Download presentation

Presentation is loading. Please wait.

1
**Solving Equations How to Solve Them**

Handling linear equations is a basic skill that all students of algebra should have. This module addresses what is required. Equations 7/8/2013

2
**What Are Equations ? Equation Statement that two expressions are equal**

Solving Equations Equation Statement that two expressions are equal What do we DO with expressions ? Simplify them Evaluate them What do we DO with equations ? Solve them What Are Equations ? Let us consider what we mean by an equation. An equation is a binary relation setting two algebraic expressions equal to each other. Generally speaking the problem is to solve the equation. Generally speaking, if the expressions are formed from a real variable, then solving the equation simply means finding all values of the variable that make the equation a true statement. When all variable values are known, the equation is either TRUE or FALSE – the only values an equation can have. We note that the process of solving an equation may involve evaluating one or more algebraic expressions. In its simplest form, an expression represents a number, whose value is known only when all variables are given values. 7/8/2013 Equations Equations 7/8/2013

3
**What Are Equations ? Three equation categories**

Solving Equations Three equation categories Identity: Logically True Example: 2(x + 1) = 2x + 2 True for ALL values of x Logically False Example: x + 3 = x NOT true for ANY x ( would imply 3 = 0, a contradiction ! ) What Are Equations ? There are three general categories of equations: Identity equation: This is a logically true statement. That is, there is no value of any variable that will make the statement false. For example: 2(x + 1) = 2x TRUE for all x Logically False equation: This is equation is false for any value of the variables. We think of this as a contradiction. For example: x + 3 = x FALSE for all x IF this were true, it would be equivalent to 3 = 0, which contradicts something we already know, namely that 3 and 0 are not the same number. These equations are not actually solvable in the normal sense of solvability. 7/8/2013 Equations Equations 7/8/2013

4
**What Are Equations ? The third equation category Conditional Equations**

Solving Equations The third equation category Conditional Equations Example: 3x + 1 = 2x – 7 True for SOME values of x (x = – 6 in this case) False for other values of x What Are Equations ? There are three general categories of equations: Conditional equation: This equation is true only if some condition is met. This usually means only certain values of x (the solutions) are allowed. Thus the equation is a true statement for some values and not for others. For example: 3x + 1 = 2x – TRUE only if x = – FALSE if x ≠ – 8 When we solve equations we are dealing with conditional equations to find the value, or values, of the variables that make the equation a true statement. The others are not actually solvable. 7/8/2013 Equations Equations 7/8/2013

5
**Linear Equations in 1 Variable**

Solving Equations Standard Form: for some constants a, b with a ≠ 0 Solutions Solution is a value of x that makes the equation TRUE A solution is a number ax + b = 0 WHY a ≠ 0 ? Linear Equations in One Variable Any equation that can be written in form ax + b = 0 is a linear equation in one variable. This means that many equations, not initially in this form, are also linear equations in one variable. Why is the equation called linear? The graph of the equation is always a line. Note also that the highest power of the variable is 1. Later we will talk about polynomials and their classification by powers, or degrees. The linearity of the equation in standard form (sometimes referred to as the homogeneous form because of the 0 on the right side) guarantees a unique solution, r. There are several ways to see this. First, the expression ax + b can be thought of as the functional value of a linear function f(x). So, the equation can be written as f(x) = 0. If r is a solution, then the ordered pair (r, 0) is in the function, bearing in mind that a function is a set of ordered pairs, no two of which have the same first component. This means that the function cannot contain both (r, 0) and (r2, 0) for some r2 ≠ r. Thus r is a unique for the equation f(x) = 0. Second, we can study the graph of the equation (same as the graph of f(x) ) to observe that the graph (a line) crosses the x-axis at only one point, namely (r, 0). A third view derives from assuming there are two solutions, say r1 and r2, and then showing that r1 = r2 by putting each of them into the linear equation. 7/8/2013 Equations Equations 7/8/2013

6
**Linear Equations in 1 Variable**

Solving Equations Equivalent Equations Equations are equivalent if and only if they have the same solutions Solving an equation transforms it into an equivalent equation of form: The number r is the solution x = r Equivalent Equations Two equations are equivalent if and only if they have exactly the same solutions. The process of solving an equation typically involves transforming it into an equivalent that is easier to solve, in fact, typically one in which a solution is isolated on one side of the equation. An equation in one variable will have at most one solution. An easy way to show this is to assume there are at least two solutions, say r1 and r2. Then show that in fact r1 = r2 , that is, there is only one number that satisfies the equation. – there is only one WHY ? 7/8/2013 Equations Equations 7/8/2013

