Section 1.2 Applications of Linear Equations in One Variable 63 Answer 6. 30 oz of the 30% solution is needed. The information can also be organized in a table. Final Solution: 40% Antifreeze 10% Antifreeze 35% Antifreeze Number of liters of solution x 4 (4 x) Number of liters of pure antifreeze 0.40x 0.10(4) 0.35(4 x) Notice that an algebraic equation is obtained from the second row of the table relating the number of liters of pure antifreeze in each container. 0.40x 0.10142 0.3514 x2 0.40x 0.10142 0.351x 42 Mathematical equation 0.4x 0.4 0.35x 1.4 Apply the distributive property. 0.4x 0.35x 0.4 0.35x 0.35x 1.4 Subtract 0.35x from both sides. 0.05x 0.4 1.4 0.05x 0.4 0.4 1.4 0.4 Subtract 0.4 from both sides. 0.05x 1.0 Divide both sides by 0.05. x 20 Therefore, 20 L of a 40% antifreeze solution is needed. Skill Practice 6. Find the number of ounces (oz) of 30% alcohol solution that must be mixed with 10 oz of a 70% solution to obtain a solution that is 40% alcohol. 6. Applications Involving Distance, Rate, and Time The fundamental relationship among the variables distance, rate, and time is given by Distance 1rate21time2 or d rt For example, a motorist traveling 65 mph (miles per hour) for 3 hr (hours) will travel a distance of d a65 mi hr b 13 hr2 195 mi 0.05x 0.05 1.0 0.05 aPure antifreeze from solution 1 b apure antifreeze from solution 2 b a pure antifreeze in the final solution b
miller_intermediate_algebra_4e_ch1_3
To see the actual publication please follow the link above