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Table of Content
The substitution calculator makes the solution of algebraic equations simple and fast by substituting one variable into another equation.
“It is a process of algebraic equations do involve gathering one variable value from one and inserting into the other linear equation to find the values of all the variables”
The solution of the system of equations by substitution process provides the simple solution of the linear algebraic variable. It is possible to know the roots “x” and “y” of linear algebraic equations.
The algebraic equation calculator can be used to solve linear equations by substitution with the following major methods:
There are 3 basic steps involved in the process:
Now solve the system of linear values.
2x + 4y = 4———-(Eq1)
x + y = 3———-(Eq2)
Find values of “x” from the (Eq2)
x = 3 – y
Input values of the variable “x” in the (Eq1)
2 ⋅ (3 – 1y) + 4y = 4
Now the value of variables “y”
2y = -2
The value of “y” in (Eq1)
y = -1
y = -1 in the
2x + 4 ⋅ (-1) = 4
The value of “x”
x = 4
The final values of the “x” and “y”
x = 4 , y = -1
In the substitution calculator, take the value of one variable from one to the other.
System of linear equation | Results |
x+y=5 ; x+2y=7 | x=3, y =2 |
-x-7y=-1 ; x-2y=-8 | x=−6, y=1 |
2x + y = 7 ;3x – 4y = 1 | x=38/11, y=1/11 |
2x – y = 3 ; x + 2y = 2 | x=8/5, y=1/5 |
x-3y = -43 ; 2x+2y=-41 | x=-209/8, y=45/8 |
You simply cut down the “x” and “y” by making their coefficients the same.
You simply cut down the “x” and “y” by making their coefficients the same.
The substitution method calculator is efficient in solving a system of equations.
Let’s have a look at its working procedure!
Input:
Output:
Implement the substitution process to solve the values of the variables “x” and “y” in a system of linear values. The substitution method calculator with steps elaborates all the steps of the process and the simple for the users
From Wikipedia: (logic), Propositional logic
From Calc-Workshop: What is the process?