Solving Linear Equations: In solving an equation, we are trying to find the value of the variable(s) that would make the quation a true equation. E.g. x + 3 = 5 In this equation, if x = 2, then the equation is true. The basic idea in solving a linear equation is to isolate the variable. I.e. we need the variable on one side of the equation and the rest of the numbers in the other side. Central Idea in solving equations For any equation, whatever we do to one side of the equation, if we do the same action to the other side of the equation, the two sides will remain equal. For example, if we add −3 to one side of the equation, as long as we add −3 to the other side of the equation, the two sides are still equal. Using this idea, we can perform operations on both sides of the equation to try to isolate the variable. E.g. 2x + 4 = 10 To solve this equation, we need to isolate the 2x term. We add −4 to both sides of the equation to eliminate the 4 on the right side. (You may also think of this as subtracting 4 from both sides) (1) 2x + 4 = 10 (2) 2x + 4 + (−4) = 10 + (−4) (3) 2x = 6 After adding −4 on both sides of the equation, the left side is simplified to just 2x, and we get equation (3). Equations (1) and (3) are said to be equivalent equations because they have the same solution(s). The idea of solving an equa- tion is to using operations to simplify an originally complicated equation to an equivalent equation whose solution is easy to determine. To solve equation (3), we need to isolate x by getting rid of the factor of 2. Remember that 2x means 2 times x, to get rid of the 2, we multiply both sides of the equation by 1/2 (you may think of it as dividing by 2), we have: 1 1 (4) ( )2x = ( )6 2 2 (5) x = 3 The solution is now clear to us, x must be equal to 3 for the equation to be true. We can plug the value of 3 into the original equation to check: 2(3) + 4 = 6 + 4 = 10 The left and right hand side are both equal to 10, therefore 3 is a proper solution. In solving any equation, you can always do whatever operation you need to the equation to try to isolate the variable, just keep in mind that whatever you do to one side of the equation, you must do the same operation to the other side of the equation. E.g. 4x + 3 = 2x − 9 4x+3+(−3) = 2x−9+(−3) (add −3 (subtract 3) on both sides of the equation) 4x = 2x − 12 (collect like terms) 4x + (−2x) = 2x − 12 + (−2x) (add −2x (subtract 2x) on both sides of the quation because we only want x on one side of the quation) 2x = −12 (collect like terms) 1 1 1 (2x) = (−12) (multiply both sides by (divide by 2)) 2 2 2 x = −6 Sometimes the distributive property needs to be applied to get ride of parenthesis involved in an equation: 5(x + 2) − 5 = 2(x − 1) + x 5x + 10 − 5 = 2x − 2 + x (distributive property) 5x + 5 = 3x − 2 (simplify by collecting like terms) 5x + 5 + (−5) = 3x − 2 + (−5) (adding −5 to both sides) 5x = 3x − 7 (simplify) 5x − 3x = 3x − 7 − 3x (subtract 3x from both sides) 2x = −7 (simplify by collecting like terms) 1 1 (2x) = (−7) (multiply both sides by 1/2) 2 2 7 x = − 2 Example: Solve the equation: 12 − 3(x + 2) = 5 + 2(2x − 1) − 7x We solve this equation using similar methods as the examples before: 12 − 3x − 6 = 5 + 4x − 2 − 7x (distribute) −3x + 6 = −3x + 3 (collect like-terms) −3x + 6 + (3x) = −3x + 3 + (3x) (add 3x to both sides of equation) 6 = 3 For this equation, it turns out that the x variable has been cancelled on both sides of the equation, and we reached an equation that can be never true, 6 = 3. This means that, in order to solve the equation, we will need to be able to make 6 equal to 3. This is impossible, so our equation has no solution. We say that an equation like this is inconsistent. Example: Solve the equation: 4x + 2(x − 3) − 6 = 10x − 4(x + 3) We solve using the same approach: 4x + 2x − 6 − 6 = 10x − 4x − 12 (distribute) 6x − 12 = 6x − 12 (collect like-terms) 6x − 12 − (6x) = 6x − 12 − (6x) (subtract 6x from both sides of equation) −12 = −12 Again the x variable is cancelled on both sides of the equation, but this time the equation reduces to −12 = −12. This