7
**Techniques Solving Equations**

Cancellation Rule for Addition a + c = b + c if and only if a = b for any numbers a, b and c Cancellation Rule for Multiplication ac = bc if and only if a = b for any numbers a, b and c with c ≠ 0 Techniques for Solving Equations Removing a term or factor in an equation is a standard type of approach to solving equations. The basic caveat in doing this is: Whatever you do to one side of the equation you must also do to the other side Note that we when we say “subtract” and “divide” we are talking about the arithmetic “shortcuts” defined by addition of negatives and multiplication by reciprocals. Thus 7 – 3 gives the same result as 7 + (-3), where -3 is just the negative of 3. Even your calculator does this – try it ! Similarly, (½)(2) is the same as 2/2. Try this on your calculator. 7/8/2013 Equations Equations 7/8/2013

8
**Techniques for Solving Examples**

Solving Equations Examples Example 1. 2x + 3 = 7 Example 2. = 4 + 3 Cancellation Rule for Addition then 2x = 4 Techniques for Solving Equations Removing a term or factor in an equation is a standard type of approach to solving equations. The basic caveat in doing this is: Whatever you do to one side of the equation you must also do to the other side In the first example we add –3 to (or “subtract 3” from) both sides of the equation. This isolates the term containing x (i.e., the 2x term). Then, in the second example, we multiply by ½ (or “divide” by 2) on both sides to isolate x. Thus we see that 2 is the only value of x that satisfies the original equation, and is thus the (only) solution of the original conditional equation. Note that the whole expression on the left side of the equation gets the same treatment as the whole expression on the right side. This is done to conform to the two cancellation rules listed above. Also note that we when we say “subtract” and “divide” we are talking about the arithmetic “shortcuts” defined by addition of negatives and multiplication by reciprocals. Thus 7 – 3 gives the same result as 7 + (-3), where -3 is just the negative of 3. Even your calculator does this – try it ! Similarly, (½)(2) is the same as 2/2. Try this on your calculator). If 2x = 4 • = 2 2 Cancellation Rule for Multiplication then x = 2 7/8/2013 Equations Equations 7/8/2013

9
**Same Rules in Reverse Addition Rule for any numbers a, b and c**

Solving Equations Addition Rule a = b if and only if a + c = b + c for any numbers a, b and c Multiplication Rule a = b if and only if ac = bc for any numbers a, b and c with c ≠ 0 Same Rules in Reverse We can in essence reverse the cancellation rules with the addition and multiplication rules (although using inverses they are the same rules). Given any two expression a and b, with a = b, we can add to each side of the equation any other expression without changing the truth of equation statement. Hence, as the illustration shows, a = b if and only if a + c = b + c for any expression c. Note that the simplest expression is just a number. Similarly, for any two expression a and b, with a = b, we can multiply both sides by any nonzero expression c without changing the truth value of the statement. Hence, as shown, a = b if and only if a c = b c for any nonzero c. Why must c be nonzero? If we start with a false statement, such as x + 3 = x and multiply both sides by 0, we produce a true statement; that is, multiplication by zero changed the truth value of the equation. Question: Why can we add 0 but not multiply by 0 ? 7/8/2013 Equations Equations 7/8/2013

10
**Same Rules in Reverse Example 1. If 2x – 3 = 7 then 2x – 3 + 3 = 7 + 3**

Solving Equations Example 1. If 2x – 3 = 7 Example 2. If 2x = 10 then 2x – = 7 + 3 Addition Rule so 2x = 10 then (2x) = (10) 2 1 Multiplication Rule Same Rules in Reverse We can in essence reverse the cancellation rules with the addition and multiplication rules (although using inverses they are the same rules). Given any two expression a and b, with a = b, we can add to each side of the equation any other expression without changing the truth of equation statement. Hence, as the illustration shows, a = b if and only if a + c = b + c for any expression c. Note that the simplest expression is just a number. Similarly, for any two expression a and b, with a = b, we can multiply both sides by any nonzero expression c without changing the truth value of the statement. Hence, as shown, a = b if and only if ac = bc for any nonzero c. Why must c be zero? If we start with a false statement, such as x + 3 = x and multiply both sides by 0, we produce a true statement; that is, we have changed the truth value of the equation. so x = 5 Question: Why can we add 0 but not multiply by 0 ? 7/8/2013 Equations Equations 7/8/2013

