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Transcript

Solving Equations With Variables on Both Sides Section 2-4 Goals Goal • To solve equations with variables on both sides. • To identify equations that are identities or have no solution. Rubric Level 1 – Know the goals. Level 2 – Fully understand the goals. Level 3 – Use the goals to solve simple problems. Level 4 – Use the goals to solve more advanced problems. Level 5 – Adapts and applies the goals to different and more complex problems. Vocabulary • Identity Equations With Variables on Both Sides To solve an equation with variables on both sides, use inverse operations to "collect" variable terms on one side of the equation. Helpful Hint Equations are often easier to solve when the variable has a positive coefficient. Keep this in mind when deciding on which side to "collect" variable terms. Example: Solving Equations with Variables on Both Sides Solve 7n – 2 = 5n + 6. 7n – 2 = 5n + 6 –5n –5n 2n – 2 = +2 2n = To collect the variable terms on one side, subtract 5n from both sides. 6 +2 8 Since n is multiplied by 2, divide both sides by 2 to undo the multiplication. n=4 Your Turn: Solve 4b + 2 = 3b. 4b + 2 = 3b –3b –3b b+2= 0 –2 –2 b = –2 To collect the variable terms on one side, subtract 3b from both sides. Your Turn: Solve 0.5 + 0.3y = 0.7y – 0.3. 0.5 + 0.3y = 0.7y – 0.3 –0.3y –0.3y 0.5 = 0.4y – 0.3 +0.3 + 0.3 0.8 = 0.4y 2=y To collect the variable terms on one side, subtract 0.3y from both sides. Since 0.3 is subtracted from 0.4y, add 0.3 to both sides to undo the subtraction. Since y is multiplied by 0.4, divide both sides by 0.4 to undo the multiplication. Using the Distributive Property To solve more complicated equations, you may need to first simplify by using the Distributive Property or combining like terms. Example: Simplifying Both Sides Solve 4 – 6a + 4a = –1 – 5(7 – 2a). 4 – 6a + 4a = –1 –5(7 – 2a) Distribute –5 to the expression in parentheses. 4 – 6a + 4a = –1 –5(7) –5(–2a) 4 – 6a + 4a = –1 – 35 + 10a 4 – 2a = –36 + 10a +36 +36 40 – 2a = + 2a 40 = 10a +2a 12a Combine like terms. Since –36 is added to 10a, add 36 to both sides. To collect the variable terms on one side, add 2a to both sides. Example: Continued 40 = 12a Since a is multiplied by 12, divide both sides by 12. Your Turn: Solve . 1 Distribute to the expression in 2 parentheses. 3=b–1 +1 +1 4=b To collect the variable terms on one side, subtract 1 b from 2 both sides. Since 1 is subtracted from b, add 1 to both sides. Your Turn: Solve 3x + 15 – 9 = 2(x + 2). 3x + 15 – 9 = 2(x + 2) Distribute 2 to the expression 3x + 15 – 9 = 2(x) + 2(2) in parentheses. 3x + 15 – 9 = 2x + 4 3x + 6 = 2x + 4 –2x –2x x+6= –6 x = –2 4 –6 Combine like terms. To collect the variable terms on one side, subtract 2x from both sides. Since 6 is added to x, subtract 6 from both sides to undo the addition. Summary Solving linear equations in one variable 1) Clear the equation of fractions by multiplying both sides of the equation by the LCD of all denominators in the equation. 2) Use the distributive property to remove grouping symbols such as parentheses. 3) Combine like terms on each side of the equation. 4) Use the addition property of equality to rewrite the equation as an equivalent equation with variable terms on one side and numbers on the other side. 5) Use the multiplication property of equality to isolate the variable. 6) Check the proposed solution in the original equation. Conditional Equations & Contradictions A conditional equation is an equation that is true for some values of the variable and false for other values of the variable. Normal equations with a finite number of solutions. A contradiction is an equation that is false for every replacement value of the variable. Example: Solve the equation 4x – 8 = 4x – 1. 4x – 8 = 4x – 1 –8=–1 Subtract 4x from both sides. This is a false statement so the equation is a contradiction. The solution set is the empty set, written { } or . Contradictions WORDS NUMBERS ALGEBRA Contradiction When solving an equation, if you get a false equation, the original equation is a contradiction, and it has no solutions. 1=1+2 1=3 x= x+3 –x –x 0=3 Identities An identity is an equation that is satisfied for all values of the variable for which both sides of the equation are defined. Example: Solve the equation 5x + 3 = 2x + 3(x + 1). 5x + 3 = 2x + 3(x + 1). 5x + 3 = 2x + 3x + 3 Distribute to remove parentheses. 5x + 3 = 5x + 3 Combine like terms. 3=3 Subtract 5x from both sides. This is a true statement for all real numbers x. The solution set is all real numbers. Identities WORDS Identity When solving an equation, if you get an equation that is always true, the original equation is an identity, and it has infinitely many solutions. NUMBERS 2+1=2+1 3=3 ALGEBRA 2+x=2+x –x –x 2=2 Summary 1) A conditional equation is an equation in one variable that has a finite (normally one solution) number of solutions. 