means that, in order to solve the equation, we need to make −12 equal to −12. Since this is always true, x can be any value and the equation will still be true, so our solution set is all real numbers. An equation like this is called an identity. In solving an equation that involve fractions, multiply both sides of the equation by the least common multiple of the denominators of the fractions to get rid of the fraction first. Remember to distribute the factor to each term of the expression. E.g. 9 2 33 y + y = 5 5 5 The LCM of the denominators is 5, therefore we multiply 5 to both sides of the equation, we get: 9 2 33 5( y + y) = 5( ) 5 5 5 9 2 33 5( y) + 5( y) = 5( ) (distribute 5 to each of the terms in both sides) 5 5 5 9y + 2y = 33 (simplify and get rid of the fractions) 11y = 33 (collect like terms) y = 3 (divide by 11) E.g. 4 1 3 x − = x + 2 3 5 15 The LCM of the denominators is 15, so we multiply this number to both sides of the equations: 4 1 3 15( x − ) = 15( x + 2) 3 5 15 4 1 3 15( x) − 15( ) = 15( x) + 15(2) (distribute 15 to each term on each side) 3 5 15 5(4)x − 3(1) = 3x + 30 (simplify the fractions) 20x − 3 = 3x + 30 20x = 3x + 33 (add 3 to both sides) 17x = 33 (subtract 3x from both sides) 33 x = (divide by 17) 17 It is important that you do not selectively multiply the LCM to just the term you need. If you multiply the LCM to one side of the equation, it must be distributed to every term of the expression. A formula is an equation that represents a relationship between two or more quantities. The quantities involved are represented by variables. Sometimes it is useful to try to change a formula from one form the the other. E.g. C = 2πr This is the formula that expresses the relationship between the circumference of a circle (C) and its radius (r). What if we want to isolate r? We can solve the equation for r just like we did with numbers: To isolate r, we divide both sides by 2 and π, we get: 1 1 ( )C = ( )2πr 2π 2π C = r 2π C r = 2π E.g. (b + b )h A = 1 2 2 This is the formula that expresses the area of a trapazoid (A) in terms of its two bases (b1, b2) and height (h). If we want to isolate b1, we can do: (b + b )h (2)A = (2)( 1 2 ) (multiply both sides by 2) 2 2A = (b1 + b2)h (simplify) 1 1 2A = ((b + b )h) (divide by h) h h 1 2 2A = b + b (simplify) h 1 2 2A − b = b (subtract b from both sides) h 2 1 2 2A b = − b (subtract b from both sides) 1 h 2 2 We may use the techniques of equation solving to solve some application prob- lems. The most difficult task in any of these problems is to set up the correct equation. Some guidelines in trying to solve the problem: Read the problem carefully, ask yourself, what (quantity) does the problem ask for? Use a variable to represent the quantity you are looking for. When choosing the variable name, try use meaningful names instead of generic letters like x or y. For example, if the problem askes for the age of a person, try using a as the variable for the unknown. If the problem askes for the amount of time, try using t. Always lable your variable. In your solution, there should always be descrip- tions as to what the variable represent. It should never be just an equation with variables. For example, in your solution, write something like: ”Let t = the amt of time to reach city A”. Check your answer for sensibility. If the question askes for the age of a person and you get 208, or if the question askes for the number of coins in a pocket, and you get a decimal number, something is wrong. E.g. The sum of the page numbers on the facing pages of a book is 373, what are the page numbers? Ans: What does the question askes for? We randomly turn a book to a page, the sum of the two pages facing us, is 373, the question askes for what is each of the number of the page. Let p = the number of the smaller of the two pages. Then what is the number of the larger of the two pages? The two pages are next to each other, so the larger of the two pages must be one more than the smaller one, so the large one is p + 1, the sum of them is p + (p + 1), which is 373. This is the equation we can set up: p + (p + 1) = 373 Solving inequalities: To solve an inequality like x + 3 < 10 means we want to find the values of x that would make the inequality true. Unlike an equation, however, there are usually more (a lot more) values that could solve an inequality. For the above inequality, x could be 4, could be 6, could be 0, or any value that is less than 7. To express the solution to an inequality, we usually use a graph or an interval: (−∞, 7)
