Introduction to linear elementary algebra:
A linear elementary algebra equation is constructed by constants and one or more variables by using only the arithmetic operations like, addition, subtraction, multiplication and division.
The linear equations are most often used in linear elementary algebra.
A general form with two variables x and y is
y= mx +c
Where, m and b values are constants.
Example problem for linear elementary algebra 1
Example problem for linear elementary algebra 1 is given below:
Find four different solutions of the linear equation x + 2y = 6.
Solution:
x = 2, y = 2 is a solution because for x = 2, y=2
x + 2y = 6, 2 + 2(2) = 6, 2+4=6, 6=6.
Now, let us choose x = 0. With this value of x, the specified equation reduces to 2y = 6, which has the unique solution y = 3.
So x = 0, y = 3 is also a solution of x + 2y = 6.
Similarly, plug y = 0, the given equation reduces to x = 6.
So, x = 6, y = 0 is a solution of x + 2y = 6 as well.
Finally, let us take y = 1.
The given equation now reduces to x + 2 = 6, whose solution is given by x = 4. Then, (4, 1) is also a solution of the given equation so four of the infinitely many solutions of the given linear equation are:
(2, 2), (0, 3), (6, 0) and (4, 1).
Example problem for linear elementary algebra 2:
Example problem for linear elementary algebra 2 is given below:
The price of 2 pencils and 3 erasers is Rs 9 and the price of 4 pencils and 6 erasers is Rs 18. Find the price of each pencil and each eraser.
Solution:
The pair of linear equations formed was:
2x + 3y = 9 ……………..… (1)
4x + 6y = 18 …………….... (2)
We first write the value of x in terms of y from the equation
2x + 3y =9, to get x = (9-3y)/2 ………………. (3)
Now we plug this value of x in Equation (2),
4(9-3y)/2+ 6y = 18
18 – 6y + 6y = 18
18 = 18
This statement is true for all values of y.
However, there is no specific value for y as a solution.
Therefore, we cannot get a specific value of x.
Therefore, Equations (1) and (2) have infinitely lots of solutions.
Here an exact cost of a pencil and an eraser can't be found, because there are many common solutions, to the given situation.
A linear elementary algebra equation is constructed by constants and one or more variables by using only the arithmetic operations like, addition, subtraction, multiplication and division.
The linear equations are most often used in linear elementary algebra.
A general form with two variables x and y is
y= mx +c
Where, m and b values are constants.
Example problem for linear elementary algebra 1
Example problem for linear elementary algebra 1 is given below:
Find four different solutions of the linear equation x + 2y = 6.
Solution:
x = 2, y = 2 is a solution because for x = 2, y=2
x + 2y = 6, 2 + 2(2) = 6, 2+4=6, 6=6.
Now, let us choose x = 0. With this value of x, the specified equation reduces to 2y = 6, which has the unique solution y = 3.
So x = 0, y = 3 is also a solution of x + 2y = 6.
Similarly, plug y = 0, the given equation reduces to x = 6.
So, x = 6, y = 0 is a solution of x + 2y = 6 as well.
Finally, let us take y = 1.
The given equation now reduces to x + 2 = 6, whose solution is given by x = 4. Then, (4, 1) is also a solution of the given equation so four of the infinitely many solutions of the given linear equation are:
(2, 2), (0, 3), (6, 0) and (4, 1).
Example problem for linear elementary algebra 2:
Example problem for linear elementary algebra 2 is given below:
The price of 2 pencils and 3 erasers is Rs 9 and the price of 4 pencils and 6 erasers is Rs 18. Find the price of each pencil and each eraser.
Solution:
The pair of linear equations formed was:
2x + 3y = 9 ……………..… (1)
4x + 6y = 18 …………….... (2)
We first write the value of x in terms of y from the equation
2x + 3y =9, to get x = (9-3y)/2 ………………. (3)
Now we plug this value of x in Equation (2),
4(9-3y)/2+ 6y = 18
18 – 6y + 6y = 18
18 = 18
This statement is true for all values of y.
However, there is no specific value for y as a solution.
Therefore, we cannot get a specific value of x.
Therefore, Equations (1) and (2) have infinitely lots of solutions.
Here an exact cost of a pencil and an eraser can't be found, because there are many common solutions, to the given situation.
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