11
**{ } Solving Symbolically Solve: –3(2x – 1) = 2x**

Solving Equations Solve: –3(2x – 1) = 2x – 6x + 3 = 2x distributive property – 6x x = 2x + 6x addition rule 3 = 8x simplification (1/8)(3) = (1/8)(8x) multiplication rule 3/8 = x simplification Solution is Solving Equations Symbolically We apply the rules we have just learned to solving an equation. The goal is to isolate x on one side of the equation, with an arithmetic expression on the other. We start by isolating a term that contains x as a factor. Applying the distributive property to break apart the binomial 2x – 1 and then applying the addition rule and simplifying, we isolate the term 8x on one side and the arithmetic expression 3 on the other side of the equation. We can now apply the multiplication rule to find x = 3/8. The rational number 3/8 is thus the only solution of the equation. NOTE: The equation “x = 3/8” is not the solution but simply an equivalent equation to the initial equation, i.e. an equation with the same solution. The solution set is simply {3/8} , the set of all solutions. The solutions themselves are numbers. We can check the accuracy of our solution by substituting it back into the original equation to see that we do indeed have a true statement. { } 3 8 3 8 Solution Set: 3 8 x = Note: The solution is NOT WHY ? 7/8/2013 Equations Equations 7/8/2013

12
**( ) ( ) ( ) ( ) Solving Symbolically – 2 – Solve: = – 2 3x – 1 5 2 – x**

Solving Equations – 2 3x – 1 5 = 2 – x 3 Solve: Simplify by clearing fractions ( ) – 2 3x – 1 5 15 = ( ) 2 – x 3 15 ( ) 3x – 1 5 15 15 2 – ( ) = ( ) 3 15 2 – x ( ) Solving Equations Symbolically We try our hand at solving a slightly more complex equation that requires more simplification. We begin by applying the multiplication rule with a multiplier that has factors from all the denominators – in this case 15 which has factors of 5 and 3. Larger multipliers, such 30, will also work but the most efficient is the least common multiple of the denominators. This clears out the fractions leaving a simpler equation. Simplifying and gathering like terms we apply the addition rule to isolate a term with x, namely 14x. Applying the multiplication rule we isolate x and identify the solution from the equivalent equation x = 43/14. Thus the solution is 43/14 and solution set is {43/14}. Again, we can check the solution by substituting it back into the original equation to verify that it makes the equation a true statement. 3(3x – 1) – 30 = 5(2 – x) 7/8/2013 Equations Equations 7/8/2013

13
**{ } Solving Symbolically Solve: = – 2 3x – 1 5 2 – x 3**

Solving Equations – 2 3x – 1 5 = 2 – x 3 Solve: 3(3x – 1) – 30 = 5(2 – x) 9x – 33 = 10 – 5x 14x = 43 Equivalent Equation = 14 43 x Solving Equations Symbolically We try our hand at solving a slightly more complex equation that requires more simplification. We begin by applying the multiplication rule with a multiplier that has factors from all the denominators – in this case 15 which has factors of 5 and 3. Larger multipliers, such 30, will also work but the most efficient is the least common multiple of the denominators. This clears out the fractions leaving a simpler equation. Simplifying and gathering like terms we apply the addition rule to isolate a term with x, namely 14x. Applying the multiplication rule we isolate x and identify the solution from the equivalent equation x = 43/14. Thus the solution is 43/14 and solution set is {43/14}. Again, we can check the solution by substituting it back into the original equation to verify that it makes the equation a true statement. 14 43 { } 14 43 Solution : Solution set : 7/8/2013 Equations Equations 7/8/2013

