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MODULE 2 - EQUATIONS
OVERVIEW
This module will be discussing the concepts governing equations and inequalities. We are going to learn on how to solve equations, how to establish equations, and how to construct equations that reflects real-world phenomena and problems. We shall learn the importance of equation and inequalities in modelling as it is part of engineering skills.
This module will be discussing the concepts governing equations and inequalities. We are going to learn on how to solve equations, how to establish equations, and how to construct equations that reflects real-world phenomena and problems. We shall learn the importance of equation and inequalities in modelling as it is part of engineering skills.
LEARNING OBJECTIVES:
At the end of this module, you should be able to:1. Solve linear equations.2. Solve one variable in terms of others.3. Provide models of real-life applications using equation.4. Solve quadratic equations.
At the end of this module, you should be able to:1. Solve linear equations.2. Solve one variable in terms of others.3. Provide models of real-life applications using equation.4. Solve quadratic equations.
An equation is a statement that two mathematical expressions are equal. The majority of the equations we study in algebra contain variables which are numbers represented by symbols (usually letters). In the equation 8x + 9 = 17 the letter x is the variable. We think of x as the “unknown” in the equation, and our goal is to find the value of x that makes the equation true. The values of the unknown that make the equation true are called the solutions or roots of the equation, and the process of finding the solutions is called solving the equation. Two equations with exactly the same solutions are called equivalent equations. To solve an equation, we try to find a simpler, equivalent equation in which the variable stands alone on one side of the equal sign. Here are the properties that we use to solve an equation. (In these properties, A, B, and C stand for any algebraic expressions, and the symbol ≍ means “is equivalent to.”(Watson, 2012).
Property Description
A = B ↔ A + C = B + C Adding the same quantity to both sides of an equation gives an equivalent equation.
A = B ↔ C(A) = C(B) (C ≠ 0) Multiplying both sides of an equation by the same nonzero quantity gives an equivalent equation.
Property Description
A = B ↔ A + C = B + C Adding the same quantity to both sides of an equation gives an equivalent equation.
A = B ↔ C(A) = C(B) (C ≠ 0) Multiplying both sides of an equation by the same nonzero quantity gives an equivalent equation.
LINEAR EQUATIONS
A linear equation is an equation in which the highest power of the variable is always 1. It is also known as a one-degree equation. The standard form of a linear equation in one variable is of the form Ax + B = 0. Here, x is the real variable, and A and B are real numbers. The standard form of a linear equation in two variables is of the form Ax + By = C. Here, x and y are variables, and A, B and C are constants.
A linear equation is an equation in which the highest power of the variable is always 1. It is also known as a one-degree equation. The standard form of a linear equation in one variable is of the form Ax + B = 0. Here, x is the real variable, and A and B are real numbers. The standard form of a linear equation in two variables is of the form Ax + By = C. Here, x and y are variables, and A, B and C are constants.
SOLUTIONS OF CONDITIONAL EQUATIONS
Solutions of conditional equations are values of the unknowns which make both expressions equal. The solutions are said to satisfy the equation. If only one unknown is involved the solutions are also called roots. However, there are times that when the solution/root is checked, it results to a number divided by zero, thus, the equation has no solution.
The given problem on the left taken from my quiz 2 is a conditional equation because it is only true for r is equal to -5/4. Note that other values of r does not satisfy the equation, hence, -5/4 is the root of the equation. Though it was not shown in the figure, when you substitute the root to the r's of the equation, you will get -21/4 = -21/4, thus, it satisfies the equation.
Additional information: In case we arrive to a false solution, for example x = x + 1, then it is called contradiction equation. In short, the equation has no solution since a contradiction is never true. Meanwhile, when we end up with a true statement and no variable such as 0 = 0, we can stop and say our equation has "an infinite number of solutions". It is therefore called an identity equation because an identity is an equation which is always true, no matter what values are substituted.
Solutions of conditional equations are values of the unknowns which make both expressions equal. The solutions are said to satisfy the equation. If only one unknown is involved the solutions are also called roots. However, there are times that when the solution/root is checked, it results to a number divided by zero, thus, the equation has no solution.
The given problem on the left taken from my quiz 2 is a conditional equation because it is only true for r is equal to -5/4. Note that other values of r does not satisfy the equation, hence, -5/4 is the root of the equation. Though it was not shown in the figure, when you substitute the root to the r's of the equation, you will get -21/4 = -21/4, thus, it satisfies the equation.
Additional information: In case we arrive to a false solution, for example x = x + 1, then it is called contradiction equation. In short, the equation has no solution since a contradiction is never true. Meanwhile, when we end up with a true statement and no variable such as 0 = 0, we can stop and say our equation has "an infinite number of solutions". It is therefore called an identity equation because an identity is an equation which is always true, no matter what values are substituted.
