1. 6-5 Applying Systems Additional Example 2: Solving Mixture Problems A chemist mixes a 20% saline solution and a 40% saline solution to get 60 milliliters of a 25% saline solution. How many milliliters of each saline solution should the chemist use in the mixture? Let t be the milliliters of 20% saline solution and f be the milliliters of 40% saline solution. Use a table to set up two equations–one for the amount of solution and one for the amount of saline.

2. 6-5 Applying Systems Additional Example 2 Continued 20% + 40% = 25% Solution t +f = 60 Saline 0.20t + 0.40f= 0.25(60) = 15 Solve the system t + f = 60 0.20t + 0.40f = 15. Use substitution. Saline

3. 6-5 Applying Systems Additional Example 2 Continued Step 1 Solve the first equation for t by subtracting f from both sides. t + f = 60 – f t = 60 – f Step 2 Substitute 60 – f for t in the second equation. 0.20t + 0.40f = 15 0.20(60 – f) + 0.40f = 15 Distribute 0.20 to the expression in parentheses. 0.20(60) – 0.20f + 0.40f = 15

4. 6-5 Applying Systems Additional Example 2 Continued 12 – 0.20f + 0.40f = 15Step 3 Simplify. Solve for f. 12 + 0.20f = 15 – 12 0.20f = 3 Subtract 12 from both sides. Divide both sides by 0.20. f = 15

5. 6-5 Applying Systems Additional Example 2 Continued Step 4 Write one of the original equations. t + f = 60 Substitute 15 for f. t + 15 = 60 Subtract 15 from both sides. –15 –15 t = 45 Step 5 Write the solution as an ordered pair. (15, 45) The chemist should use 15 milliliters of the 40% saline solution and 45 milliliters of the 20% saline solution.

6. 6-5 Applying Systems Check It Out! Example 2 Suppose a pharmacist wants to get 30 g of an ointment that is 10% zinc oxide by mixing an ointment that is 9% zinc oxide with an ointment that is 15% zinc oxide. How many grams of each ointment should the pharmacist mix together? 9% Ointment + 15% Ointment = 10% Ointment Ointment (g) s +t = 30 Zinc Oxide (g) 0.09s+ 0.15t= 0.10(30) = 3 Solve the system s + t = 30 0.09s + 0.15t = 3. Use substitution.

7. 6-5 Applying Systems Step 1 Solve the first equation for s by subtracting t from both sides. s + t = 30 – t s = 30 – t Check It Out! Example 2 Continued Step 2 Substitute 30 – t for s in the second equation. 0.09(30 – t)+ 0.15t = 3 Distribute 0.09 to the expression in parentheses. 0.09(30) – 0.09t + 0.15t = 3 0.09s + 0.15t = 3

8. 6-5 Applying Systems Check It Out! Example 2 Continued Step 3 Simplify. Solve for t. 2.7 + 0.06t = 3 – 2.7 0.06t = 0.3 Subtract 2.7 from both sides. Divide both sides by 0.06. t = 5 2.7 – 0.09t + 0.15t = 3

9. 6-5 Applying Systems Check It Out! Example 2 Continued Step 4 Write one of the original equations. s + t = 30 Substitute 5 for t. s + 5 = 30 Subtract 5 from both sides. –5 –5 s = 25 Step 5 Write the solution as an ordered pair. (25, 5) The pharmacist should use 5 grams of the 15% ointment and 25 grams of the 9% ointment.

10. 6-5 Applying Systems Additional Example 3: Solving Number-Digit Problems The sum of the digits of a two-digit number is 10. When the digits are reversed, the new number is 54 more than the original number. What is the original number? Let t represent the tens digit of the original number and let u represent the units digit. Write the original number and the new number in expanded form. Original number: 10t + u New number: 10u + t

11. 6-5 Applying Systems Additional Example 3 Continued Now set up two equations. The sum of the digits in the original number is 10. First equation: t + u = 10 The new number is 54 more than the original number. Second equation: 10u + t = (10t + u) + 54 Simplify the second equation, so that the variables are only on the left side.

12. 6-5 Applying Systems Additional Example 3 Continued 10u + t = 10t + u + 54 Subtract u from both sides. – u 9u + t = 10t + 54 Subtract 10t from both sides. – 10t =–10t 9u – 9t = 54 Divide both sides by 9. u – t = 6 –t + u = 6 Write the left side with the variable t first.

13. 6-5 Applying Systems Additional Example 3 Continued Now solve the system t + u = 10 –t + u = 6. Use elimination. Step 1 –t + u = + 6 t + u = 10 2u = 16 Add the equations to eliminate the t term. Step 2 Divide both sides by 2. u = 8 Step 3 Write one of the original equations. Substitute 8 for u. t + u = 10 t + 8 = 10 Subtract 8 from both sides. – 8 t = 2

14. 6-5 Applying Systems Additional Example 3 Continued Step 4 Write the solution as an ordered pair. (2, 8) The original number is 28. Check Check the solution using the original problem. The sum of the digits is 2 + 8 = 10. When the digits are reversed, the new number is 82 and 82 – 54 = 28.

15. 6-5 Applying Systems Check It Out! Example 3 The sum of the digits of a two-digit number is 17. When the digits are reversed, the new number is 9 more than the original number. What is the original number? Let t represent the tens digit of the original number and let u represent the units digit. Write the original number and the new number in expanded form. Original number: 10t + u New number: 10u + t

16. 6-5 Applying Systems Check It Out! Example 3 Continued Now set up two equations. The sum of the digits in the original number is 17. First equation: t + u = 17 The new number is 9 more than the original number. Second equation: 10u + t = (10t + u) + 9 Simplify the second equation, so that the variables are only on the left side.

17. 6-5 Applying Systems Check It Out! Example 3 Continued 10u + t = 10t + u + 9 Subtract u from both sides. – u 9u + t = 10t + 9 Subtract 10t from both sides. – 10t =–10t 9u – 9t = 9 Divide both sides by 9. u – t = 1 –t + u = 1 Write the left side with the variable t first.

18. 6-5 Applying Systems Check It Out! Example 3 Continued Now solve the system t + u = 17 –t + u = 1. Use elimination. Step 1 –t + u = + 1 t + u = 17 2u = 18 Add the equations to eliminate the t term. Step 2 Divide both sides by 2. u = 9 Step 3 Write one of the original equations. Substitute 9 for u. t + u = 17 t + 9 = 17 Subtract 9 from both sides. – 9 t = 8

19. 6-5 Applying Systems Check It Out! Example 3 Continued Step 4 Write the solution as an ordered pair. (9, 8) The original number is 98. Check Check the solution using the original problem. The sum of the digits is 9 + 8 = 17. When the digits are reversed, the new number is 89 and 89 + 9 = 98.