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System of Equations Solver
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A System of Linear Equations or systems is a set of linear equations, which contains same set of variables. A system of linear equations in three variables is mostly need to use more variable than single variable to form a condition in many fields. In such a situation we need to use system of equation in order to get solution for the given condition.

When a linear equation has at least one answer, it is reliable. If the system has no solution, then it is non – reliable. When the system has more number of answers, then it is dependent, else it is independent.

A system of linear equations in two variables is given by:
Ax + By = C
Dx + Ey = F
A system of linear equations in three variables is given by:
Ax + By + Cz = D
Ex + Fy + Gz = H
Ix + Jy + Kz = L
In a two systems of linear equations we have two unknown variables and we need to calculate those two unknown variables. Similarly in a three systems of linear equations we have three unknown variables and we need to calculate those three unknown variables. A system of linear equations can be solved in the following ways:
  • Substitution method
  • Elimination method
  • Cramer's rule
  • Graphing method
Solving systems of equations calculator or System of Equations Calculator will help us to find the solution for system of linear equations. If the systems of equation are given in the block provided it solves and gives you the value of variable. Hence it acts as a System of Equations Solver  or ordered pair calculator or linear system calculator.
 

System of Equations Solver Step by Step

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Step 1 :  

Observe the given equations and select any easiest equation to solve for x = or y =.



Step 2 :  

Now, substitute the equation comes from the step1 into other equation and then solve for the variable. After getting the value for either of the variable, just put it on the equation ofstep1.



Step 3 :  

The solution of the variables can be written is an order pair.



System of Equations Solver Show Work

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  1. Solve the system of equations

    x - y = 2

    2x - y = 2


    Step 1 :  

    Consider the first equation: x - y = 2
    $\Rightarrow$ x = 2 + y



    Step 2 :  

    Now consider second equation, 2x - y = 2

    Substitute x = 2 + y in the above equation.

    So, 2(2 + y) - y = 2

    $\Rightarrow$ 4 + 2y - y = 2

    $\Rightarrow$ 4 + y = 2


    $\Rightarrow$ y = 2 - 4 = -2


    If we substitute y = -2 in x = 2 + y


    then, x = 2 - 2 = 0



    Step 3 :  

    So, x = 0 and y = -2


    The solution is (x, y) = (0, -2).



    Answer  :  

    The solution is (x, y) = (0, -2).



  2. Solve the system of equations

    x + 3y = 2

    x + 4y = 3


    Step 1 :  

    Consider the second equation: x + 4y = 3
    $\Rightarrow$ x = 3 - 4y



    Step 2 :  

    Now consider first equation, x + 3y = 2


    Substitute x = 3 - 4y in the above equation.

    So, 3 - 4y + 3y = 2

    $\Rightarrow$ 3 -y = 2

    $\Rightarrow$ y = 3 - 2


    $\Rightarrow$ y = 1


    If we substitute y = 1 in x = 3 - 4y


    then, x = 3 - 4(1) = 3 - 4 = -1



    Step 3 :  

    So, x = -1 and y = 1


    The solution is (x, y) = (-1, 1).



    Answer  :  

    The solution is (x, y) = (-1, 1).



  3. A school planned to conduct admission tests for class A and Class B. The admission fee for class A is 5 dollar and 6 dollars for class B. On a specific day, 2000 children attended the test and 11000 dollars collected by school. How many children appeared for the written test for class A and class B.


    Step 1 :  

    Let us choose two different variables for the two different unknowns.
    Number of children appeared for written test for class A is x, and for class B is y.



    Step 2 :  

    Total number of children = 2000
    i.e. x + y = 2000  .........(1)
    Total amount collected by school = 11000
    i.e. 5x + 6y = 11000  .........(2)



    Step 3 :  

    Now solve for x and y
    Apply substitution method, to find the value of x and y
    (1) => x = 2000 - y
    Substitute the value of x in equation (2), we get
    5(2000 - y) + 6y = 11000
    10000 - 5y + 6y = 11000
    y = 1000, again substitute this value in equation (1), we obtain
    x + 1000 = 2000  or x = 1000



    Answer  :  

    1000 children appeared for each class.



  4. A school has organized an educational trip to Disney world. 148 students went for the trip. There were ten drivers and two types of vehicles including cabs and buses. Each cab has 10 seating capacity and each bus has 22 seating capacity, including driver seat. How many buses and cabs did school hired for the educational trip?


    Step 1 :  

    Let company hired x number of buses and y number of cabs.



    Step 2 :  

    Total number of drivers = 10
    x + y = 10  .....(1)
    Number of employees going for city trip = 148
    22x + 10y = 148  ....(2)



    Step 3 :  

    Find the value of y from (1) and substitute in (2)
    (1)=> y = 10 - x   ....(3)
    Now 22x + 10(10 - x) = 148   (Substitution method)
    22x + 100 - 10x = 148
    12x = 48
    x = 4 (on solving)
    Put x = 4 in (3), we get
    y = 10 - 4 = 6



    Answer  :  

    Company hired 4 buses and 6 cabs.



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