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Ch 2 Sec 5: Slide #1 Columbus State Community College Chapter 2 Section 5 Solving Equations with Several Steps

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Ch 2 Sec 5: Slide #2 Solving Equations with Several Steps 1.Solve equations, using the addition and division properties of equality. 2.Solve equations, using the distributive, addition, and division properties.

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Ch 2 Sec 5: Slide #3 Solving an Equation Using the Addition and Division Properties Step 1 Add ( or subtract ) the same amount to ( or from ) both sides of the equation so that the variable term ( the variable and its coefficient ) ends up by itself on one side of the equal sign. Step 2Divide both sides by the coefficient of the variable term to find the solution. Step 3Check the solution by going back to the original equation.

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Ch 2 Sec 5: Slide #4 21= 21 Solving an Equation with Several Steps EXAMPLE 1 Solving an Equation with Several Steps 3w + 6 = 21 Solve this equation and check the solution:3w + 6 = 21. – 6 3w + 6 = 15 3w + 6 = 21 = · = 21 Balance statement Step 1Get the variable term by itself by subtracting 6 from both sides. 33 Step 2Divide both sides by the coefficient. 3w + 6 = 5 Step 3Check the solution using the original equation. Solution

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Ch 2 Sec 5: Slide #5 Solving an Equation with Variable Terms on Both Sides EXAMPLE 2 Solving Equations – Variable Terms on Both Sides 3a – 5 = 5a + 9 Solve this equation and check the solution:3a – 5 = 5a + 9. – 3a – 5 = 2a + 9 First, “get rid of ” the “ + 3a” by subtracting it from both sides. 22 Next, “get rid of ” the “ + 9” by subtracting it from both sides. + 6 – 7 = aSolution – 9 3a – 14 = 2aFinally, “get rid of ” the coefficient 2, by dividing both sides by it.

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Ch 2 Sec 5: Slide #6 Solving an Equation with Variable Terms on Both Sides EXAMPLE 2 Solving Equations – Variable Terms on Both Sides 3a – 5 = 5a + 9 Solve this equation and check the solution:3a – 5 = 5a + 9. – 3a – 5 = 2a + 9 To “get rid of ” a term, use addition or subtraction. The operation you choose depends on the operation in front of the term you wish to remove – 7 = aSolution – 9 3a – 14 = 2a To “get rid of ” a coefficient, use division.

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Ch 2 Sec 5: Slide #7 Solving an Equation with Variable Terms on Both Sides EXAMPLE 2 Solving Equations – Variable Terms on Both Sides 3a – 5 = 5a + 9 Solve this equation and check the solution:3a – 5 = 5a + 9. Check the solution: a = – 7. 3( – 7) – 5 – 21 – 5 – 26 = 5( – 7) + 9 = – = – 26Balance statement NOTE Checking Your Solution: Be sure to substitute the solution into the original equation in order to check the entire problem.

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Ch 2 Sec 5: Slide #8 Note on Solving Equations NOTE More than one sequence of steps will work to solve complicated equations. The basic approach is the following: If possible, simplify each side of the equation by removing parentheses and combining like terms. Get the variable term by itself on one side of the equation and the constant term by itself on the other side. Remember, use either addition or subtraction to “get rid of ” a term. Divide both sides of the equation by the coefficient of the variable term.

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Ch 2 Sec 5: Slide #9 Solving an Equation Using the Distributive Property EXAMPLE 3 Solving Equations Using the Distributive Property 4 ( n + 2 ) = – 8 ( n – 1 ) Solve and check: 4 ( n + 2 ) = – 8 ( n – 1 ). First, simplify each side distribute. Next, “get rid of ” the “ – 8n” by adding it to both sides. Finally, “get rid of ” the coefficient by dividing both sides by it. Next, “get rid of ” the “+ 8” by subtracting it from both sides. 4n + 8 = – 8n n 12n + 8 = 8 12 n + 8 = 0Solution – 8 12n + 8 = 0

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Ch 2 Sec 5: Slide #10 Solving an Equation Using the Distributive Property EXAMPLE 3 Solving Equations Using the Distributive Property 4 ( n + 2 ) = – 8 ( n – 1 ) Solve and check: 4 ( n + 2 ) = – 8 ( n – 1 ). Check the solution: n = 0. 4 ( ) 4 ( 2 ) 8 = – 8 ( 0 – 1 ) = – 8 ( – 1 ) = 8Balance statement

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Ch 2 Sec 5: Slide #11 Solving an Equation Step 1 If possible, use the distributive property to remove parentheses. Step 2Combine like terms on each side of the equation. Step 3Add (or subtract) the same amount to both sides to get the variable term and constant terms on opposite sides of the equal sign. Step 4Divide both sides by the coefficient of the variable term to find the solution. Step 5Check your solution by substituting it into the original equation and following the order of operations to arrive at a balance statement.

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Ch 2 Sec 5: Slide #12 Solving an Equation EXAMPLE 4 Solving an Equation 6 ( x + 3 ) – 2 = 3x + 1 – 2x Solve and check: 6 ( x + 3 ) – 2 = 3x + 1 – 2x. Step 1 Distribute. Step 2 Combine like terms. 6x + 18 – 2 = 3x + 1 – 2x 6x + 16 = 1x + 1 – 1x 5x + 16 = 1x x + 8 = – 3Solution – 16 5x + 16 = 1x – 15 Step 3 Get variable terms to one side; constant terms to the opposite side. Step 4 Divide by the coefficient.

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Ch 2 Sec 5: Slide #13 Solving an Equation EXAMPLE 4 Solving an Equation 6 ( x + 3 ) – 2 = 3x + 1 – 2x Solve and check: 6 ( x + 3 ) – 2 = 3x + 1 – 2x. Check the solution: x = – 3. 6 ( – ) – 2 6 ( 0 ) – 2 0 – 2 = 3 ( – 3 ) + 1 – 2 ( – 3 ) = – = – Balance statement –2–2= – 2

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Ch 2 Sec 5: Slide #14 Solving Equations with Several Steps Chapter 2 Section 5 – Completed Written by John T. Wallace

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