# Ch 6 Sec 3: Slide #1 Columbus State Community College Chapter 6 Section 3 More on Solving Linear Equations.

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Ch 6 Sec 3: Slide #1 Columbus State Community College Chapter 6 Section 3 More on Solving Linear Equations

Ch 6 Sec 3: Slide #2 More on Solving Linear Equations 1.Learn the four steps for solving a linear equation, and apply them. 2.Solve equations with fractions or decimals as coefficients. 3.Solve equations that have no solution or infinitely many solutions. 4.Write expressions for two related unknown quantities.

Ch 6 Sec 3: Slide #3 Solving Linear Equations Step 1Simplify each side separately. Clear parentheses using the distributive property, if needed, and combine terms. Step 2Isolate the variable term. Use the addition property if necessary so that the variable term is on one side of the equation and a number is on the other. Step 3Isolate the variable. Use the multiplication property if necessary to get the equation in the form x = a number. Step 4Check. Substitute the proposed solution into the original equation to see if a true statement results.

Ch 6 Sec 3: Slide #4 9 ( x – 5 ) = 4 ( x – 1 ) – 6 Using the Four Steps to Solve an Equation EXAMPLE 1 Using the Four Steps to Solve an Equation Solve9 ( x – 5 ) = 4 ( x – 1 ) – 6. 9x – 45 = 4x – 4 – 6 9x – 45 = 4x – 10 + 45 5x5x= 35 – 4x 5x – 45= – 10 Step 1 Simplify each side separately. Step 2 Isolate the variable term on one side.

Ch 6 Sec 3: Slide #5 Using the Four Steps to Solve an Equation EXAMPLE 1 Using the Four Steps to Solve an Equation Solve9 ( x – 5 ) = 4 ( x – 1 ) – 6. 5x5x= 35 Step 3 Isolate the variable. 5x5x35 55 = x= 7

Ch 6 Sec 3: Slide #6 Using the Four Steps to Solve an Equation EXAMPLE 1 Using the Four Steps to Solve an Equation Solve9 ( x – 5 ) = 4 ( x – 1 ) – 6.Check 9 ( x – 5 ) = 4 ( x – 1 ) – 6 Step 4 Check. 9 ( 7 – 5 ) 9 ( 2 ) 18 = 4 ( 7 – 1 ) – 6 = 4 ( 6 ) – 6 = 24 – 6 = 1818True The solution of the equation is 7.

Ch 6 Sec 3: Slide #7 31 – ( 4 – n ) = – 3 ( 2n + 5 )31 – 1( 4 – n ) = – 3 ( 2n + 5 ) Using the Four Steps to Solve an Equation EXAMPLE 2 Using the Four Steps to Solve an Equation Solve31 – ( 4 – n ) = – 3 ( 2n + 5 ). 31 – 4 + n = – 6n – 15 27 + n = – 6n – 15 – 27 7n7n= – 42 + 6n 27 + 7n= – 15 Step 1 Simplify each side separately. Step 2 Isolate the variable term on one side.

Ch 6 Sec 3: Slide #8 Using the Four Steps to Solve an Equation EXAMPLE 2 Using the Four Steps to Solve an Equation Solve31 – ( 4 – n ) = – 3 ( 2n + 5 ). 7n7n= – 42 Step 3 Isolate the variable. 7n7n– 42 77 = n= – 6

Ch 6 Sec 3: Slide #9 Using the Four Steps to Solve an Equation EXAMPLE 2 Using the Four Steps to Solve an Equation Solve31 – ( 4 – n ) = – 3 ( 2n + 5 ).Check 31 – ( 4 – n ) = – 3 ( 2n + 5 ) Step 4 Check. 31 – ( 4 – ( – 6 ) ) 31 – ( 4 + 6 ) 31 – 10 = – 3 ( 2 ( – 6 ) + 5 ) = – 3 ( – 12 + 5 ) = – 3 ( – 7) = 2121True The solution of the equation is – 6.

