Solving Algebraic Equations in one variable

Solving Algebraic Equations in one variable Written by Dave Didur August 19, 014 -- Webster s defines algebra as the branch of mathematics that deals with general statements of relations, utilizing letters and other symbols to represent specific sets of numbers, values, vectors, etc. in the description of such relations. Most people would say that it had something to do with finding the value of x and that would be right! This article will focus on the topic of solving equations, one of the fundamental skills in the field of algebra. What is an equation? It is a mathematical statement that expresses an equality. 5+3=8 is an equation that expresses a truth. The expression on the left side of the equal sign is equivalent in value to the expression on the right side of the equal sign. An arithmetic expression may be a single value (like 8) or many values connected by arithmetic operators (such as addition, subtraction, multiplication, division, square root). Here s a more complicated equation, which contains arithmetic expressions on each side: 3 X 4 (+1) = 7 8 There are four arithmetic operators in this equation: multiplication, subtraction, addition and division. Brackets are also included. An equation may also be formed by including algebraic expressions. An algebraic expression will include a variable (this is just a letter, such as x, which represents a number). An algebraic expression may be a single variable (like x) or a longer expression such as x 5. When a number immediately precedes a variable (like the in front of the x in this example), it is a multiplier. x means times x. Mathematicians give such a multiplier a fancy name a coefficient. We say that the coefficient of x is. An algebraic equation is an equation that contains one or more variables. A simple example is x + 3 = 8. When we determine the numerical value represented by x that makes the equation true, we are said to have solved the equation. In this case, the solution is 5. We write x = 5. Youngsters in early grades of elementary school are presented with the concept of solving equations without specific reference to variables or algebra. Instead, they are asked what number goes into the box to make a true statement. For example: 4 + = 9. What is? In Middle School, the same question would be posed in an algebraic fashion: Solve for x: 4 + x = 9 This instruction is often shortened to just: Solve 4 + x = 9

In elementary school, students are presented with algebraic equations containing one variable. Examples: 3x 5 = 16 The solution is x = 7 5y - 1 = 3y + 9 The solution is y = 5 (m 4) = 10 m The solution is m = 6 I ll deal with how equations are solved in the second half of this article. For now, I want to discuss how we determine whether a given value is actually a solution for an equation. To check a solution means to determine whether a value makes the equation into a true statement. Checking is properly done by substituting the numerical value in place of the variable and evaluating each side of the equation separately and comparing the results afterwards. Example 1: Example : Check if m = 6 is the solution to (m 4) = 10 m. Left Side (m 4) = (6 4) = () = 4 Right Side 10 m = 10 6 = 4 L.S. = R.S. (short for Left Side = Right Side ) m = 6 (is the math symbol meaning therefore ) Check if y = 4 is the solution to 5y 1 = 3y + 9 Left Side 5y - 1 = 5(4) - 1 = 0-1 = 19 Right Side 3y + 9 = 3(4) + 9 = 1 + 9 = 1 Recall that 5y and 3y mean multiplication L.S. R.S. (short for Left Side does not equal Right Side) y 4 (short for Therefore y does not equal 4 ) NOTE: In this article, we are looking at equations that are part of the classification called polynomial equations. In their simplest forms, they contain terms, which are constants or variables and the variable terms consist of a numerical coefficient multiplying a power such as x 1 (which is the same as x), x, x 3, x 4, etc. Examples: 5x 9 = 0 3x + 4x + 7 = 0 The largest exponent of x is 1. This is a first-degree polynomial equation that is also called a linear equation. The largest exponent of x is. This is a second degree polynomial equation that is also called a quadratic equation. -6 x 3 + x - 5x - = 0 The largest exponent of x is 3. This is a third degree polynomial equation.

Solving Linear Equations in One Variable An analogy that is often used to describe an equation is a pan balance. Whatever is on the left side of an equal sign can be considered to be weights that are on the left pan. Whatever is on the right side of the equal sign can be considered to be weights on the right pan. When they are equal, the pans balance perfectly. Let s suppose that we had weights of 1 gram, grams, 3 grams and 4 grams. We would get a balance if we put the weights of gm and 3 gm on one pan and the weights of 1 gm and 4 gm in the other, as shown below. Since the weights combine (or add up) to 5 grams on each side of the pan balance, we could represent this balance by the equation + 3 = 1 + 4. The pan balance will remain level as long as we make the SAME changes to both pans. For example, we could add 8 grams to each side of the balance, giving us a total of 13 grams in each pan. We d still have a balance. We could triple the size of the weights in each pan giving us 15 grams per side and we d still have a balance. We could remove grams from each side and we d still have a balance with 3 grams per side. Taking 1 gm from each side still leaves a balance (with 4 grams per side) Doubling the weights on each side still leaves a balance with 10 gm per side When solving algebraic equations, the same principle is used: make the same changes to both sides of the equation in order to maintain equality. Let s see how this is done by considering an example.