SOLVING FOR ONE VARIABLE IN TERMS OF OTHERS
Solving equations in terms of other variables is very similar to solving any other equation. You must manipulate the equation to isolate the variable you are trying to solve for. You will not get a constant answer such as x = 3 or y = 1/2. Instead, you'll get one variable defined by another variable, such as x = 2y.
Let us take this example from my quiz #2: A farmer has a rectangular plot surrounded by 200 feet of fence. Find the length and width of the garden if its area is 2400 square feet.
Perimeter is defined as the total distance around a shape, hence, for a rectangular plot, the formula will be 2 lengths + 2 widths. Meanwhile, area is the amount of space taken up by a 2D shape and for a rectangle, the formula is length times width. Then, we can make two equations from that statements. As shown in my solution, solving for one variable in terms of others was applied when I equate L = 100 - W. Then, I substituted that value of length to the second equation which resulted to a quadratic equation. To obtain of W, I just factored out the equation and resulted to two values. These two values will be the length and width of the rectangular plot in which the higher value will be the former and the lower value will be the latter one. Therefore, the dimensions of the rectangular plot is 60 feet by 40 feet.
Solving equations in terms of other variables is very similar to solving any other equation. You must manipulate the equation to isolate the variable you are trying to solve for. You will not get a constant answer such as x = 3 or y = 1/2. Instead, you'll get one variable defined by another variable, such as x = 2y.
Let us take this example from my quiz #2: A farmer has a rectangular plot surrounded by 200 feet of fence. Find the length and width of the garden if its area is 2400 square feet.
Perimeter is defined as the total distance around a shape, hence, for a rectangular plot, the formula will be 2 lengths + 2 widths. Meanwhile, area is the amount of space taken up by a 2D shape and for a rectangle, the formula is length times width. Then, we can make two equations from that statements. As shown in my solution, solving for one variable in terms of others was applied when I equate L = 100 - W. Then, I substituted that value of length to the second equation which resulted to a quadratic equation. To obtain of W, I just factored out the equation and resulted to two values. These two values will be the length and width of the rectangular plot in which the higher value will be the former and the lower value will be the latter one. Therefore, the dimensions of the rectangular plot is 60 feet by 40 feet.
MODELING WITH LINEAR EQUATIONS
Mathematical modeling is essential in solving problems in real-life situation. Mathematical Modelling is the process of translating phrases or sentences into algebraic expression or equations. Modeling with linear equations typically has three simple steps. First is verbal description of the problem, followed by verbal model and then creating algebraic equation from the model created.
For example: A plumber and his assistant work together to replace the pipes in an old house. The plumber charges $45 an hour for his own labor and $25 an hour for his assistant’s labor. The plumber works twice as long as his assistant on this job, and the labor charge on the final bill is $4025. How long did the plumber and his assistant work on this job? (taken from quiz 2)
Now, the problem itself will be the verbal description and from which we will make the verbal model. Looking at my solution at the left, it was just simply equating the plumber to 45, assistant to 25, and final bill to 4025. Additionally, it was stated that the plumber works twice as long as his assistant, therefore, I let plumber be 2x and the assistant to x itself. Then, from the model and representation, I derived the algebraic expression which is 90x + 25x = 4025. Finally, I combined like terms and perform some operations to obtain the value of x which is 35. The assistant therefore worked for 35 hours while the plumber worked twice as the assistant or 2 time 35 which is 70 hours.
Mathematical modeling is essential in solving problems in real-life situation. Mathematical Modelling is the process of translating phrases or sentences into algebraic expression or equations. Modeling with linear equations typically has three simple steps. First is verbal description of the problem, followed by verbal model and then creating algebraic equation from the model created.
- Verbal Description
- Verbal Model
- Algebraic Expression
For example: A plumber and his assistant work together to replace the pipes in an old house. The plumber charges $45 an hour for his own labor and $25 an hour for his assistant’s labor. The plumber works twice as long as his assistant on this job, and the labor charge on the final bill is $4025. How long did the plumber and his assistant work on this job? (taken from quiz 2)
Now, the problem itself will be the verbal description and from which we will make the verbal model. Looking at my solution at the left, it was just simply equating the plumber to 45, assistant to 25, and final bill to 4025. Additionally, it was stated that the plumber works twice as long as his assistant, therefore, I let plumber be 2x and the assistant to x itself. Then, from the model and representation, I derived the algebraic expression which is 90x + 25x = 4025. Finally, I combined like terms and perform some operations to obtain the value of x which is 35. The assistant therefore worked for 35 hours while the plumber worked twice as the assistant or 2 time 35 which is 70 hours.
