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Linear equation calculator and the world of mathematics

Linear equation calculator – the modern method of solving linear equations

What is a linear equation? This is simply an algebraic equation or expression in which each term is either a constant of the product of a constant but it must have one or more variables. Constants can be numbers and parameters. This equation is widely used in different aspects of mathematics most especially in applied mathematics. Graphically, linear equation can be defined as an equation for a straight line. A quick instance to shed more light: Plotting a graph of y= 2x + 1, we realize that when x increases, y also increases but twice as faster, hence 2x. Therefore, when our x becomes zero (0), y is already one (1). Hence, +1 is needed so we can have a linear equation like this: y= 2x + 1.

On the other hand, linear equation calculato r is used in mathematics to solve or find a variable in any linear equation. At the very beginning of algebra class, students will be introduced to linear equations. At this stage, just the way students are taught how to write a philosophy paper, they will be asked to rearrange or rewrite any given linear equation and find the answer to any of the unknown variables found in the equation.

Any linear equation calculator must be able to evaluate from simple equations to any complex linear equations. This calculator has many functions which speed up the solving of linear equations as easy as it can be. When you input an equation into the calculator, it will begin by first expanding or simplifying the equation. Just as essay proofreading is being carried out in details, for the purpose of linear equation calculator, one of the following functions can be performed:

  • Square root taking
  • Addition
  • Factoring
  • Division
  • Subtraction
  • Completing the square

Exponents – exponents can be supported on variables using the caret symbol (^). For instance, to express x2, we simply input: x^2. Exponents must be a positive integer but cannot be either: negative, decimal, or variable.

Parentheses and Brackets – parentheses ( ) and brackets [ ] can also be used to group terms as in a standard equation or expression.

Subtraction, Multiplication and Addition – the standards + and – symbols can be used in case of addition and subtraction. For multiplication, use the asterisk symbol (*). E.g. 2 * x can also be entered as 2x. Similarly, 2 * (x + 5) can as well be imputed as: 2(x + 5); 2x * (5) can be imputed as 2x(5). The * is also optional when we want to multiply with parentheses, example: (x + 1)(x – 1).

Order of Operations – the calculator follows a standard order of operations taught by most algebra books – Exponents, Parentheses, Multiplication, Addition, Subtraction and Division. The only exception is: division is not currently supported; attempts to use the division symbol (/) will result in an error.

Linear equation calculator – history and the inventor

Let’s set it out as a 1000 words essay, it would be explained and described in short as how the linear equation was discovered. The study of linear algebra includes vector algebra, the theory of vector spaces and matrix algebra. Linear algebra started as the study of linear equations as well as the solution of simultaneous linear equations. An equation is called linear if the variable in its equation is multiplied by itself or any other unknown. The equation 3x + 2y + z = 0 is a linear equation with three variables. The equation x3 + 6y + z + 5 = 0 is not linear, because the variable x is raised to the power 3. The equation 5x – xy + 6z = 7 is not a linear equation as well, because the product of two unknowns (xy) appears in it. Thus linear equations are always of degree 1.

Sir William Rowan Hamilton was an Irish physicist, a mathematician and an astronomer as well. Unlike how easy it is now to carry out a thesis generator, his outstanding studies of mechanical and optical systems brought about his discovery of a new mathematical concepts and the techniques used.

His best known contribution to mathematical physics is the redevelopment of the Newtonian mechanics also called Hamiltonian mechanics in today’s mathematical physics. Sir Williams book report has proven central to the modern study of the field theories like: electromagnetism, and to the development of quantum mechanics. While in pure mathematics, he is best known as the inventor of quaternions.

The inventor of linear equations is William Rowan Hamilton. He was the first to introduce linear equations in the year 1843. These equations describe mechanics in 3 dimensional space and are an important historical discovery.

Linear equation calculator – researches and different methods

A system of linear equations can be solved using 4 ways and they are:

  • Method of Substitution
  • Method of Elimination
  • Cramer’s rule
  • Graphing method

Substitution method – the substitution method is most useful in solving systems of two equations where there are two unknowns. The aim of this method of substitution is to solve one of the equations for the unknowns, and then use the answer in the other equation. There are 4 steps to solving any linear equation using the substitution method and they include:

  • Step 1: Solve one of the equations for either x = or y =
  • Step 2: Substitute the solution from step 1 into the other equation
  • Step 3: Solve this new equation
  • Step 4: Solve for the second variable

Example 1: solve the following equations by the substitution method: 3x+2y=−2x− y=3−1:

