# Al13 EO3 TSW6 Linear Eq. - PowerPoint PPT Presentation

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Al13 EO3 TSW6 Linear Eq. Ye Rin. Shelby. TSW 6. Linear Equation, Solving!!. Using a Table as a Check. Long. 400 students take Spanish and has been increasing 20 students per year. 98 students take German and has been decreasing 5 students per year.

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Al13 EO3 TSW6 Linear Eq.

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## Al13 EO3 TSW6 Linear Eq.

Ye Rin

Shelby

TSW 6.LinearEquation,

Solving!!

### Using a Table as a Check.

Long

• 400 students take Spanish and has been increasing 20 students per year.

• 98 students take German and has been decreasing 5 students per year.

• When will there be five times as many students taking Spanish as taking German?

Making it

Shorter

### verbal + label

taking Spanish (400)

+

Increase (20)

x

Years (x)

=

Number taking German (98)

Decrease (5)

Years (x)

5 (

-

x

)

Algebraic

400+20x=5(98-5x)

400+20x=490-25x

400+20x+25x=490-25x+25x

400+45x=490

400-400+45x490-400

45x=90

X=2

Students taking Spanish will be five times than students taking German in two years!!

Check Table

400+20=420

equal

98*5=490

Using a Graph as a Check

• A gazelle can run 73 ft per sec. For several minutes

• A cheetah can run faster(88 ft per sec.) but can only sustain its top speed for about 20 sec.

• How Far away from the cheetah does the gazelle need to stay for it to be safe?

### Solution

Use a diagram.

• If the gazelle is to far away for the cheetah to catch it within 20 seconds, the gazelle is probably safe.

X

Gazelle caught in 20 seconds

If the gazelle is closer to the cheetah than x feet, the cheetah can catch the gazelle

If the gazelle is farther away than x feet, the gazelle is in the safety zone

### …

• To find the “Safety zone,” first find the starting distance for which the cheetah reaches the gazelle in 20 seconds.

• To find the distance each animal runs in 20 seconds use the formula d=rt

Distance Gazelle runs in 20 sec.

Gazelle’s starting distance from cheetah

Distance cheetah runs in 2o seconds

Verbal Model

+

=

Gazelle’s distance = 73 x 2o (feet)

Gazelle’s starting distance from cheetah = x (feet)

Cheetah’s distance = 88 x 20 (feet)

73 x 20 + x = 88 x 20

Labels

Algebraic Model

### …

• You find that x= 300, so the gazelle needs to stay more than 300 ft away from the cheetah to be safe.

Why?

• A graph helps you see relationships

How?

• First, make a table

• Then, plot the points

• Next, find and mark the points for every 5 sec.

Thank you for watching!