Applications of Linear Systems

1. Applications of Linear Systems Now that you know how to solve a linear system, you can use it to solve real-life problems.

2. Methods we can use… • Graphing- Use this method when after both equations are in slope-intercept form. • Substitution-Use this method when one of the variables is isolated. • Elimination-Use this method when both equations are in Standard Form

3. Selling Shoes • A store sold 28 pairs of cross-trainer shoes for a total of $2200. Style A sold for $70 per pair and Style B sold for $90 per pair. How many of each style were sold? • Keep in mind that a system has two equations. We need an equation for the quantity of shoes sold and one for the total price.

4. Given Information • Total number of shoes 28 • Total receipts 2220 • Price of Style A $70 • Price of Style B $90 • Assign variables to unknowns • Number of style A x • Number of style B y

5. Equation 1: Number of style A + Number of style B = Total number sold • x + y = 28 • Equation 2: Price A*Quantity A + Price B*Quantity B = Total Price • 70x + 90y = 2220

6. Choose a method & solve • I will use substitution… • x + y = 28 • y = 28 – x • Substitute into 2nd equation • 70x + 90*(28 – x) = 2220 • 70x + 2520-90x = 2220 • -20x = -300 • x = 15 pairs of Style A

7. continued • y = 28 – x • Substitute x = 15 to find y • y = 28 – 15 • y = 13 pairs of Style B • Solution (15 pairs of Style A, 13 pairs of Style B)

8. Mixture Problem • Your car’s manual recommends that you use at least 89-octane gasoline. Your car’s 16-gallon gas tank is almost empty. How much regular gasoline (87-octane) do you need to mix with premium gasoline (92-octane) to produce 16 gallons of 89-octane gasoline? • You need to know that an octane rating is the percent of isooctane in the gasoline, so 16 gallons of 89-octane gasoline contains 89% of 16, or 14.24, gallons of isooctane.

9. Given information • Unknowns • Volume of regular gas x • Volume of premium gas y • Volume of 89-octane 16 gallons • Isooctane in regular .87x • Isooctane in premium .92y • Isooctane in 89-octane 16*.89 = 14.24

10. Equations • Volume of regular + volume of premium = total volume x + y = 16 • Isooctane in regular + isooctane in premium = Isooctane in 89-octane. .87x + .92y = 14.24

11. Solve the system • x + y = 16 • y = 16 – x • Substitute into 2nd equation • 0.87x + 0.92*(16 – x) = 14.24 • 0.87x + 14.72 – 0.92x = 14.24 • -.05x = -0.48 • x = 9.6 gallons of 87 octane • y = 16 – 9.6 = 6.4 gallons of 92 octane

12. Making a decision • You are offered two different jobs. Job A offers an annual salary of $30,000 plus a year-end bonus of 1% of your total sales. Job B offers an annual salary of $24,000 plus a year-end bonus of 2% of your total sales. • How much would you have to sell to earn the same amount in each job? • If you believe you can sell between $500,000 and $800,000 of merchandise per year, which job should you choose?

13. Given information • If you pay attention to the wording, the problems gives you an initial amount (b) and percent of sales (m). Both equations can be written in y = mx + b form. • Job 1: y = 0.01x + 30,000 • Job 2: y = 0.02x + 24,000

14. Solve the system • Use your graphing calculator to solve the system since both equations are in slope-intercept form. • Solution (break even point) • (x = 600,000, y = 36,000) • This represents the break even point.

15. continued • If you believe you can sell between $500,000 and $800,000 of merchandise per year, which job should you choose? • If you looked at the graph of the linear system, you can see that if your sales are greater than $600,000, Job B would pay you better than Job A.

16. Assignment • Algebra book • Page 422 • Problems 31 – 45 odd, 46 - 56