"SLC-S22W3//Equations and Systems of equations."
Steemian Friends,
Today, I will write the homework for @khursheedanwar's brother's learning lesson for week 3 of Steemit Learning Challenge 22. The lesson name is SLC S22W3//Equations and Systems of Equations. One of my favorite subjects is Mathematics. I have liked doing algebra since childhood. I hope everyone likes my homework.
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Linear Equation:
An equation whose power of the variable is at most one is called a linear equation. Here, Ax + By = C is a linear equation.
A and B are coefficients,
x, y are variables, and
C is constant.
Example:
2x + 3y = 5 is a linear equation.
Here, x and y are variables.
2 and 3 are coefficients,
and 5 is constant.
A linear equation is an algebraic equation in which each term has exponent one and is always a straight line when graphed.
Nonlinear equation:
An algebraic equation with two or more variable powers is called a nonlinear equation.
Here, Ax^2 + By^2 = C is a nonlinear equation.
A and B are coefficients,
x and y are variables,
C is constant.
Here, the degree of x is 2.
Our nonlinear equation says that if the degree is two or more in the equation, then it is called a nonlinear equation.
Example:
x^2 + y^2 = 10 is a linear equation.
Here, x and y are variables. Degrees 2 and 10 are constant.
Difference between linear equation and nonlinear equation:
linear equation | nonlinear equation |
---|---|
Algebraic equations that have degree one are called linear equations. | Algebraic equations that have degree two or more are called nonlinear equations. |
Example: 2x + y = 8 | Example: x^3 + y^2 = 5 |
Graphing linear equations results in a straight line. | Graphing nonlinear equations results in a curve. |
The representation of the linear equation is, y = mx + c where x and y are variables, m is the slope of the line. c is a constant. | The representation of the nonlinear equation is, Ax^2 + By^2= C where x and y are variables, degree is 2. C is a constant. |
I have shown how to solve a linear equation using Cramer's rule. First, I created two linear equation formats, as shown in Equations 1 and 2.
Equation:
- a1x + b1y = c1 and
- a2x + b2y = c2
Here, the determinants consist of the coefficients of x and y.
D= a1b2-a2b1
The determinants consist of constant terms instead of the coefficients of x.
D= b2c1-b1c2
The determinants consist of constant terms instead of the coefficients of y.
D= a1c2-a2c1
Now multiply the equations 1 and 2. I get
x = Dx/D,
y = Dy/D
according to cremars rule. I have shown the calculation on the paper below.
I have done an example with a linear equation using the cremars rule above.
Example:
I get the values of x and y from the linear equation as -2/5 and 8/5, respectively. I have shown the math technique on paper.
(a)
x + 2y = 7
3x - 2y = 5
(b)
4x + 6y = 2
x - 2y = 3
(You are required to solve these problems at paper and then share clear photographs for adding a touch of your creativity and personal effort which should be marked with your username)
Solution:
I solved the linear equations in three ways. I can solve them in another way if I want. Today, I found the values of x and y using the rules of matrices, determinants, differentials, and algebra. In each of my rules, the values of x and y came out the same. I have given each method's solution image and my username below.
Using Cramer's Rule:
Rule of Algebra:(First Way)
Rule of Algebra:(Second Way)
Scenario number 1
Suppose there's a company producing two products, A and B.If cost of producing x units of A and y units of B is given by system then;
2x + 3y = 130 (cost of materials)
x + 2y = 110 (cost of labor)
If company wants for producing 50 units of product A then calculate how much units of product B they may produce?
(Solve the above scenerio based questions and share step by step that how you reach to your final outcome)
Solution:
When x = 50 is put into equation 1, y is 10. Again, when we put x = 50 in equation 2, we get the value of y as 30. That is, for x = 50, the value of y is different.
As above, the two equations give different results, which means that the system of equations is inconsistent. We also understand that no solution will satisfy both equations simultaneously.
According to the above equation, producing 50 units of product, materials, and labor will be inconsistent. Product production is disproportionate to product labor and material costs.
Final report:
I read your lesson and understood that the equations have no solution because they have different values.
Scenario number 2
• Suppose there's a bakery producing two types of cakes which are vanilla and chocolate.If cost of producing x cakes of vanilla and y cakes of chocolate is given by system then;
x + 2y = 80 (cost of ingredients)
2x + y = 70 (cost of labor)
If bakery wants for producing 30 cakes of vanilla then calculate how much cakes of chocolate can they produce?
Solution:
When we plug the value of x = 30 into equation 1, we get the value of y as 25. Again, when we put x = 30 in equation 2, we get the value of y as 10. For x = 30, the two equations have different values for y.
When x = 30, the two equations have different values for y. From this, we understand that the equations are inconsistent. Production costs are inconsistent with vanilla cake and chocolate cake.
Final report:
I read your lesson and understood that the equations have no solution because they have different values.
SL No. | My Invited Steemit Friends |
---|---|
1 | @memamun |
2 | @solaymann |
3 | @lirvic |
This is my Twitter share link :
https://twitter.com/mahadih83660186/status/1874397274916417813?t=ZUhGZFoGsHpi8H_hJ78_lw&s=19
Hi, @mahadisalim,
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