1. Holt CA Course 1 4-6 Solving Equations Containing Fractions AF1.1 Write and solve one-step linear equations in one variable. Also covered: NS2.1, NS2.2, NS2.4 California Standards

2. Holt CA Course 1 4-6 Solving Equations Containing Fractions The goal when solving equations that contain fractions is the same as when working with other kinds of numbers—to get the variable by itself on one side of the equation.

3. Holt CA Course 1 4-6 Solving Equations Containing Fractions Solve. Write the answer in simplest form. Teacher Example 1A: Solving Equations by Adding or Subtracting x – 3737 = 5757 3737 = 5757 3737 + 3737 = 5757 + 3737 x= 8787 = 1 1717 Simplify. 3737 Since is subtracted from x, 3737 add to both sides.

4. Holt CA Course 1 4-6 Solving Equations Containing Fractions Solve. Write the answer in simplest form. Teacher Example 1B: Solving Equations by Adding or Subtracting 5 12 +t = 3838 3838 + t = 5 12 Find a common denominator. Subtract. 3838 + t –= 5 12 – 3838 3838 t = 10 24 – 9 24 t = 1 24 Since is added to t, 3838 3838 subtract from both sides.

5. Holt CA Course 1 4-6 Solving Equations Containing Fractions Solve. Write the answer in simplest form. Teacher Example 1C: Solving Equations by Adding or Subtracting 4949 + r = 1212 4949 + r = 1212 4949 + r – 4949 = 1212 – 4949 Find a common denominator. Subtract. r = 1 18 – r = 9 18 8 18 Since is added to r, 4949 4949 subtract from both sides.

6. Holt CA Course 1 4-6 Solving Equations Containing Fractions Check. Teacher Example 1C Continued: 4949 + r = 1212 = 9 18 9 18 Find a common denominator. Add. You can check that a value is a solution to an equation by substituting the value for the variable. Helpful Hint Substitute for r. 1 18 4949 + 1 18 1212 = ? 8 18 + 1 18 9 18 = ?

7. Holt CA Course 1 4-6 Solving Equations Containing Fractions Solve. Write the answer in simplest form. Student Practice 1A: x – 3838 = 7878 3838 = 7878 3838 + 3838 = 7878 + 3838 x = 10 8 = 1 1414 Simplify. Since is subtracted from x, add to both sides. 3838 3838

8. Holt CA Course 1 4-6 Solving Equations Containing Fractions Solve. Write the answer in simplest form. Student Practice 1B: 1414 + y = 3838 1414 + y = 3838 1414 3838 1414 + y – = – 1414 3838 y = – 2828 y = 1818 Find a common denominator. Subtract. Since is added to y, subtract from both sides. 1414 1414

9. Holt CA Course 1 4-6 Solving Equations Containing Fractions Check. Student Practice 1B: 1414 + y = 3838 = 3838 3838 Find a common denominator. Add. Substitute for y. 1818 1414 + 1818 3838 = ? 2828 + 1818 3838 = ?

10. Holt CA Course 1 4-6 Solving Equations Containing Fractions Solve. Write the answer in simplest form. Student Practice 1C: 3 14 +t = 2727 3 14 + t = 2727 Find a common denominator. Subtract. 3 14 + t – = 2727 – 3 14 3 14 t = – 3 14 4 14 t = 1 14 Since is added to t, 3 14 subtract from both sides. 3 14

11. Holt CA Course 1 4-6 Solving Equations Containing Fractions Solve. Write the answer in simplest form. Teacher Example 2A: Solving Equations by Multiplying 3838 x = 1414 3838 x=  8383 1414  8383 2 1 =x 2323 Multiply by the reciprocal of. 3838 Then simplify. 3838 = 1414 x = To undo multiplying by Caution! 3838, you can divide by 3838 or multiplyby its reciprocal, 8383.

12. Holt CA Course 1 4-6 Solving Equations Containing Fractions Teacher Example 2B: Solving Equations by Multiplying 4y = 8989 8989 4y4y =  1414 8989  1414 1 2 y = 2929 4y = Multiply by the reciprocal of 4. Then simplify. Solve. Write the answer in simplest form.

13. Holt CA Course 1 4-6 Solving Equations Containing Fractions Solve. Write the answer in simplest form. Student Practice 2A: 3434 x = 1212 3434 x=. 4343 1212.4343 2 1 =x 2323 Multiply by the reciprocal of. 3434 Then simplify. 3434 = 1212 x =

14. Holt CA Course 1 4-6 Solving Equations Containing Fractions Student Practice 2B: 3y = 6767 6767 y=  1313 6767  1313 1 2 y = 2727 3y = 3 Multiply by the reciprocal of 3. Then simplify. Solve. Write the answer in simplest form.

15. Holt CA Course 1 4-6 Solving Equations Containing Fractions Teacher Example 3: Physical Science Application The amount of copper in brass is of the total weight. If a sample contains 4 ounces of copper, what is the total weight of the sample? 3434 1515 Let w represent the total weight of the sample. 3434 w = 4 1515 3434 w  4343 = 4 1515  4343 w = 21 5  4343 7 1 w = 28 5 or 5 3535 Write an equation. Multiply by the reciprocal of 3434 · Write 4 1515 as an improper fraction. Then simplify. The sample weighs 5 3535 ounces.

16. Holt CA Course 1 4-6 Solving Equations Containing Fractions The amount of copper in zinc is of the total weight. If a sample contains 5 ounces of copper, what is the total weight of the sample? Student Practice 3: 1414 1313 Let w represent the total weight of the sample. 1414 w = 5 1313 1414 w  4141 = 5 1313  4141 w = 16 3  4141 w = 64 3 or 21 1313 Write an equation. Multiply by the reciprocal of 1414 · Write 5 1313 as an improper fraction. Multiply. The sample weighs 21 1313 ounces.