  • Solution
  • Step1: first find the unknown variables in the equation above: 3x+2y=−2x+1=3y
  • Step2: Substitute y into first equation: 3x+2(−2x+1)=−2x+1=3y
  • Step3: Solve first equation for x: x=−2x+1=−1y
  • Step4: To find y, substitute −1 for x into second equation: x=y=−1−2⋅(−1)+1
  • The answer is: x=y=−13

Example 2: Solve the following system by substitution

  • 2x + 3y = 5 

    X + y = 5

  • Solution
  • Step 1: Solve either of the equations for x = or y =, then solve the other equation for y: 

    x + y = 5 

    y = 5 – x

  • Step 2: Substitute the answer from step 1 in the second equation: 

    2x + 3y = 5 

    2x + 3 (5 – x) = 5

  • Step 3: Solve the new equation above: 

    2x + 3 (5 – x) = 5 

    2x + 15 – 3x = 5 

    -x + 15 = 5 

    -x = 5 – 15 

    x = 10

  • Step 4: Solve for the other variable: 

    y = 5 – x 

    y = 5 – 10 

    y = -5

  • The answer is: (x, y) = (10, -5)

Example 3: Solve by substitution

2x + 5y = 12


4x – y = 2:


  • Solution
  • Step 1: Solve the equations for x = or y =. Since the coefficient of y in the second equation is -1, solve for y in equation 2: 

    4x – y = 2 

    -y = 2 – 4x 

    y = 4x – 2

  • Step 2: Substitute the answer from step 1 into the second equation: 

    2x + 5y = 12 

    2x +5(4x – 2) = 12

  • Step 3: Solve this new equation for x: 

    2x +5(4x – 2) = 12 

    2x +20 – 10 = 12 

    22x = 22 

    x = 1

  • Step 4: Solve for the second variable: 

    y = 4x – 2 

    y = 4•1 – 2 

    y = 2

  • The solution is: (x, y) = (1, 2)

Elimination method – another way of solving a linear equation is to use the elimination method. Using this method, you can add or subtract the equations to get an equation with one variable. When the coefficients of that variable are opposites, you add the equations in order to eliminate that variable. But when the coefficients of the variables are equal, you subtract the equations to eliminate the variable.

Example 1: Solve the system of equations using the elimination method: 3x+2y=4x− 5y=−114

      • Solution
      • Step1: Multiply first equation by 5 and second by 2: 3⋅5⋅x+2⋅5⋅y=4⋅2⋅x− 5⋅2⋅y=−1⋅514⋅2. Then, simplifying to have: 15x+10y=8x−10y=−528
      • Step2: add the 2 equations together to eliminate y from the system: (15x+10y)+(8x−10y)=15x+10y+8x−10y=23x=x=−5+2823231
      • Step 3: Substitute the value for x into the first equation to solve for y: 3x+2y=3⋅1+2y=3+2y=2y=y=−1−1−1−4−2
      • The answer is: x=y=1−2

Example 2: Solve the system of equations by elimination:

      • 3x – y = 5


      • X + y = 3


        • Solution
        • Cancel out y by adding both equations together: 

          3x – y =5 

          x + y =3 

          4x = 8 

          x = 2

        • In order to solve for y, use the value for x and then substitute it into either one of the original equations: 

          x + y = 3 

          2 + y = 3 

          y = 1

        • The solution is (x, y) = (2, 1)

Example 3: Solve the system using elimination:

        • x+3y=-5


        • 4x-y=6


          • Solution
          • Multiply the first equation by -4, so that we will be able to cancel out the x-coefficients: 


            4x-y=6) Multiply them 


            4x-y=6) Add them 



          • Take the value for y and substitute it into either one of the original equations: 





          • The solution is (x, y) = (1, -2)

Example 4: Solve the system of equation using elimination method:

          • 2x-5y=11


          • 3x=2y=7


            • Solution
            • First multiply the first row by -3 and the second row by 2; then we will add the equations together: 

              (2x-5y=11) multiply by -3 

              (3x=2y=7) multiply by 2 


              6x+4y=14) add both equations 



            • Substitute y = -1 back into first equation: 





            • The solution is (x, y) = (3, -1)

Cramer’s Rule – uses determinants of a matrix to solve any linear equation. Linear equation calculator using Cramer’s rule can only work for square systems of linear equations where the determinants of the coefficient matrix is not zero. This rule is absolutely flawless on computer systems as it is very fast in handling any equation we want to find a solution to, slow on the calculator without a program as every determinant must be entered manually but can be used when you need to find one of the unknown variables. Cramer’s rule can be used in finding the unknown variable, without having to solve the whole system of the given equation.

Don’t let all the equation confuse you. The Rule is really simple and easy to use. Just pick the variable to solve for, change the variable column of values in the coefficient determinant in the answer column’s values in order to assess the determinant, and divide by the coefficient determinant.

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