14
**Solving Equations Graphically**

Solve: 3x + 1 = –2x + 11 Consider this as the equality of two functions y1 and y2 with and Lines intersect where (x, y1) = (x, y2) y x y1 = 3x + 1 y1 y2 y1 = 3x + 1 x y2 = –2x + 11 y2 = –2x + 11 Solving Equations Graphically To “see” a solution of given equation assume that each side is an expression in x satisfying an expression of form y = f(x) or y = g(x). So, we formally define lines y1 = 3x and y2 = -2x + 11 If there is a solution it will be the point where y1 = y2 , that is where 3x + 1 = -2x + 11 The point of intersection (x, y1) = (x, y2) = (2,7) identifies the solution, which is 2. 7/8/2013 Equations Equations 7/8/2013

15
**Solving Equations Graphically**

Solve: 3x + 1 = –2x + 11 Lines intersect where (x, y1) = (x, y2) In this case (x, y1) = (x, y2) = (2, 7) y x y1 = 3x + 1 y2 = –2x + 11 y1 y2 So x = 2 Solving Equations Graphically To “see” a solution of given equation assume that each side is an expression in x satisfying an expression of form y = f(x) or y = g(x). So, we formally define lines y1 = 3x and y2 = -2x + 11 If there is a solution it will be the point where y1 = y2 , that is where 3x + 1 = -2x + 11 The point of intersection (x, y1) = (x, y2) = (2,7) identifies the solution, which is 2. Now let us consider what we have just done: found the value of x for which While it is true that the graph illustrates what we have found, exactly how did we find this value? If this is to be a truly graphical solution, then all we can use are the graphs of y1 and y2. Did we actually measure the coordinates of the intersection point (2, 7) ? The point is that what we probably did is solve the equation shown for the desired value of x and then illustrated the solution graphically. This is not the same thing as solving the given equation graphically. The algebraic solution is much easier and more precise (e.g., what if the solutions are 8-decimal-place numbers?). Solution is 2 Solution set is { 2 } Question: How do we find 2 and 7 graphically ? 7/8/2013 Equations Equations 7/8/2013

16
**Solving Equations Numerically**

Solve: 14x – 36 = 7 This can be written as y(x) = 14x – 36 = 7 Want x value where y = 7 Desired x between 3 and 4 Increase resolution between 3 and 4 x y 0 –36 1 –22 –8 Solving Equations Numerically: Approximation Solutions can sometimes be approximated (or with luck, identified accurately) by building tables of values. One difficulty is deciding how many values of x to include. In the example we build a table beginning with x = 0 and incrementing by 1. For each entry we calculate the value of the expression on the left to see how close to 7 it is. We note that for x = 3 we get 6 and for x = 4 we get 20. We infer that the correct x lies between 3 and 4, probably much closer to 3 than to 4. Question: What if the first digit is not between 3 and 4? This allows us to build a second table starting with 3.0 and incrementing by 0.1, again calculating the value of the left hand expression. We see that 3.0 give 6.0 and 3.1 gives 7.4, so the x we want is between 3.0 and The table shown gives a few more (unnecessary) entries to show that 4.0 gives the 20.0 value from the first table. Question: What if the second digit is not less than 5? If we want greater accuracy, we then build a third table for x values between 3.0 and 3.1, starting with 3.00 and incrementing by Thus for x = 3.01, 3.02, 3.03, 3.04, 3.05, 3.06, 3.07, produces values 6.14, 6.28, 6.42, 6.56, 6.70, 6.84, 6.98, 7.12, ... Thus the value of x we want is between 3.07 and 3.08 and if we need still more accuracy we must build yet another table with x starting with and incrementing by This process is continued with a new table for each decimal place of accuracy. Clearly this is a great deal of work for a little bit of accuracy. 7 7/8/2013 Equations Equations 7/8/2013

