This online 2×2 System of Linear Equations Calculator solves a system of 2 linear equations with 2 unknowns.Enter the coefficients values for each linear equation of the system in the appropriate fields of the calculator. All the fields left blank will be interpreted as coefficients with zero values. After clicking the ‘Calculate’ button you will get the values of the unknowns as well as the graphical representation of the linear equations and the solution to the system as the point of intersection of the lines.
Solving a system of linear equations
In mathematics, a system of linear equations is a set of one or more linear equations with the same number of variables (or unknowns). The simplest nontrivial linear system involves two equations with two unknowns:
$${ a }_{ 1 }x+{ b }_{ 1 }y={ c }_{ 1 }$$ $${ a }_{ 2 }x+{ b }_{ 2 }y={ c }_{ 2 },$$ where \(x, y\) are the unknowns, \(a_1, a_2, b_1, b_2\) are the coefficients of the system, and \(c_1, c_2\) are the constant terms.
Solving linear system of equations is a search for the values of the unknowns \(x, y\) such that each of the equations is satisfied. A linear system of equations may behave in any one of three possible ways:
• The system has a single unique solution.
• The system has no solution (the linear system is inconsistent).
• The system has infinitely many solutions (the equations are equivalent).
There are a number of methods for solving a system of linear equations. This linear system of equations calculator uses the substitution method – the simplest one that can be easily applied to a system of two linear equations with two unknowns.
According to this method, we use the first equation of the system to express one of the two variables, say \(x\), through the second variable \(y\). Then we substitute the first variable \(x\) in the second equation with that expression and thus get an equation with only one unknown \(y\) which can be easily solved. And finally we plug the found value for the second unknown \(y\) into the first equation and solve the respective equation for one unknown and, as a result, find the solution for the first variable \(x\).
The solution to the above system of linear equations is as follows:
$$x=\frac { { c }_{ 1 }{ b }_{ 2 }-{ b }_{ 1 }{ c }_{ 2 } }{ { a }_{ 1 }{ b }_{ 2 }-{ b }_{ 1 }a_{ 2 } }$$ $$y=\frac { { a }_{ 1 }{ c }_{ 2 }-c_{ 1 }{ a }_{ 2 } }{ { a }_{ 1 }{ b }_{ 2 }-{ b }_{ 1 }a_{ 2 } }.$$
In case the denominator \({ a }_{ 1 }{ b }_{ 2 }-{ b }_{ 1 }a_{ 2 }\) equals to zero, the system, obviously, has no solution. In case both the denominator and the numerator \({ c }_{ 1 }{ b }_{ 2 }-{ b }_{ 1 }{ c }_{ 2 }\) (or \({ a }_{ 1 }{ c }_{ 2 }-c_{ 1 }{ a }_{ 2 }\)) equal to zero, the system has infinitely many solutions.
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