Algebra substitution elimination
Introduction :
In algebra, the linear systems of equations contain two methods. They are the elimination and the substitution. The equations can be solved by eliminating one of the variables and the value of the variables can be determined. In algebra substitution, one of the equations can be solved by separating the variable values and the other values can be found by substituting the variable values in other equation.
About algebra substitution elimination:
Now we see how the values can be determined for substitution as follows,
The given algebra equations should be taken and one variable should be found by separating the variables. The other variable value can be determined by using the substitution method and can be found.
For the algebra elimination method, the variables can be found by eliminating one of the variables by adding two equations. Then, the variable value can be determined and the other variable found by substituting in another equation.
Problems for algebra substitution elimination:
Example 1:
Solve the two equations by elimination method as follows,
x + 3y = 10
- x + 3y =14
Solution:
Take the two equations as follows,
x + 3y = 10
- x + 3y =14
-----------------
6y = 24
-----------------
Divide the equation by 6 and we get,
6y = 24
y = 4
Now we put the value in the equation,
x + 3y = 10
x + 3(4) = 10
x + 12 = 10
x = 10 -12
x = -2
Thus, x = -2 and y = 4.
Example 2:
Solve the algebra linear equations by substitution method:
x + y = 4
x - y = 2
Solution:
First take the equation x + y = 4
x = 4 - y
Substitute x in the equation,
x – y = 2
4 - y - y = 2
- 2y = -2
Divide the equation by - 2 we get,
y = 1.
Substitute the value of y in x = 4 - y
x = 4 - y
x = 4 - 1
x = 3.
The x and y value are x = 3 and y = 1.
In algebra, the linear systems of equations contain two methods. They are the elimination and the substitution. The equations can be solved by eliminating one of the variables and the value of the variables can be determined. In algebra substitution, one of the equations can be solved by separating the variable values and the other values can be found by substituting the variable values in other equation.
About algebra substitution elimination:
Now we see how the values can be determined for substitution as follows,
The given algebra equations should be taken and one variable should be found by separating the variables. The other variable value can be determined by using the substitution method and can be found.
For the algebra elimination method, the variables can be found by eliminating one of the variables by adding two equations. Then, the variable value can be determined and the other variable found by substituting in another equation.
Problems for algebra substitution elimination:
Example 1:
Solve the two equations by elimination method as follows,
x + 3y = 10
- x + 3y =14
Solution:
Take the two equations as follows,
x + 3y = 10
- x + 3y =14
-----------------
6y = 24
-----------------
Divide the equation by 6 and we get,
6y = 24
y = 4
Now we put the value in the equation,
x + 3y = 10
x + 3(4) = 10
x + 12 = 10
x = 10 -12
x = -2
Thus, x = -2 and y = 4.
Example 2:
Solve the algebra linear equations by substitution method:
x + y = 4
x - y = 2
Solution:
First take the equation x + y = 4
x = 4 - y
Substitute x in the equation,
x – y = 2
4 - y - y = 2
- 2y = -2
Divide the equation by - 2 we get,
y = 1.
Substitute the value of y in x = 4 - y
x = 4 - y
x = 4 - 1
x = 3.
The x and y value are x = 3 and y = 1.