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10 Real Numbers, Equations, and Inequalities Copyright © 2014, 2010, 2007 Pearson Education, Inc. Slide 1 10.2 More on Solving Linear Equations Objectives 1. Solve more difficult linear equations. 2. Solve equations that have no solution or infinitely many solutions. 3. Solve equations by first clearing fractions and decimals. Copyright © 2014, 2010, 2007 Pearson Education, Inc. Slide 2 Solve More Difficult Linear Equations Copyright © 2014, 2010, 2007 Pearson Education, Inc. Slide 3 Solve More Difficult Linear Equations Example 1 Solve this equation and check the solution: –2 + 3(2x + 7) = 7 + 3x Step 1 Use the distributive property. –2 + 3(2x + 7) = 7 + 3x Step 2 Combine like terms. –2 + 6x + 21 = 7 + 3x 6x + 19 = 7 + 3x Step 3 Subtract 19 from both sides. –19 –19 6x + 0 = –12 + 3x 6x = –12 + 3x Subtract 3x from both sides. –3x –3x 3x = –12 + 0 3 x 12 Step 4 Divide both sides by 3. 3 3 x 4 The solution is –4. Copyright © 2014, 2010, 2007 Pearson Education, Inc. Slide 4 Solve More Difficult Linear Equations Step 5 Check the solution by replacing each x with –4 in the original equation. –2 + 3(2x + 7) = 7 + 3x –2 + 3[2(–4) + 7] = 7 + 3(–4) –2 + 3[(–8) + 7] = 7 + –12 –2 + 3(–1) = –5 –2 + –3 = –5 –5 = –5 ✓ Balances When x is replaced with –4, the equation balances, so 4 is the correct solution (not –5). Copyright © 2014, 2010, 2007 Pearson Education, Inc. Slide 5 Solve More Difficult Linear Equations Example 2 Solve this equation: 8a – (3 + 2a) = 3a + 1 Start by removing the parentheses on the left side of the equation. In this situation, the – sign acts like a factor of –1, changing the sign of every term inside the parentheses. 8a (3 2a) 3a 1 8a 1(3 2a) 3a 1 8a 3 2a 3a 1 6a 3 3a 1 6a 3 3 3a 1 3 6a 3a 4 6a 3a 3a 4 3a 3a 4 3a 4 3 3 4 a 3 The solution is 4/3. You can check the solution by going back to the original equation and replacing each a with 4/3. Copyright © 2014, 2010, 2007 Pearson Education, Inc. Slide 6 Solve More Difficult Linear Equations Example 3 Solve: 4(8 – 3t) = 32 – 8(t + 2) 4(8 3t ) 32 8(t 2) 32 12t 32 8t 16 32 12t 32 32 8t 16 32 12t 16 8t 12t 8t 16 8t 8t 4t 16 4t 16 4 4 t4 The solution is 4. The check is left to you. Copyright © 2014, 2010, 2007 Pearson Education, Inc. Slide 7 Solving Equations With Infinitely Many Solutions Example 4 Solve: 5x – 15 = 5(x – 3) 5 x 15 5( x 3) 5 x 15 5 x 15 5 x 15 15 5 x 15 15 5x 5x 5x 5x 5x 5x 00 When both sides of an equation are exactly the same, the equation is called an identity. An identity is true for all replacements of the variables. The solution for 5x – 15 = 5(x – 3) is all real numbers. Copyright © 2014, 2010, 2007 Pearson Education, Inc. Slide 8 Solving Equations With No Solution Example 5 Solve: 2w – 7(w + 1) = –5w + 4 2w 7( w 1) 5w 4 2w 7 w 7 5w 4 5w 7 5w 4 5w 7 5w 5w 4 5w 7 4 As in Example 4, the variable has “disappeared.” This time, however, we are left with a false statement. Whenever this happens in solving an equation, it is a signal that the equation has no solution. So, you write “no solution.” Copyright © 2014, 2010, 2007 Pearson Education, Inc. Slide 9 Solve Equations with Fraction or Decimal Coefficients Example 6 Solve the equation. 5 m – 10 = 8 8 8 5 m 8 3 m + 1 m 4 2 5 m – 10 8 = 8 3 m + 1 m 4 2 – 8 10 = 8 3 m 4 + 8 Multiply by LCD, 8. 1 m 2 5m – 80 = 6m + 4m Distribute. Multiply. Now use the four steps to solve this equivalent equation. Copyright © 2014, 2010, 2007 Pearson Education, Inc. Slide 10 Solve Equations with Fraction or Decimal Coefficients Example 6 (continued) Solve the equation. 5m – 80 = 6m + 4m Step 1 5m – 80 = 10m Step 2 5m – 80 – 5m = 10m – 5m – 80 = 5m Step 3 – 80 = 5m 5 5 – 16 = m Copyright © 2014, 2010, 2007 Pearson Education, Inc. Combine terms. Subtract 5m. Combine terms. Divide by 5. Slide 11 Solve Equations with Fraction or Decimal Coefficients Example 6 (continued) Solve the equation. Step 4 Check by substituting –16 for m in the original equation. 5 m – 10 = 8 5 (–16) – 10 = 8 3 m + 1 m 4 2 3 (–16) 1 (–16) + 4 2 –10 – 10 = –12 – 8 –20 = –20 ? Let m = –16. ? Multiply. True The solution to the equation is –16. Copyright © 2014, 2010, 2007 Pearson Education, Inc. Slide 12 Solve Equations with Fraction or Decimal Coefficients Note Multiplying by 10 is the same as moving the decimal point one place to the right. Example: 1.5 ( 10 ) = 15. Multiplying by 100 is the same as moving the decimal point two places to the right. Example: 5.24 ( 100 ) = 524. Multiplying by 10,000 is the same as moving the decimal point how many places to the right? Answer: 4 places. Copyright © 2014, 2010, 2007 Pearson Education, Inc. Slide 13 Solve Equations with Fraction or Decimal Coefficients Example 7 Solve the equation. 0.2v – 0.03 ( 11 + v ) = – 0.06 ( 31 ) 20v – 3 ( 11 + v ) = – 6 ( 31 ) Step 1 20v – 33 – 3v = – 186 17v – 33 = – 186 Copyright © 2014, 2010, 2007 Pearson Education, Inc. Multiply by 100. Distribute. Combine terms. Slide 14 Solve Equations with Fraction or Decimal Coefficients Example 7 (continued) Solve the equation. 17v – 33 = – 186 Step 2 Step 3 From Step 1 17v – 33 + 33 = – 186 + 33 Add 33. 17v = – 153 Combine terms. 17v – 153 = 17 17 Divide by 17. v = –9 Check to confirm that – 9 is the solution. Copyright © 2014, 2010, 2007 Pearson Education, Inc. Slide 15