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# 2x2 system of equations solver

This calculator solves system of two equations with two unknowns. The solver provides a step-by-step explanation using an addition/elimination method or Cramer's rule.

2x2 System of equations solver
Two solving methods + detailed steps
show help ↓↓ examples ↓↓
 Solve by using an addition/elimination method (default) Solve by using Cramer's rule
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examples
example 1:ex 1:
Use the substitution method to solve the system of equations.: \begin{aligned} -5x + 3y & = 1 \\[2ex] 3x - y & = 3 \end{aligned}
example 2:ex 2:
Solve by using an addition method: \begin{aligned} \frac{1}{2}x + 3y & = 2 \\[2ex] -2 x - \frac{4}{3} y & = - 3 \end{aligned}
example 3:ex 3:
Solve by using a Cramer's rule: \begin{aligned} -\frac{3}{2}x - 0.4 y & = 2/3 \\[2ex] -2 x + y & = - 3/7 \end{aligned}

## 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 $D_y$.

\begin{aligned} & D~~ = \left|\begin{array}{cc} {\color{blue}{3}} & {\color{red}{~12}} \\ {\color{blue}{7}} & {\color{red}{-8}} \end{array}\right| = {\color{blue}{3}}\cdot{\color{red}{(-8)}} - {\color{blue}{7}}\cdot{\color{red}{12}} = -24 - 84 = -108\\ & D_x = \left|\begin{array}{cc} -4 & {\color{red}{~12}} \\ ~3 & {\color{red}{-8}} \end{array}\right| = -4\cdot{\color{red}{(-8)}} - 3\cdot{\color{red}{12}} = 32 - 36 = -4\\ & D_y = \left|\begin{array}{cc} {\color{blue}{3}} & -4 \\ {\color{blue}{7}} & ~3 \end{array}\right| = {\color{blue}{3}}\cdot3 - {\color{blue}{7}}\cdot(-4) = 9 + 28 = 37\\ \end{aligned}

Therefore,

\begin{aligned} & {\color{blue}{ x = \frac{D_x}{D} = \frac{-4}{-108} = \frac{1}{27} }}\\ & {\color{blue}{ y = \frac{D_y}{D} = \frac{37}{-108} = -\frac{37}{108} }} \end{aligned}
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