Example 1: Solve 3x 1 = x + 17. Add 1 to each side: 3x 1 + 1 = x + 17 + 1 Simplifying, we get: 3x = x + 18 Now subtract x from each side: 3x x = x + 18 x x is the same as 1x, so simplifying 3x x = 3x 1x = x. We now have x = 18 Finally, divide both sides by : x = 18 x = 9 We tend to write the division step by placing the divisor in the denominator (the bottom) of each side, as follows: x 18 x 18 x 9 The objective in our manipulations is to isolate the variable on one side of the equal sign and the fixed values on the other. This requires appropriate planning in the order of steps. Example : Solve 4(x 7) 3x = 5(6 x) + In general, we get rid of any brackets by multiplying, as required. 4(x 7) means that the 4 must multiply BOTH the x and the -7 which are the terms inside the brackets. 5(6 x) means that the 5 multiples BOTH the 6 and x. 8x 8 3x = 30 5x + Then we simplify whatever we can on each side. We refer to this process as collecting like terms. Expressions that include the variable x are called x-terms and the numerical values are called constant terms (or just constants). If there is more than one x-term on any side of the equation, then we collect them into one term by adding or subtracting, as required. The same is true for the constant terms. On the left, we have two x-terms 8x and -3x -- so we can simplify 8x 3x = 5x (the sign that applies to the 3x is the one that is IMMEDIATELY in front of the 3x; we say that the coefficient of x is -3, not just 3). The sign that is in front of the 8 applies to the 8 (i.e. it is -8). On the right side, there are two constant terms -- 30 and + (remember, the - sign belongs to the 5x). Simplifying, 30 + = 3. 8x 3x 8 = 30 + 5x 5x 8 = 3 5x We may eliminate the x-term from either side. Usually we eliminate it from the right side, but that is not a rule. It works either way. I ll demonstrate both methods.

Eliminating variables from R.S. Eliminating variables from L.S. 5x 8 = 3 5x On the right, we have -5x. To eliminate it, we will put a +5x at the end of the expression. We must do the same thing to the left side: 5x 8 + 5x = 3 5x + 5x Simplify the x-terms on each side: 10x 8 = 3 Now eliminate the constant term from the left side: 10x 8 + 8 = 3 + 8 10x = 60 Divide both sides by 10 (the coefficient of the x-term): 10x 60 10 10 x = 6 5x 8 = 3 5x On the left, we have 5x (which is positive the same as +5x). To eliminate it, we will put a -5x at the end of the expression (on each side). 5x 8 5x = 3-5x - 5x Simplify the x-terms on each side: -8 = 3 10x Now eliminate the constant term from the right side: -8 3 = 3 10x -3-60 = -10x Divide both sides by the coefficient of x (which is -10). A negative divided by a negative yields a positive answer. 60 10x 10 10 6 = x It doesn t matter whether the variable ends up on the left side or the right side: the answer will be the same. The solution for x is 6. Check if x = 6 is a solution for 4(x 7) 3x = 5(6 x) + L.S. = 4(x 7) 3x R.S. = 5(6 x) + = 4((6) 7) 3(6) = 5(6 (6)) + = 4(1 7) 18 = 5(0) + = 4(5) 18 = 0 + = 0 18 = = L.S. = R. S. x = 6 Solving Quadratic Equations in One Variable In secondary school, students will be introduced to quadratic (second degree) equations. One of the simplest quadratic equations is: x 9 This equation has TWO solutions! x can be either 3 or -3. () 3 ()() 3 3 9 and ( 3) ( 3)( 3) 9 A negative number multiplied by a negative number yields a positive answer. Sometimes the solution is written this way:

