6-5 Applying Systems

Warm UpWarm Up

Lesson Presentation

California Standards

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6-5 Applying Systems

Warm UpSimplify each expression.

1. 3(10a + 4) – 2

2. 5(20 – t) + 8t

3. (8m + 2n) – (5m + 3n)

30a + 10

100 + 3t

3m – n

4. y – 2x = 4x + y = 7

Solve by using any method.

(1, 6) 5. 2x – y = –1y = x + 5

(4, 9)

6-5 Applying Systems

9.0 Students solve a system of two linear equations in two variables algebraically and are able to interpret the answer graphically. Students are able to solve a system of two linear inequalities in two variables and to sketch the solution sets. 15.0 Students apply algebraic techniques to solve rate problems, work problems, and percent mixture problems.

California Standards

6-5 Applying Systems

When a kayaker paddles downstream, the river’s current helps the kayaker move faster, so the speed of the current is added to the kayaker’s speed in still water to find the total speed. When a kayaker is going upstream, the speed of the current is subtracted from the kayaker’s speed in still water.

You can use these ideas and a system of equations to solve problems about rates of speed.

6-5 Applying Systems

Additional Example 1: Solving Rate Problems

With a tailwind, an airplane makes a 900-mile trip in 2.25 hours. On the return trip, the plane flies against the wind and makes the trip in 3 hours. What is the plane’s speed? What is the wind’s speed?

Let p be the rate at which the plane flies in still air, and let w be the rate of the wind.

Use a table to set up two equations–one for against the wind and one for with the wind.

6-5 Applying Systems

rate time = distance

Remember!

6-5 Applying Systems

Rate Time = Distance

Upwind p – w 3 = 900

Downwind p + w 2.25 = 900

Solve the system3(p – w) = 900 2.25(p + w) = 900.

First write the system as3p – 3w = 900 2.25p + 2.25w = 900,

and then use elimination.

Additional Example 1 Continued

6-5 Applying SystemsAdditional Example 1 Continued

Step 1 –0.75(3p – 3w = 900) 2.25p + 2.25w = 900

Multiply each term in the first equation by –0.75 to get opposite coefficients of p.

Add the new equation to the second equation.

–2.25p + 2.25w = –675 + 2.25p + 2.25w = 900

4.5w = 225 Simplify and solve for w.w = 50

Step 2

6-5 Applying SystemsAdditional Example 1 Continued

Write one of the original equations.

Substitute 50 for w.

Step 3 3p – 3w = 900

3p – 3(50) = 900

Add 150 to both sides.3p – 150 = 900 + 150 + 150

3p = 1050

Divide both sides by 3.

p = 350

The plane’s speed is 350 mi/h and the wind’s speed is 50 mi/h.

Step 4 (350, 50)Write the solution as an

ordered pair.

6-5 Applying SystemsCheck It Out! Example 1

Ben paddles his kayak along a course on a different river. Going upstream, it takes him 6 hours to complete the course. Going downstream, it takes him 2 hours to complete the same course. What is the rate of the current and how long is the course?

Let d be the distance traveled, and let c be the rate of the current.

Use a table to set up two equations–one for the upstream trip and one for the downstream trip.

6-5 Applying Systems

Rate Time = Distance

Upstream 3 – c 6 = d

Downstream 3 + c 2 = d

Solve the system6(3 – c) = d 2(3 + c) = d.

First write the system as18 – 6c = d 6 + 2c = d,

and then use elimination.

Check It Out! Example 1 Continued

6-5 Applying Systems

Step 1 18 – 6c = d 3(6 + 2c = d)

Multiply each term in the second equation by 3 to eliminate the c term.

Add the new equation to the first equation.

18 – 6c = d + 18 + 6c = 3d

Check It Out! Example 1 Continued

36 = 4d

Simplify and solve for d.9 = dStep 2

6-5 Applying SystemsCheck It Out! Example 1 Continued

Write one of the original equations.

Substitute 9 for d.

Subtract 6 from both sides.

Divide both sides by 2.

c = 1.5

The course is 9 miles long and the current’s speed is 1.5 mi/h.

Step 4 (9, 1.5)Write the solution as an

ordered pair.

Step 3 6 + 2c = d

6 + 2c = 9

– 6 – 6

2c = 3

6 + 2c = 9

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.

6-5 Applying SystemsAdditional Example 2 Continued

20% + 40% = 25%

Solution t + f = 60

Saline 0.20t + 0.40f = 0.25(60) = 15

Solve the systemt + f = 60 0.20t + 0.40f = 15.

Use substitution.

Saline Saline Saline

6-5 Applying SystemsAdditional Example 2 Continued

Step 1 Solve the first equation for t by subtracting f from both sides.

t + f = 60– f – ft = 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

6-5 Applying SystemsAdditional Example 2 Continued

12 – 0.20f + 0.40f = 15Step 3

Simplify. Solve for f. 12 + 0.20f = 15– 12 – 12

0.20f = 3Subtract 12 from both

sides.

Divide both sides by 0.20.

f = 15

6-5 Applying SystemsAdditional 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.

6-5 Applying SystemsCheck 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 systems + t = 30 0.09s + 0.15t = 3.

Use substitution.

6-5 Applying Systems

Step 1 Solve the first equation for s by subtracting t from both sides.

s + t = 30– t – ts = 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

6-5 Applying SystemsCheck It Out! Example 2 Continued

Step 3

Simplify. Solve for t. 2.7 + 0.06t = 3– 2.7 – 2.7

0.06t = 0.3Subtract 2.7 from both

sides.

Divide both sides by 0.06.

t = 5

2.7 – 0.09t + 0.15t = 3

6-5 Applying SystemsCheck 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.

6-5 Applying SystemsAdditional 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

6-5 Applying SystemsAdditional 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.

6-5 Applying SystemsAdditional Example 3 Continued

10u + t = 10t + u + 54 Subtract u from both sides.– u – u

9u + t = 10t + 54Subtract 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.

6-5 Applying SystemsAdditional Example 3 Continued

Now solve the systemt + 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 – 8

t = 2

6-5 Applying SystemsAdditional 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.

6-5 Applying SystemsCheck 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

6-5 Applying SystemsCheck 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.

6-5 Applying SystemsCheck It Out! Example 3 Continued

10u + t = 10t + u + 9 Subtract u from both sides.– u – u

9u + t = 10t + 9Subtract 10t from both

sides.– 10t =–10t

9u – 9t = 9

Divide both sides by 9.

u – t = 1

–t + u = 1Write the left side with the

variable t first.

6-5 Applying SystemsCheck It Out! Example 3 Continued

Now solve the systemt + 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 = 9Step 3 Write one of the original

equations.Substitute 9 for u.

t + u = 17 t + 9 = 17

Subtract 9 from both sides.– 9 – 9

t = 8

6-5 Applying SystemsCheck 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.

6-5 Applying SystemsLesson Quiz: Part I

1. Allyson paddles her canoe 9 miles upstream in 4.5 hours. The return trip downstream takes her 1.5 hours. What is the rate at which Allyson paddles in still water? What is the rate of the current? 4 mi/h, 2mi/h

2. A pharmacist mixes Lotion A, which is 5% alcohol, with Lotion B, which is 10% alcohol, to make 50 mL of a new lotion that is 8% alcohol. How many milliliters of Lotions A and B go into the mixture? 20 mL of Lotion A and 30 mL of Lotion B.

6-5 Applying SystemsLesson Quiz: Part II

3. The sum of the digits of a two digit number is 13. When the digits are reversed, the new number is 9 less than the original number. What is the original number? 76