How to Find a System of Equations has No Solution or Infinitely Many Solutions

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.

System of Equations has No Solution or Infinitely Many Solutions

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.

Related Links:

• Linear equations
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• Techniques To Solve System Of Equations
• Solving Linear Equations using Matrix