1. Solving Equations With Variables on Both Sides Section 2-4

2. 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.

3. Vocabulary Identity

4. 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. Equations With Variables on Both Sides

5. Solve 7n – 2 = 5n + 6. To collect the variable terms on one side, subtract 5n from both sides. 7n – 2 = 5n + 6 –5n 2n – 2 = 6 Since n is multiplied by 2, divide both sides by 2 to undo the multiplication. 2n = 8 + 2 n = 4 Example: Solving Equations with Variables on Both Sides

6. Solve 4b + 2 = 3b. To collect the variable terms on one side, subtract 3b from both sides. 4b + 2 = 3b –3b b + 2 = 0 b = –2 – 2 – 2 Your Turn:

7. Solve 0.5 + 0.3y = 0.7y – 0.3. To collect the variable terms on one side, subtract 0.3y from both sides. 0.5 + 0.3y = 0.7y – 0.3 –0.3y 0.5 = 0.4y – 0.3 0.8 = 0.4y +0.3 + 0.3 2 = y 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. Your Turn:

8. To solve more complicated equations, you may need to first simplify by using the Distributive Property or combining like terms. Using the Distributive Property

9. Solve 4 – 6a + 4a = –1 – 5(7 – 2a). Combine like terms. Distribute –5 to the expression in parentheses. 4 – 6a + 4a = –1 –5(7 – 2a) 4 – 6a + 4a = –1 –5(7) –5(–2a) 4 – 6a + 4a = –1 – 35 + 10a 4 – 2a = –36 + 10a +36 40 – 2a = 10a + 2a +2a 40 = 12a 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: Simplifying Both Sides

10. 40 = 12a Since a is multiplied by 12, divide both sides by 12. Example: Continued

11. Solve. Since 1 is subtracted from b, add 1 to both sides. Distribute to the expression in parentheses. 1 2 + 1 3 = b – 1 To collect the variable terms on one side, subtract b from both sides. 1 2 4 = b Your Turn:

12. Solve 3x + 15 – 9 = 2(x + 2). Combine like terms. Distribute 2 to the expression in parentheses. 3x + 15 – 9 = 2(x + 2) 3x + 15 – 9 = 2(x) + 2(2) 3x + 15 – 9 = 2x + 4 3x + 6 = 2x + 4 –2x x + 6 = 4 – 6 – 6 x = –2 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. Your Turn:

13. 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.

14. A 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 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 . Conditional Equations & Contradictions

15. Contradiction When solving an equation, if you get a false equation, the original equation is a contradiction, and it has no solutions. WORDS x = x + 3 –x –x 0 = 3  1 = 1 + 2 1 = 3  ALGEBRA NUMBERS Contradictions

16. An 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). Distribute to remove parentheses. This is a true statement for all real numbers x. The solution set is all real numbers. 5x + 3 = 2x + 3(x + 1). 5x + 3 = 2x + 3x + 3 Combine like terms. 5x + 3 = 5x + 3 Subtract 5x from both sides. 3 = 3 Identities

17. 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 Identities

18. Summary 1)A 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.

19. Solve 10 – 5x + 1 = 7x + 11 – 12x. Add 5x to both sides. Identify like terms. 10 – 5x + 1 = 7x + 11 – 12x 11 – 5x = 11 – 5x 11 = 11 + 5x + 5x True statement. Combine like terms on the left and the right. 10 – 5x + 1 = 7x + 11 – 12x 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: Identity

20. Solve 12x – 3 + x = 5x – 4 + 8x. Subtract 13x from both sides. Identify like terms. 12x – 3 + x = 5x – 4 + 8x 13x – 3 = 13x – 4 –3 = –4 –13x False statement. Combine like terms on the left and the right. 12x – 3 + x = 5x – 4 + 8x  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. Example: Contradiction

21. Solve 4y + 7 – y = 10 + 3y. Subtract 3y from both sides. Identify like terms. 4y + 7 – y = 10 + 3y 3y + 7 = 3y + 10 7 = 10 –3y –3y False statement. Combine like terms on the left and the right. 4y + 7 – y = 10 + 3y  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:

22. Solve 2c + 7 + c = –14 + 3c + 21. Subtract 3c both sides. Identify like terms. 2c + 7 + c = –14 + 3c + 21 3c + 7 = 3c + 7 7 = 7 –3c –3c True statement. Combine like terms on the left and the right. 2c + 7 + c = –14 + 3c + 21 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). Your Turn:

23. 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.

24. 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. Example: Continued Finding an Unknown Number

25. The product of · twice a number 2x2x and 3 3 is the same as = 5 times the number 5x5x and ¾ ¾ the difference of – Example continued: 2.) Translate Continued

26. 3.) Solve 2x · 3 = 5x – ¾ 6x = 5x – ¾ (Simplify left side) x = – ¾ (Simplify both sides) 6x + (– 5x) = 5x + (– 5x) – ¾ (Add –5x to 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. Example continued:

27. 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 Example: Your Turn:

28. 2.) Translate Continued Daily costs 2(24.95) mileage costs 0.29x plus + is equal to = 100 maximum budget Continued:

29. 3.) Solve Continued 2(24.95) + 0.29x = 100 49.90 + 0.29x = 100 (Simplify left side) 0.29x = 50.10 (Simplify both sides) (Divide both sides by 0.29) x  172.75 (Simplify both sides) (Subtract 49.90 from both sides) 49.90 – 49.90 + 0.29x = 100 – 49.90 Continued:

30. 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. Continued:

31. 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.

32. Assignment 2.4 Exercises Pg. 114 – 116: #10 – 48 even