no solution to the system.

Upshot: We will have no solutions whenever we end up with one or more rows of all $0$s except in the last column as we reduce the augmented matrix.

By contrast, if we'd started with the system $$\begin{cases}x+3y-z=4\\4x-y+2z=8\\2x-7y+4z=0,\end{cases}$$ which has corresponding matrix $$\left[\begin{array}{ccc|c}1 & 3 & -1 & 4\\4 & -1 & 2 & 8\\2 & -7 & 4 & 0\end{array}\right],$$ then our reduction process will get us $$\left[\begin{array}{ccc|c}1 & 3 & -1 & 4\\0 & -13 & 6 & -8\\0 & 0 & 0 & 0\end{array}\right].$$ Again, we must have no solution or infinitely many. Continuing to row-reduce as much as possible gets us to $$\left[\begin{array}{ccc|c}1 & 0 & 5/13 & 28/13\\0 & 1 & -6/13 & 8/13\\0 & 0 & 0 & 0\end{array}\right].$$ Translated back into terms of $x,y,z$ this is the equivalent system $$\begin{cases}x+0y+\frac5{13}z=\frac{28}{13}\\0x+y-\frac6{13}z=\frac8{13}\\0=0.\end{cases}$$ One of these equations is always true, so one of our variables can take on any value. We might as well let $z$ take on any value, at which point the other two equations will tell us the values that $x$ and $y$ must take. Hence, we have infinitely-many solutions.

Upshot: We will have infinitely-many solutions whenever we end up with one or more rows of all $0$s as we reduce the augmented matrix, so long as we don't have any rows with all $0$s except in the last column.

Added: Simply taking the determinant of the unaugmented matrix of the system--meaning of $$\begin{bmatrix}1 & 3 & -1\\4 & -1 & 2\\2 & -1 & -3\end{bmatrix}$$ in the first example and of $$\begin{bmatrix}1 & 3 & -1\\4 & -1 & 2\\2 & -7 & 4\end{bmatrix}$$ in the other two examples--will give us part of the answer. If the determinant is $0$ (as in the second and third example), then the system either has no solution or infinitely-many, but we cannot (by this method alone) say which. Otherwise, the system has a unique solution, but we cannot (by this method alone) say what it might be. That's why I tend to prefer the first method I suggested, at least when dealing with only a few equations and a few variables: it tells us the whole story.

How do you tell if a system has no solution or infinitely many?

Give the condition for a system of linear equations that has no solution. If a₁/a₂ = (b₁/b₂) ≠ (c₁/c₂), then there will be no solution.

How do you tell if a system of equations has infinitely many?

Conditions for Infinite Solution 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.