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C h a p t e r 3

C h a p t e r 3. Solving Linear Equations. By Kristen Sanchez and Emma Nuss. INVERSE OPERATIONS: Two operations that undo each other, such as addition and subtraction are called inverse operations. These can help you isolate the variable on one side of the equation.

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C h a p t e r 3

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  1. Chapter3 Solving Linear Equations By Kristen Sanchez and Emma Nuss

  2. INVERSE OPERATIONS: • Two operations that undo each other, such as addition and subtraction are called inverse operations. These can help you isolate the variable on one side of the equation. • To solve an equation with a fractional coefficient, such as 10 = 2/3m, multiply each side of the equation by the reciprocal of the fraction (in this case 3/2). The product of a nonzero number and its reciprocal is 1.

  3. Summary: Properties of Equality • Addition Property of Equality: If a = b, then a + c = b + c • Subtraction Property of Equality: If a = b, then a – c = b – c • Multiplication Property of Equality: If a = b, then ca = cb • Division Property of Equality: If a = b and c does not = 0, then a/c = b/c

  4. 3.1: Solving Equations Using Addition and Subtraction

  5. 3.2: Solving Equations Using Multiplication and Division

  6. Solving Multi-Step Equations • When solving a multi-step equation, the first step is isolate the variable, then solve for x. Example: 3x+7=-8 -7 -7 3x=-15 x=-5

  7. Solving Equations with Variables on Both Sides • Some equations have variables on both sides. To solve these, you collect the variable terms on one side of the equation. • Example: 7x+19=-2x+55 +2x +2x 9x+19=55 -19 -19 9x=36 9 9 x=4

  8. Tips on Solving Linear Equations • Simplify: combine like terms or distribution • Collect: put the variable terms on the side with the larger coefficient • Inverse operations: use inverse operations to isolate the variable • Check: check your solution with the original equation

  9. 3.6: Solving Decimal Equations • If you have a long decimal answer to an equation, it is often more practical to round to an approximate answer • Example: -38x-39=118 +39 +39 -38x=157 x=157/-38 x≈-4.131578947 x≈-4.13 • The rounded answer may not be exactly equal to the original equation, but the symbol ≈ is used to show that the answer approximately balances the two sides of the equation.

  10. 3.7: Formulas • Formulas are algebraic equations that relate two or more quantities. • Formulas can be used to solve problems with variables other than x and y. • Example: Celsius and Fahrenheit are related by the equation C=5/9(F-32) where C is Celsius and F is Fahrenheit. Solve the formula for degrees Fahrenheit F. C=5/9(F-32) *9/5 *9/5 9/5C=F-32 +32 +32 9/5C+32=F

  11. 3.8: Ratios and Rates • The ratio of a to b is a/b. • If a and b are measured in different units, then a/b is called the rate of a per b. • Rates are often expressed as unit rates, or the rate per one given unit (ex. 60 miles per 1 gallon)

  12. Unit Analysis • Writing the units when comparing each quantity of a rate is called unit analysis. • You can multiply and divide units just as you can multiply and divide numbers. • Example: 60 mins= 1 hour convert 3 hours to minutes 3 hours x 60 mins/1 hour= 180 mins

  13. 3.9: Percents • A percent is a ratio that compares a number to 100. For example, forty percent can be written as 40/100, 0.40, or 40%. • Number being compared to base= a percent= p/100 base number = b a = p/100 * b • Example: Fourteen dollars is 25% of what amount of money? Percent = 25/100 =1/4 14 = ¼ * x 4(14) = 4(1/4) * x 56 = x

  14. Using Percents to find Discounts • Discount is the difference between the regular price of and item and its sale price. To find the discount percent, use the regular price as the base number in the percent equation. • Example A portable CD player has a regular price of $90 and a sale price of $72. What is the discount percent? Discount=regular price-sale price =90-72=18 Percent=p/100 Regular price=90 18=p/100 * (90) 18/90=p/100 .20=p/100 20=p

  15. Summary of Percents a = p/100 * b

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