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- 2x2 Systems Solver

This calculator solves system of two equations with two unknowns.
Calculator will generate a **step by step** explanation using an addition/elimination method or Cramer's rule.

examples

example 1:

Solve by using an elimination method:
$$
\begin{aligned}
-5x + 3y & = 1 \\[2ex]
3x - y & = 3
\end{aligned}
$$

example 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:

Solve by using an Cramer's rule:
$$
\begin{aligned}
-\frac{3}{2}x - 0.4 y & = 2/3 \\[2ex]
-2 x + y & = - 3/7
\end{aligned}
$$

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

**Example:** Solve the system of equations by the substitution method.

**Solution:**

**Step1: **Solve one of the equations for one of the variables. We note that is
simplest to solve the second equation for $ y $.

**Step2: **SUBSTITUTE $y$ into first
equation.

**Step3: **Solve first equation for $x$.

**Step4: **To find $y$,
substitute $-1$ for
$x$ into second equation.

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.

**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.

**Solution:**

**Step1: **Multiply first equation by 5 and second by 2.

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.

**Step 3:**substitute the value for x into the
original equation to solve for y.

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.

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

**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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