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- 2x2 Systems Solver
2x2 system of equations solver
This calculator solves a system of two equations with two unknowns. The calculator applies the addition/elimination approach and Cramer's rule to solve a system of equations. The solver explains in detail how the work was done.
About solving system of two equations with two unknown
A system of linear equations can be solved in four different ways
1. Substitution method
2. Elimination method
3. Cramer's rule
4. Graphing method
1. Substitution method
Example: Solve the system of equations by the substitution method.
$$ \begin{aligned} 3x + 2y =& 3 \\ -2x - ~y =& -1 \end{aligned} $$Solution:
Step1: Solve one of the equations for one of the variables. We note that is simplest to solve the second equation for $ y $.
$$ \begin{aligned} 3x + 2 y = & 3 \\ {\color{red}{-2x + 1 =}} & {\color{red}{ y }} \end{aligned} $$Step2: SUBSTITUTE $y$ into first equation.
$$ \begin{aligned} 3x + 2({\color{red}{-2x + 1}}) =& 3 \\ -2x + 1 = & y \end{aligned} $$Step3: Solve first equation for $x$.
$$ \begin{aligned} {\color{blue}{ x =}} & {\color{blue}{-1}} \\ -2x + 1 = & y \end{aligned} $$Step4: To find $y$, substitute $-1$ for $x$ into second equation.
$$ \begin{aligned} x = & -1 \\ y = & -2\cdot(-1) + 1 \end{aligned} $$The solution is:
$$ \begin{aligned} {\color{blue}{ x =}} & {\color{blue}{-1}} \\ {\color{blue}{ y =}} & {\color{blue}{3}} \end{aligned} $$You can check the solution using the above calculator.
2. Elimination method
Note: This method is implemented in above calculator. The calculator follows steps which are explained in following example.
Example: Solve the system of equations by the elimination method.
$$ \begin{aligned} 3x + 2y = & -1 \\ 4x - ~5y = & 14 \end{aligned} $$Solution:
Step1: Multiply first equation by 5 and second by 2.
$$ \begin{aligned} 3\cdot{\color{red}{5}}\cdot x + 2\cdot{\color{red}{5}}\cdot y = & -1\cdot {\color{red}{5}} \\ 4\cdot{\color{red}{2}}\cdot x - ~5\cdot{\color{red}{2}}\cdot y = & 14\cdot{\color{red}{2}} \end{aligned} $$After simplifying we have:
$$ \begin{aligned} {\color{blue}{ 15x + 10y }} = & {\color{blue}{ -5 }} \\ {\color{red}{ 8x - 10y }} = & {\color{red}{ 28 }} \end{aligned} $$Step2: add the two equations together to eliminate $y$ from the system.
$$ \begin{aligned} ({\color{blue}{ 15x + 10y }}) + ({\color{red}{ 8x - 10y }}) = & {\color{blue}{ -5 }} + {\color{red}{ 28 }}\\ 15x + 10y + 8x - 10y = & 23 \\ 23x = & 23 \\ x = & 1 \end{aligned} $$Step 3:substitute the value for x into the original equation to solve for y.
$$ \begin{aligned} 3x + 2y = & -1 \\ 3\cdot1 + 2y = & -1 \\ 3 + 2y = & -1 \\ 2y = & -4 \\ y = & -2 \end{aligned} $$The solution is:
$$ \begin{aligned} {\color{blue}{ x = }} & {\color{blue}{ 1 }} \\ {\color{blue}{ y = }} & {\color{blue}{ -2 }} \end{aligned} $$Check the solution by using the above calculator.
3. Cramer's rule
Given the system:
$$ \begin{aligned} a_1x + b_1y = c_1 \\ a_2x + b_2y = c_2 \end{aligned} $$with
| $$ D = \left|\begin{array}{cc} a_1 & b_1 \\ a_2 & b_2 \end{array}\right| \ne 0 $$ | $$ D_x = \left|\begin{array}{cc} c_1 & b_1 \\ c_2 & b_2 \end{array}\right| $$ | $$ D_y = \left|\begin{array}{cc} a_1 & c_1 \\ a_2 & c_2 \end{array}\right| $$ |
then the solution of this system is:
| $$ x = \frac{D_x}{D} $$ | $$ y = \frac{D_y}{D} $$ |
Example: Solve the system of equations using Cramer's rule
$$ \begin{aligned} {\color{blue}{3}}x + {\color{red}{12}}y = & -4 \\ {\color{blue}{7}}x {\color{red}{- ~8}}y = & 3 \end{aligned} $$Solution: First we compute $D,~ D_x$ and $D_y$.
Therefore,