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.

2x2 System of equations solver

two solving methods + detailed steps
help ↓↓ examples ↓↓ tutorial ↓↓

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Enter system of equations (empty fields will be replaced with zeros)

y =

Choose solving method:

Solve using an addition/elimination method (default)

Solve using Cramer's rule

Settings:

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Examples

ex 1:

Use the substitution method to solve the system of equations.:

- 5 x + 3 y 3 x - y = 1 = 3

ex 2:

Solve using an addition method:

\frac{1}{2} x + 3 y - 2 x - \frac{4}{3} y = 2 = - 3

ex 3:

Solve using a Cramer's rule:

- \frac{3}{2} x - 0.4 y - 2 x + y = 2/3 = - 3/7

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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.

3 x + 2 y = - 2 x - y = 3 - 1

Solution:

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

3 x + 2 y = - 2 x + 1 = 3 y

Step2: SUBSTITUTE $y$ into first equation.

3 x + 2 (- 2 x + 1) = - 2 x + 1 = 3 y

Step3: Solve first equation for $x$ .

x = - 2 x + 1 = - 1 y

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

x = y = - 1 - 2 \cdot (- 1) + 1

The solution is:

x = y = - 1 3

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.

3 x + 2 y = 4 x - 5 y = - 1 14

Solution:

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

3 \cdot 5 \cdot x + 2 \cdot 5 \cdot y = 4 \cdot 2 \cdot x - 5 \cdot 2 \cdot y = - 1 \cdot 5 14 \cdot 2

After simplifying we have:

15 x + 10 y = 8 x - 10 y = - 5 28

Step2: add the two equations together to eliminate $y$ from the system.

(15 x + 10 y) + (8 x - 10 y) = 15 x + 10 y + 8 x - 10 y = 23 x = x = - 5 + 28 23231

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

3 x + 2 y = 3 \cdot 1 + 2 y = 3 + 2 y = 2 y = y = - 1 - 1 - 1 - 4 - 2

The solution is:

x = y = 1 - 2

Check the solution by using the above calculator.

3. Cramer's rule

Given the system:

a_{1} x + b_{1} y = c_{1} a_{2} x + b_{2} y = c_{2}

with

D = a_{1} a_{2} b_{1} b_{2} \neq = 0

D_{x} = c_{1} c_{2} b_{1} b_{2}

D_{y} = a_{1} a_{2} c_{1} c_{2}

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

3 x + 12 y = 7 x - 8 y = - 4 3

Solution: First we compute $D, D_{x}$ and $D_{y}$ .

D = 37 12 - 8 = 3 \cdot (- 8) - 7 \cdot 12 = - 24 - 84 = - 108 D_{x} = - 4 3 12 - 8 = - 4 \cdot (- 8) - 3 \cdot 12 = 32 - 36 = - 4 D_{y} = 37 - 4 3 = 3 \cdot 3 - 7 \cdot (- 4) = 9 + 28 = 37

Therefore,

x = \frac{D _{x}}{D} = \frac{- 4}{- 108} = \frac{1}{27} y = \frac{D _{y}}{D} = \frac{37}{- 108} = - \frac{37}{108}

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