17
**Solving Equations Numerically**

Solve: 14x – 36 = 7 Increased resolution x y y = 7 for x between 3.0 and 3.1 * 7 Expand 3.0 – 3.1 into new table (3.00 – 3.09) for next decimal on y Solving Equations Numerically: Approximation Solutions can sometimes be approximated (or with luck, identified accurately) by building tables of values. One difficulty is deciding how many values of x to include. In the example we build a table beginning with x = 0 and incrementing by 1. For each entry we calculate the value of the expression on the left to see how close to 7 it is. We note that for x = 3 we get 6 and for x = 4 we get 20. We infer that the correct x lies between 3 and 4, probably much closer to 3 than to 4. Question: What if the first digit is not between 3 and 4? This allows us to build a second table starting with 3.0 and incrementing by 0.1, again calculating the value of the left hand expression. We see that 3.0 give 6.0 and 3.1 gives 7.4, so the x we want is between 3.0 and The table shown gives a few more (unnecessary) entries to show that 4.0 gives the 20.0 value from the first table. Question: What if the second digit is not less than 5? If we want greater accuracy, we then build a third table for x values between 3.0 and 3.1, starting with 3.00 and incrementing by Thus for x = 3.01, 3.02, 3.03, 3.04, 3.05, 3.06, 3.07, produces values 6.14, 6.28, 6.42, 6.56, 6.70, 6.84, 6.98, 7.12, ... Thus the value of x we want is between 3.07 and 3.08 and if we need still more accuracy we must build yet another table with x starting with and incrementing by This process is continued with a new table for each decimal place of accuracy. Clearly this is a great deal of work for a little bit of accuracy. Continue refining x 7/8/2013 Equations Equations 7/8/2013

18
**Solving Equations Numerically**

Solve: 14x – 36 = 7 x y Continue refining x to force y closer to 7 Question: 7 … …. How accurate is this method ? * Solving Equations Numerically: Approximation Solutions can sometimes be approximated (or with luck, identified accurately) by building tables of values. One difficulty is deciding how many values of x to include. In the example we build a table beginning with x = 0 and incrementing by 1. For each entry we calculate the value of the expression on the left to see how close to 7 it is. We note that for x = 3 we get 6 and for x = 4 we get 20. We infer that the correct x lies between 3 and 4, probably much closer to 3 than to 4. Question: What if the first digit is not between 3 and 4? This allows us to build a second table starting with 3.0 and incrementing by 0.1, again calculating the value of the left hand expression. We see that 3.0 give 6.0 and 3.1 gives 7.4, so the x we want is between 3.0 and The table shown gives a few more (unnecessary) entries to show that 4.0 gives the 20.0 value from the first table. Question: What if the second digit is not less than 5? If we want greater accuracy, we then build a third table for x values between 3.0 and 3.1, starting with 3.00 and incrementing by Thus for x = 3.01, 3.02, 3.03, 3.04, 3.05, 3.06, 3.07, produces values 6.14, 6.28, 6.42, 6.56, 6.70, 6.84, 6.98, 7.12, ... Thus the value of x we want is between 3.07 and 3.08 and if we need still more accuracy we must build yet another table with x starting with and incrementing by This process is continued with a new table for each decimal place of accuracy. Clearly this is a great deal of work for a little bit of accuracy. How long does it take ? 7/8/2013 Equations Equations 7/8/2013

19
**Solving Equations: Review**

Graphical Solution Least accurate, visual solution Can be automated via computer/calculator Makes trends more obvious Numerical Solution Approximate solution but refinable Natural for collected data Easily automated Solving Equations: Comparisons Thus far we have considered symbolic, graphical and numerical techniques for solving linear equations. The illustration compares the three methods. Each method has advantages and drawbacks, as indicated. 7/8/2013 Equations Equations 7/8/2013

20
**Solving Equations: Review**

Symbolic Solution The most accurate Purely algebraic Good for predictions Solving Equations: Comparisons Thus far we have considered symbolic, graphical and numerical techniques for solving linear equations. The illustration compares the three methods. Each method has advantages and drawbacks, as indicated. 7/8/2013 Equations Equations 7/8/2013

21
**Think about it ! Solving Equations Equations 7/8/2013 Equations**

Similar presentations

OK

Copyright © 2003 Pearson Education, Inc. Slide 1 Computer Systems Organization & Architecture Chapters 8-12 John D. Carpinelli.

Copyright © 2003 Pearson Education, Inc. Slide 1 Computer Systems Organization & Architecture Chapters 8-12 John D. Carpinelli.

© 2018 SlidePlayer.com Inc.

All rights reserved.

Ads by Google

Ppt on tourism in pakistan Ppt on revolt of 1857 leaders Ppt on earth dam failure Ppt on paintings and photographs related to colonial period of american Ppt on suspension type insulation tool Ppt on acute and chronic renal failure Ppt on switching devices and timers Running message display ppt on tv Ppt on kingdom monera phylum Ppt on places in our neighbourhood download