x 9 x 9 x 3 It is not my intention to go too deeply into the methods that are taught for solving quadratic equations. I will mention some of them briefly. More detailed explanations for the various methods can be found in the Resources section that follows the article. One method is factoring. This is a complicated topic on its own and too lengthy to go into great detail here. However, I ll include a few notes. Factoring is the opposite of expanding. The word 'expand' means to make larger. In algebra we use this term whenever there is an expression that involves a multiplication that involves brackets. Example 1: Expand 3(4x 3y + 1) We say that the 3 is 3(4x 3y + 1) = 3(4x) 3(3y) + 3(1) distributed into the bracket = 1x 9y + 3 (it multiples each of the terms that are inside the brackets Example : Expand and simplify x(x 5) 7(x x + 3) (8 x) x(x 5) 7(x x + 3) (8 x) The term in front of = x(x) x(5) -7(x ) -7(-x) -7(+3) -1(8) -1(-x) each bracket -- along = x 10x - 7x + 14x 1 8 + x with its sign is the = x - 7x 10x + 14x + x 1 8 multiplier for each = - 5x + 5x 9 term inside NOTE: There is just a minus sign in front of the last bracket. This is equivalent To having a multiplier of -1 in front of the bracket. We collect like terms (the last two lines) by adding or subtracting the coefficients of terms, which have identical variables (including identical exponents). Remember that the sign in front of each term belongs to that coefficient. This final step of collecting is the simplification step. Example 3: Expand and simplify (x 3)(4x + 1) (x 3)(4x + 1) = x(4x + 1) -3(4x + 1) Each term in the 1 st = x(4x) + x(1) 3(4x) -3(1) bracket multiples = 8x + x 1x 3 every term in the nd = 8x 10x 3 bracket Factors are multipliers. The number 6 has factors of and 3 because 6 = (3)(). Of course, it also has factors of 1 and 6 because 6 = (6)(1). When we are asked to factor, we are required to write the original value or expression as a sequence of multipliers.

The factors of 1x 9y + 3 are (3)(4x 3y + 1). The final answer must be a multiplication. In this example, we recognized that each of the three terms had the factor of 3: 1x = (3)(4x) -9y = (3)(-3y) 3 =(3)(1) Whenever the same factor occurs in all of the terms, it is called a common factor. In example #3 on the previous page, we saw that (x 3)(4x + 1) = 8x 10x 3 Since the original question was all multipliers, then the expression 8x 10x 3 can be factored. Going backwards in cases like this is challenging but there are techniques for doing it. The Resources section has a few links that will take you to web sites that will try to explain the process of factoring quadratics. The factors of 8x 10x 3 are (x 3)(4x + 1). Other factored quadratic expressions: x 5x 4 = (x 8)(x + 3) x 13x +36 = (x - 9)(x - 4) x x 6 = (x + 3)(x - ) x 16 = (x 4)(x + 4) Not all quadratic expressions can be factored. Let s return to the equation that we had to solve at the beginning of this section (i.e. solve x 9 ), we ll see how that equation can be solved by factoring. x 9 x 999 x 90 x3x30 This last step is a factoring step. Once the factors are obtained AND the right side of the equation is a ZERO the solution is arrived at by using this logical deduction: When two expressions are multiplied together and the answer is zero then either one or both of the multipliers must be equal to zero. Thus, S.S. = { -3, 3 }

Example : By factoring, solve: x(x 5) -6 = 8 x As with linear equations that include brackets, we expand first and then simplify the equation. x( x 5) 6 8 x x 5x68x x 5x6x 8xx x 3x68 This equation has evolved into a second degree (quadratic) equation. With these types of equations, we eliminate ALL the terms from the right side, leaving just ZERO on the right. This is done because the factors must be multipliers that result in a zero answer (in order to use the logic explained above). x 3x6888 x 3x140 The factoring step comes next: x 7x 0 Either x 7 0 or x 0 x 7 7 0 7 x 0 x 7 x x 7 x 3 1 S.S. = 3 1, Some quadratic equations have only one solution. Example 3: Solve x -8x + 16 = 0 By factoring: (x 4)(x 4) = 0 Thus, either x 4 =0 or x- 4 = 0 x = 4 or x = 4 S.S.= { 4 } Some quadratic equations do not have any REAL solutions! We ll see an example shortly. The other principle method of solving quadratic equations is by using the quadratic formula. The formula can be derived from a factoring approach (that is why the factoring method is taught first). It is useful because not every quadratic equation can be solved by factoring! The quadratic formula, on the other hand, can be used in all cases.