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When graphing an inequality using a real line, we use the open parenthesis, ), or (, to indicate that the end point is not part of the solution. In other words, when graphing a strict inequality, x < a or x > a, we use open parenthesis. On the other hand, we use the bracket ] or [ to indicate the end point is part of the solution. In other words, when graphing x ≤ a or x ≥ a, we use bracket. When we use the interval notation to express the solution, ∞ is the infinity symbol. The interval indicates that any number to the left of 7 are all acceptable values. Another notation uses open and closed circles instead of parenthesis and brackets in graphing the solution to an inequality. So the above solution, x < 7, may also be graphed like this:
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The process for solving an inequality is almost the same as solving an equation, the only difference is that, when multiplying or dividing by a negative number, the inequality sign much be switched. E.g. 3x + 7 ≥ 19 3x ≥ 12 (subtract 7 from both sides) x ≥ 4 (divide both sides by 3) [4, ∞)
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Note that we always use open parenthesis for the infinity symbol, because infinity is not a number, it is just a symbol to express an idea. E.g. −4x − 8 ≤ 2x + 16 −4x ≤ 2x + 24 (add 8 to both sides of inequality) −6x ≤ 24 (subtract 2x from both sides) 1 1 − (−6x) ≥ − (24) (divide both sides by −6, notice that the direction of the 6 6 inequality symbol is switched) x ≥ −4 [−4, ∞)
[ −4 Compound Inequality x + 3 < 7 or 2x − 4 > 10 If we have two inequalities connected by an or conjunctive, we need to solve each inequality individually, and any values that are solution to one or the other, will be part of the solution. x < 4 or x > 7 (−∞, 4) ∪ (7, ∞)
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The union ∪ symbol means we want both of the two intervals put together. Any number that is in the first interval (numbers that are less than 4) or any number in the second interval (numbers greater than 7) are solutions. If two inequalities are connected by and conjunctive, solutions must be values that solve both inequalities. E.g. 3x − 5 < 16 and 2x + 6 > −2 3x < 21 and 2x > −8 x < 7 and x > −4 We need numbers that are less than 7 and at the same time greater than −4, the solution set written in interval notation is: (−4, 7)
−4 7 Absolute Value Equations and Inequalities |x| = 3 How many solutions does this equation have? If x = 3, this solves the equation, if x = −3, this still solves the equation. In general, the equation: |x| = c, where c is a non-negative real number, has two solutions, namely, x = c or x = −c. We can generalize this concept: |3x + 5| = 22 The 3x + 5 is the expression inside the absolutel value, if it is equal to 22 or −22, the absolute value of that will be 22, so we have two equivalent equations: 3x + 5 = 22 or 3x + 5 = −22 Solving these two equations give us: 3x = 17 or 3x = −27 17 x = or x = −9 3 E.g. |x| < 5 What kind of values must x be to solve this: If x is larger than 5, this will not work. If x is too negative, say if x < −5. If x = −6, then |x| = 6, and this still fails, so we need: x < 5 and −5 < x (−5, 5)
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What about: |x| > 3 If x > 3, this works. If x < −3, say x = −4, then |x| = 4, and 4 is still greater than 3, so the solution will be: x > 3 or x < −3
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In general, if c > 0, then the inequality |x| < c has the solution: x < c and −c < x |x| > c has solution: x > c or x < −c