2) An identity is an equation that is true for all values of the variable (ie. the variable is eliminated and results in a true statement). An equation that is an identity has infinitely many solutions. 3) A contradiction is an equation that is false for any value of the variable (ie. the variable is eliminated and results in a false statement). It has no solutions. Example: Identity Solve 10 – 5x + 1 = 7x + 11 – 12x. 10 – 5x + 1 = 7x + 11 – 12x 10 – 5x + 1 = 7x + 11 – 12x 11 – 5x = 11 – 5x + 5x + 5x 11 = 11 Identify like terms. Combine like terms on the left and the right. Add 5x to both sides. True statement. The equation 10 – 5x + 1 = 7x + 11 – 12x is an identity. All values of x will make the equation true. All real numbers are solutions. Example: Contradiction Solve 12x – 3 + x = 5x – 4 + 8x. 12x – 3 + x = 5x – 4 + 8x 12x – 3 + x = 5x – 4 + 8x 13x – 3 = 13x – 4 –13x –13x –3 = –4 Identify like terms. Combine like terms on the left and the right. Subtract 13x from both sides. False statement. The equation 12x – 3 + x = 5x – 4 + 8x is a contradiction. There is no value of x that will make the equation true. There are no solutions. Your Turn: Solve 4y + 7 – y = 10 + 3y. 4y + 7 – y = 10 + 3y 4y + 7 – y = 10 + 3y 3y + 7 = 3y + 10 –3y –3y 7 = 10 Identify like terms. Combine like terms on the left and the right. Subtract 3y from both sides. False statement. The equation 4y + 7 – y = 10 + 3y is a contradiction. There is no value of y that will make the equation true. There are no solutions. Your Turn: Solve 2c + 7 + c = –14 + 3c + 21. 2c + 7 + c = –14 + 3c + 21 2c + 7 + c = –14 + 3c + 21 3c + 7 = 3c + 7 –3c –3c 7= 7 Identify like terms. Combine like terms on the left and the right. Subtract 3c both sides. True statement. The equation 2c + 7 + c = –14 + 3c + 21 is an identity. All values of c will make the equation true. All real numbers are solutions (infinite number of solutions). Strategy for Problem Solving 1) UNDERSTAND the problem. During this step, become comfortable with the problem. Some way of doing this are: • Read and reread the problem • Propose a solution and check. • Construct a drawing. • Choose a variable to represent the unknown 2) TRANSLATE the problem into an equation. 3) SOLVE the equation. 4) INTERPRET the result. Check the proposed solution in stated problem and state your conclusion. Finding an Unknown Number Example: The product of twice a number and three is the same as the difference of five times the number and ¾. Find the number. 1.) Understand Read and reread the problem. If we let x = the unknown number, then “twice a number” translates to 2x, “the product of twice a number and three” translates to 2x · 3, “five times the number” translates to 5x, and “the difference of five times the number and ¾” translates to 5x – ¾. Continued Example continued: 2.) Translate The product of is the same as twice a number 2x the difference of and 3 · 3 = 5 times the number 5x and ¾ – ¾ Continued Example continued: 3.) Solve 2x · 3 = 5x – ¾ 6x = 5x – ¾ 6x + (– 5x) = 5x + (– 5x) – ¾ x=–¾ (Simplify left side) (Add –5x to both sides) (Simplify both sides) 4.) Interpret Check: Replace “number” in the original statement of the problem with – ¾. The product of twice – ¾ and 3 is 2(– ¾)(3) = – 4.5. The difference of five times – ¾ and ¾ is 5(– ¾) – ¾ = – 4.5. We get the same results for both portions. State: The number is – ¾. Your Turn: Example: A car rental agency advertised renting a Buick Century for $24.95 per day and $0.29 per mile. If you rent this car for 2 days, how many whole miles can you drive on a $100 budget? 1.) Understand Read and reread the problem. Let’s propose that we drive a total of 100 miles over the 2 days. Then we need to take twice the daily rate and add the fee for mileage to get 2(24.95) + 0.29(100) = 49.90 + 29 = 78.90. This gives us an idea of how the cost is calculated, and also know that the number of miles will be greater than 100. If we let x = the number of whole miles driven, then 0.29x = the cost for mileage driven Continued Continued: 2.) Translate Daily costs mileage costs plus 2(24.95) + maximum budget is equal to 0.29x = 100 Continued Continued: 3.) Solve 2(24.95) + 0.29x = 100 49.90 + 0.29x = 100 49.90 – 49.90 + 0.29x = 100 – 49.90 0.29x = 50.10 0.29 x 50.10 0.29 0.29 x 172.75 (Simplify left side) (Subtract 49.90 from both sides) (Simplify both sides) (Divide both sides by 0.29) (Simplify both sides) Continued Continued: 4.) Interpret Check: Recall that the original statement of the problem asked for a “whole number” of miles. If we replace “number of miles” in the problem with 173, then 49.90 + 0.29(173) = 100.07, which is over our budget. However, 49.90 + 0.29(172) = 99.78, which is within the budget. State: The maximum number of whole number miles is 172. Joke Time • Where do you find a dog with no legs? • Right where you left him. • What did E.T.’s Mom say when he got home? • Where on Earth have you been? • What happened when the pig pen broke? • The pigs had to use a pencil. Assignment • 2.4 Exercises Pg. 114 – 116: #10 – 48 even