Solving Quadratic Equations by Using the Quadratic Formula: Steps: 1. Simplify the original equation into the form Ax + By + C = 0. Substitute the values A, B and C into the quadratic formula: B B AC x 4 A 3. Evaluate the formula in order to get the answers. There are three possibilities: (1) Two different real number solutions () Two identical real number solutions (i.e. one solution) (3) Two different complex number solutions (not real) Let s use the formula on three different examples in order to demonstrate these possibilities. Example 1: Solve x 3x 14 0 A =, B = -3, C = -14 Example : Solve x 8x 16 0 A = 1, B = -8, C = 16 Example 3: Solve 3x x 1 0 A = 3, B = -, C = 1 B B AC x 4 A 3 3 4 14 x 3 x 9 11 4 3 x 1 4 3 x 11 4 3 x 11 3 and 11 4 4 14 8 x and 4 4 7 x and S.S.= 3 1, (Two real, distinct solutions) B B AC x 4 A x 8 8 4116 1 8 x 64 64 8 x 0 8 x 0 8 x x 4 S.S.={ 4 } (One real solution) B B AC x 4 A x x 4 1 6 x 8 6 x 4 6 x 1 x 1 x 1 x 1 1 i x 431 3

It is impossible to take the square root of a negative number, so there is no real solution for Example #3. 1 is defined as being the imaginary number i. The final answer, which is a combination of a real number and an imaginary number, is called a complex number. Complex numbers have the form A ib where A, B R, i 1. The answers for Example #3 can be written in this form: 1 i x 1 i ( where A 1 and B ) NOTE: A first degree polynomial equation has one real solution. The maximum number of real solutions that a second degree polynomial equation can have is two. The maximum number of real solutions that a third degree polynomial equation can have is three. In general, an n th degree polynomial equation can have up to n real solutions. Conclusion Solving equations is one of the most fundamental skills of algebra. It requires a solid understanding of many previous skills that are taught in the elementary and secondary schools including evaluating arithmetic expressions, simplifying algebraic expressions, factoring, and substitution. Problem solving in the sciences and engineering requires analytical skills that enable the solver to translate the situation from words into equations. Then the appropriate mathematical skills are needed to solve the equations or systems that result. This article gave a quick overview of some of the basic equation-solving situations, the methods used to solve the equations, how answers are checked, and a glimpse at what the different kinds of results mean (i.e., that a complex number answer means that there is no real solution to the problem). This topic continues to expand and develop as students learn more and more mathematics in college and university: solving higher degree polynomial equations, solving trigonometric equations, solving exponential equations, solving logarithmic equations, solving differential equations, and so on! The universe is a complicated place and intelligent skills are required by those who wish to work in it and study it. Sound mathematical skills especially in solving equations are an absolute necessity. NEXT MONTH: Solving Systems of Equations in Two Variables

This article is the sixth of a series of mathematics articles published by CHASA. Marvellous Mathematics Introduction Euclidian Geometry Article # 1 Non-Euclidean Geometry Article # Rational Numbers Fractions, Decimals and Calculators Article #3 Continued Fractions Article #4 Introduction to Fractals: The Geometry of Nature Article #5 CHASA has received many communications from concerned parents about the difficulties their children are having with the math curriculum in their schools as well as their own frustration in trying to understand the concepts - so that they can help their children. The intent of these articles is to not only help explain specific areas of history, concepts and topics in mathematics, but to also show the beauty and majesty of the subject. Dave Didur is a retired secondary school mathematics teacher with a B. Sc. degree from the University of Toronto majoring in Mathematics and Physics. He was Head of Mathematics for over twenty years, as well as the Computer Co-ordinator and consultant for the Board of Education for the City of Hamilton. He served with the Ontario Ministry of Education for three years as an Education Officer. Resources Solving Equations With Pictures and With Algebra -- Mathematics for Elementary Teachers, Sybilla Beckman (003); grades 3-5 How to Solve Linear Equations in One Variable Using Algebra from TeachersChoice.com Algebra Worksheets Math- Drills.com worksheets primarily aimed at Middle School students, covering a wide range of algebraic topics including solving linear equations How to Solve Quadratic Equations Three methods explained on the WikiHow website Factoring Quadratics The MathIsFun website presents some of the fundamentals of factoring before applying the skills to solving quadratic equations Solving Factorable Quadratic Equations 8 or 9 examples, including word problems and the use of a graphing calculator Solving Quadratic Equations by Factoring Purplemath web site Solving a Quadratic Equation by Factoring (Video) from the Khan Academy The Quadratic Formula Explained -- Purplemath web site