2. EXAMPLE 1 Rewrite a formula with two variables Solve the formula C = 2πr for r. Then find the radius of a circle with a circumference of 44 inches. SOLUTION C = 2πr C 2π = r STEP 1 Solve the formula for r. STEP 2Substitute the given value into the rewritten formula. Write circumference formula. Divide each side by 2π. r = C 2π = 44 2π 7 Substitute 44 for C and simplify. The radius of the circle is about 7 inches. ANSWER

3. GUIDED PRACTICE for Example 1 Find the radius of a circle with a circumference of 25 feet. 1. The radius of the circle is about 4 feet. ANSWER

4. GUIDED PRACTICE for Example 1 The formula for the distance d between opposite vertices of a regular hexagon is d = where a is the distance between opposite sides. Solve the formula for a. Then find a when d = 10 centimeters. 2. 2a2a 3 SOLUTION d3 a = 2 35 When d = 10cm, a = or 8.7cm

5. EXAMPLE 2 Rewrite a formula with three variables SOLUTION Solve the formula for w. STEP 1 P = 2l + 2w P – 2l = 2w P – 2l 2 = w Write perimeter formula. Subtract 2l from each side. Divide each side by 2. Solve the formula P = 2l + w for w. Then find the width of a rectangle with a length of 12 meters and a perimeter of 41 meters.

6. EXAMPLE 2 Rewrite a formula with three variables 41 – 2(12) 2 w = w = 8.5 Substitute 41 for P and 12 for l. Simplify. The width of the rectangle is 8.5 meters. ANSWER Substitute the given values into the rewritten formula. STEP 2

7. GUIDED PRACTICE for Example 2 Solve the formula P = 2l + 2w for l. Then find the length of a rectangle with a width of 7 inches and a perimeter of 30 inches. 3. Length of rectangle is 8 in. ANSWER Solve the formula A = lw for w. Then find the width of a rectangle with a length of 16 meters and an area of 40 square meters. 4. Write of rectangle is 2.5 m w = A l ANSWER

8. GUIDED PRACTICE for Example 2 Solve the formula for the variable in red. Then use the given information to find the value of the variable. A = 1 2 bhbh 5. Find h if b = 12 m and A = 84 m 2. = h 2A2A b ANSWER

9. GUIDED PRACTICE for Example 2 Find the value of h if b = 12m and A = 84m 2. Find h if b = 12 m and A = 84 m 2. = h 2A2A b h = 14m ANSWER

10. GUIDED PRACTICE for Example 2 Find b if h = 3 cm Solve the formula for the variable in red. Then use the given information to find the value of the variable. A = 1 2 bhbh 6. and A = 9 cm 2. = b 2A2A h ANSWER

11. GUIDED PRACTICE for Example 2 Find b if h = 3 cm Solve the formula for the variable in red. Then use the given information to find the value of the variable. A = 1 2 bhbh 6. and A = 9 cm 2. b = 6cm ANSWER

12. GUIDED PRACTICE for Example 2 Solve the formula for the variable in red. Then use the given information to find the value of the variable. A = 1 2 7. (b 1 + b 2 )h Find h if b 1 = 6 in., b 2 = 8 in., and A = 70 in. 2 h =h = 2A2A (b 1 + b 2 ) ANSWER

13. GUIDED PRACTICE for Example 2 Solve the formula for the variable in red. Then use the given information to find the value of the variable. A = 1 2 7. (b 1 + b 2 )h Find h if b 1 = 6 in., b 2 = 8 in., and A = 70 in. 2 h = 10 in. ANSWER

14. EXAMPLE 3 Rewrite a linear equation Solve 9x – 4y = 7 for y. Then find the value of y when x = –5. SOLUTION Solve the equation for y. STEP 1 9x – 4y = 7 –4y = 7 – 9x y = 9 4 7 4 – + x Write original equation. Subtract 9x from each side. Divide each side by –4.

15. EXAMPLE 3 Rewrite a linear equation Substitute the given value into the rewritten equation. STEP 2 y = 9 4 7 4 – + (–5) y = 45 4 7 4 – – y = –13 CHECK 9x – 4y = 7 9(–5) – 4(–13) 7 = ? 7 = 7  Substitute – 5 for x. Multiply. Simplify. Write original equation. Substitute – 5 for x and – 13 for y. Solution checks.

16. EXAMPLE 4 Rewrite a nonlinear equation Solve 2y + xy = 6 for y. Then find the value of y when x = –3. SOLUTION Solve the equation for y. STEP 1 2y + x y = 6 (2+ x) y = 6 y = 6 2 + x Write original equation. Distributive property Divide each side by (2 + x).

17. EXAMPLE 4 Rewrite a nonlinear equation Substitute the given value into the rewritten equation. STEP 2 y = 6 2 + (–3) y = – 6 Substitute –3 for x. Simplify.

18. GUIDED PRACTICE for Examples 3 and 4 Solve the equation for y. Then find the value of y when x = 2. 8. y – 6x = 7 y = 7 + 6x y = 19 ANSWER 9. 5y – x = 13 y = x 5 13 5 + y = 5 ANSWER 10. 3x + 2y = 12 y = – 3x3x 2 + 6 ANSWER y = 3

19. GUIDED PRACTICE for Examples 3 and 4 Solve the equation for y. Then find the value of y when x = 2. 11. 2x + 5y = –1 12. 3 = 2xy – x 13. 4y – xy = 28 y = 14 28 4 – x y = ANSWER y = 1 4 1 3 +x 2x y = ANSWER 2x2x 5 –1 5 – y = –1 y = ANSWER

20. EXAMPLE 5 Solve a multi-step problem Write an equation that represents the store’s monthly revenue. Solve the revenue equation for the variable representing the number of new movies rented. Movie Rental A video store rents new movies for one price and older movies for a lower price, as shown at the right. The owner wants $12,000 in revenue per month. How many new movies must be rented if the number of older movies rented is 500? 1000?

21. EXAMPLE 5 Solve a multi-step problem SOLUTION Write a verbal model. Then write an equation. STEP 1 An equation is R = 5n 1 + 3n 2. Solve the equation for n 1. STEP 2

22. EXAMPLE 5 Solve a multi-step problem R = 5n 1 + 3n 2 R – 3n 2 = 5n 1 R – 3n 2 5 = n 1 Write equation. Subtract 3n 2 from each side. Divide each side by 5. Calculate n 1 for the given values of R and n 2. STEP 3 = 2100. If n 2 = 500, then n 1 12,000 – 3 500 5 = If n 2 = 1000, then n 1 = 1800. 12,000 – 3 1000 5 = If 500 older movies are rented, then 2100 new movies must be rented. If 1000 older movies are rented, then 1800 new movies must be rented. ANSWER

23. GUIDED PRACTICE for Example 5 14. What If? In Example 5, how many new movies must be rented if the number of older movies rented is 1500 ? If 1500 older movies are rented, then 1500 new movies must be rented ANSWER

24. GUIDED PRACTICE for Example 5 15. What If? In Example 5, how many new movies must be rented if customers rent no older movies at all? If 0 older movies are rented, then 2400 new movie must be rented ANSWER

25. GUIDED PRACTICE for Example 5 16. Solve the equation in Step 1 of Example 5 for n 2. R – 5n 1 3 n 2 = ANSWER