QUADRATIC EQUATIONS
Quadratic equations or also known as quadratics can be defined as a polynomial equation of a second degree, which implies that it comprises a minimum of one term that is squared. The general form of the quadratic equation is: Ax^2 + Bx + C = 0, where A, B and C are real numbers with A ≠ 0. There are three major ways to solve a quadratic equation:1. By Factoring2. By Completing the Square3. The Quadratic Formula
Quadratic equations or also known as quadratics can be defined as a polynomial equation of a second degree, which implies that it comprises a minimum of one term that is squared. The general form of the quadratic equation is: Ax^2 + Bx + C = 0, where A, B and C are real numbers with A ≠ 0. There are three major ways to solve a quadratic equation:1. By Factoring2. By Completing the Square3. The Quadratic Formula
SOLVING QUADRATIC EQUATIONS BY FACTORING
Factoring quadratic equations is the first method of solving quadratic equations. It applies the zero-product property which states that AB = 0 if and only if A = 0 and B = 0. This means that if we can factor the left-hand side of a quadratic (or other) equation, then we can solve it by setting each factor equal to 0 in turn. This method works only when the right-hand side of the equation is 0 (Larson, 2011).
Let's take an example from my problem set #2. 6x2 - 48 = -12x is a quadratic equation since its highest degree is two. To solve it, first I noticed that A, B and C are all divisible by 6, so I divided the equation by 6 to reduce the numbers. Then, by observing the equation, we can easily say that it can be factored out. Factoring this equation requires a little knowledge of what has been discussed in the previous module in which x2 + bx + c = (x + r)(x + s) where (r)(s) = c and r + s = b. With that, we can say that c = -8 and b = 2. The factors of -8 whose sum is equal to 2 are 4 and -2. Therefore the factors are (x + 4)(x - 2). Finally, we will now apply the zero-product property in which we equate each factor to zero to get the values of x which are -4 and 2.
Factoring quadratic equations is the first method of solving quadratic equations. It applies the zero-product property which states that AB = 0 if and only if A = 0 and B = 0. This means that if we can factor the left-hand side of a quadratic (or other) equation, then we can solve it by setting each factor equal to 0 in turn. This method works only when the right-hand side of the equation is 0 (Larson, 2011).
Let's take an example from my problem set #2. 6x2 - 48 = -12x is a quadratic equation since its highest degree is two. To solve it, first I noticed that A, B and C are all divisible by 6, so I divided the equation by 6 to reduce the numbers. Then, by observing the equation, we can easily say that it can be factored out. Factoring this equation requires a little knowledge of what has been discussed in the previous module in which x2 + bx + c = (x + r)(x + s) where (r)(s) = c and r + s = b. With that, we can say that c = -8 and b = 2. The factors of -8 whose sum is equal to 2 are 4 and -2. Therefore the factors are (x + 4)(x - 2). Finally, we will now apply the zero-product property in which we equate each factor to zero to get the values of x which are -4 and 2.
SOLVING QUADRATIC EQUATION BY COMPLETING THE SQUARE
Completing the square is creating a perfect square trinomial from the quadratic equation, and then solving that trinomial by taking its square root. To make (x2 + bx) a perfect square, add (b/2)2, the square of half the coefficient of x. This gives the perfect square: x2 + bx + (b/2)2 = [x + (b/2)]2.
I got an example x2 - 18x = -72. The coefficient of x is -18 and its half is -9 and the square of -9 is 81. Thus, 81 have been added to both sides of the equation. Then, x2 - 18x + 81 is a perfect square trinomial since the first and last terms are perfect squares (x2 and 81) and the middle term (-18x) is equal to twice the product of the square roots of the first and the last terms. Given that, we can easily factor it out by getting the square root of the first and last terms which are x and +-9. We must choose -9 as the factor because its sum will be -18 which is the coefficient of x. Subsequently, to eliminate the squared (2) of the perfect square, we must get the square roots of both sides. Lastly, we will now simply combine the constants to get the values of x. Since the square root of any number is always two (positive and negative), then values of x are also two. In the example, x = 12 and x = 6.
Completing the square is creating a perfect square trinomial from the quadratic equation, and then solving that trinomial by taking its square root. To make (x2 + bx) a perfect square, add (b/2)2, the square of half the coefficient of x. This gives the perfect square: x2 + bx + (b/2)2 = [x + (b/2)]2.
