A system of equations is said to be inconsistent if
they have no solution
A system of equations consists of two or more equations that are considered together. The goal is often to find values for the variables that satisfy all equations simultaneously. These values represent the solutions to the system.
Systems of equations can be classified based on the number of solutions they have:
Based on the definitions above, an inconsistent system is specifically characterized by the absence of any solution. This means there are no points or values that lie on the graphs of all equations in the system.
Let's examine the given options:
Therefore, a system of equations is said to be inconsistent if they have no solution.
| System Type | Number of Solutions | Classification |
|---|---|---|
| One Solution | Exactly one | Consistent (Independent) |
| Infinitely Many Solutions | Infinite | Consistent (Dependent) |
| No Solution | Zero | Inconsistent |
| Term | Meaning | Example (Graphical) |
|---|---|---|
| Consistent System | Has at least one solution. The graphs of the equations intersect or are the same line. | Intersecting lines (one solution), Same line (infinite solutions) |
| Inconsistent System | Has no solution. The graphs of the equations do not intersect. | Parallel lines |
When solving a system of linear equations algebraically (using methods like substitution or elimination), you can determine the type of system based on the result:
Understanding the number of solutions helps classify systems of equations and predict their graphical or algebraic behavior.
For what value of k, the system linear equation has no solution
(3k + 1)x + 3y - 2 = 0
(k2 + 1)x + (k - 2)y - 5 = 0
The system of equations x + 2y = 13 and 3x + 6y = 9 has:
A (-3, 4), B (5, 4), C and D form a rectangle. If x - 4y + 7 = 0 is a diameter of circum circle of the rectangle ABCD then area of rectangle ABCD is
If a2 + b2 = 41 and a.b = 20, then (a + b) ÷ (a – b) = ______.
The rank of the matrix \(A = \left[ {\begin{array}{*{20}{c}} 1&1&2\\ 1&2&3\\ 0&{ - 1}&1 \end{array}} \right]\)