Ch 6 Sec 3: Slide #10 Solving an Equation with Fractions as Coefficients EXAMPLE 3 Solving an Equation with Fractions as Coefficients Solve. 1 3 v + =20– v 1 2 v 5 6 1 3 v + = – v 1 2 v 5 6 6 6 1 3 v + = – v 1 2 v 5 6 6666 2v + 5v = 120 – 3v

Ch 6 Sec 3: Slide #11 Solving an Equation with Fractions as Coefficients EXAMPLE 3 Solving an Equation with Fractions as Coefficients Solve. 1 3 v + =20– v 1 2 v 5 6 2v + 5v = 120 – 3v 10 v = 12 + 3v 10v 120 7v = 120 – 3v = =

Ch 6 Sec 3: Slide #12 Solving an Equation with Fractions as Coefficients EXAMPLE 3 Solving an Equation with Fractions as Coefficients Solve. Check 1 3 v + =20– v 1 2 v 5 6 = – ( 12 ) 1 2 1 3 + 5 6 4 + 10 1 3 v + = 20 – v 1 2 v 5 6 = 20 – 6 14= 14 True The solution of the equation is 12.

Ch 6 Sec 3: Slide #13 1.234 x10 1.234 x100 1.234 x1,000 12.34123.41,234. The decimal point moves to the right by one position for each zero that appears in each power of ten. moves decimal point 1 place moves decimal point 2 places moves decimal point 3 places Multiplying by Powers of Ten

Ch 6 Sec 3: Slide #14 Solving an Equation with Decimals as Coefficients EXAMPLE 4 Solving an Equation with Decimals as Coefficients Solve.4w –.2 ( w + 6 ) =.05w..4w –.2 ( w + 6 ) =.05w.40w –.20 ( w + 6 ) =.05w 40w – 20 ( w + 6 ) = 5w 40w – 20w – 120 = 5w 20w – 120 = 5w Move the decimal point two places to the right. 100 [.4w –.2 ( w + 6 ) ] = 100 [.05w ]Multiply by 100.

Ch 6 Sec 3: Slide #15 Solving an Equation with Decimals as Coefficients EXAMPLE 4 Solving an Equation with Decimals as Coefficients Solve.4w –.2 ( w + 6 ) =.05w. 20w – 120 = 5w – 20w – 120 = – 15w – 15 = – 120 – 15w 8 = w

Ch 6 Sec 3: Slide #16 Solving an Equation with Decimals as Coefficients EXAMPLE 4 Solving an Equation with Decimals as Coefficients Solve.4w –.2 ( w + 6 ) =.05w. Check.4 ( 8 ) –.2 ( 8 + 6 ).4w –.2 ( w + 6 ) =.05w 3.2 –.2 ( 14 ) 3.2 – 2.8.4 =.05 ( 8 ) =.4 True The solution of the equation is 8.

Ch 6 Sec 3: Slide #17 Solving an Equation with Infinitely Many Solutions EXAMPLE 5 Solving an Equation – Infinitely Many Solutions Solve9 – 2x = – ( x – 9 ) – x. 9 – 2x = – ( x – 9 ) – x 9 – 2x = – 1 ( x – 9 ) – x 9 – 2x = – 1x + 9 – x 9 – 2x = – 2x + 9 9 = 9 + 2x True  All real numbers Note: When the variable terms cancel out and a true statement remains, we say the solution contains “all real numbers.”

Ch 6 Sec 3: Slide #18 Understanding a Solution When you are solving an equation like the one in the previous example, do not write “9” as the solution. While 9 is a solution, there are infinitely many other solutions. CAUTION

Ch 6 Sec 3: Slide #19 Solving an Equation That Has No Solution EXAMPLE 6 Solving an Equation That Has No Solution Solve4n – 3 = 2 ( n + 4 ) + 2n. 4n – 3 = 2 ( n + 4 ) + 2n 4n – 3 = 2n + 8 + 2n 4n – 3 = 4n + 8 – 3 = 8 – 4n False  No Solution Note: When the variable terms cancel out and a false statement remains, we say the equation has “no solution.”

Ch 6 Sec 3: Slide #20 Translating a Phrase into an Algebraic Expression EXAMPLE 7 Translating a Phrase into an Algebraic Expression Two numbers have a sum of 38. If one of the numbers is represented by x, write an expression for the other number. To help understand the process, let’s suppose one of the numbers is 10. Since the sum was given to be 38, the other number would have to be 28. We can find this number by subtracting 38 – 10 = 28. We could use other numbers, instead of 10, and would find the process for finding the other number would be the same: 38 – the other number. Therefore, since we let x represent one of the numbers, the other number would be represented by 38 – x.

Ch 6 Sec 3: Slide #21 Checking Expressions Since the sum of the two numbers in the previous example is 38, the expression for the other number must be 38 – x, not x – 38. To check, find the sum of the two numbers: x + ( 38 – x ) = 38,as required. CAUTION

Ch 6 Sec 3: Slide #22 More on Solving Linear Equations Chapter 6 Section 3 – Completed Written by John T. Wallace

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