I got an example x2 - 18x = -72. The coefficient of x is -18 and its half is -9 and the square of -9 is 81. Thus, 81 have been added to both sides of the equation. Then, x2 - 18x + 81 is a perfect square trinomial since the first and last terms are perfect squares (x2 and 81) and the middle term (-18x) is equal to twice the product of the square roots of the first and the last terms. Given that, we can easily factor it out by getting the square root of the first and last terms which are x and +-9. We must choose -9 as the factor because its sum will be -18 which is the coefficient of x. Subsequently, to eliminate the squared (2) of the perfect square, we must get the square roots of both sides. Lastly, we will now simply combine the constants to get the values of x. Since the square root of any number is always two (positive and negative), then values of x are also two. In the example, x = 12 and x = 6.
THE QUADRATIC FORMULA
In times that the quadratic equation cannot be solved by the first two ways, we will now use the quadratic formula shown at the left. Note that the general form of quadratic equation is ax2 + bx + c = 0.
In times that the quadratic equation cannot be solved by the first two ways, we will now use the quadratic formula shown at the left. Note that the general form of quadratic equation is ax2 + bx + c = 0.
Before we proceed to the example, I will give you a brief history of quadratic formula. The 9th-century Persian mathematician Muḥammad ibn Mūsā al-Khwārizmī solved quadratic equations algebraically (Irving, 2013). The quadratic formula covering all cases was first obtained by Simon Stevin in 1594 (Struik & Stevin, 1958). In 1637 René Descartes published La Géométrie containing special cases of the quadratic formula in the form we know today.
In the example x2 - 10x + 26 = 8, the first thing I did is combine the two constants to arrive at x2 - 10x + 18 = 0. The equation can neither be solved through factoring nor completing the square, hence, I used the quadratic formula. It is important to identify the values of a, b and c to easily substitute them to the given formula. As such, a = 1, b = -10, and c = 18. Then, after entering these values, we should now perform the need operations. One must not be careless of the signs of the quantities to avoid wrong answers.
Further, familiarization of radicals which was discussed in the previous module, is a must in using the quadratic formula to simplify accurately the values of x. For instance, the √28 was simplified in the example by reducing the radicand through the perfect square which is 4. Note that there will be two solutions since the square root of all positive real numbers is two integers. In the example x = 5 + √7 and x = 5 - √7.
In the example x2 - 10x + 26 = 8, the first thing I did is combine the two constants to arrive at x2 - 10x + 18 = 0. The equation can neither be solved through factoring nor completing the square, hence, I used the quadratic formula. It is important to identify the values of a, b and c to easily substitute them to the given formula. As such, a = 1, b = -10, and c = 18. Then, after entering these values, we should now perform the need operations. One must not be careless of the signs of the quantities to avoid wrong answers.
Further, familiarization of radicals which was discussed in the previous module, is a must in using the quadratic formula to simplify accurately the values of x. For instance, the √28 was simplified in the example by reducing the radicand through the perfect square which is 4. Note that there will be two solutions since the square root of all positive real numbers is two integers. In the example x = 5 + √7 and x = 5 - √7.
Jess Marck Salazar-PS 2-Module2.pdfPROBLEM SET #2 (41/45)
Jess Marck Salazar-Activity 2-BSCE1B.pdfACTIVITY #2 (95/100)
Jess Marck Salazar-Quiz 2 Solutions-BSCE1B.pdfQUIZ #2 SOLUTIONS (45/45)
REFLECTION
Just like module 1, module 2 also serves a review since the topics were already discussed in my high school years. Nevertheless, I am happy that I was able to encounter them again since I already forgot some of the details. I was able to renew my learnings especially in the topic, modeling with linear equations. Though it was easy and simple topic, the three simple steps namely, verbal description, verbal model, and algebraic expression gave me the idea of how to solve a linear equation in a step-by-step manner. I relearned how to solve linear equations through conditional equations, one variable in terms of others, and modeling. Meanwhile, I also relearned how to solve quadratic equations through factoring, completing the square, and the quadratic formula Overall, I was glad that I was able to understand all the topics highlighting my explanations in every example given above. It marks the efficacy of the learning material as well as the teaching method of our teacher.
Just like module 1, module 2 also serves a review since the topics were already discussed in my high school years. Nevertheless, I am happy that I was able to encounter them again since I already forgot some of the details. I was able to renew my learnings especially in the topic, modeling with linear equations. Though it was easy and simple topic, the three simple steps namely, verbal description, verbal model, and algebraic expression gave me the idea of how to solve a linear equation in a step-by-step manner. I relearned how to solve linear equations through conditional equations, one variable in terms of others, and modeling. Meanwhile, I also relearned how to solve quadratic equations through factoring, completing the square, and the quadratic formula Overall, I was glad that I was able to understand all the topics highlighting my explanations in every example given above. It marks the efficacy of the learning material as well as the teaching method of our teacher.
REFERENCES:
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