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INTERPRETING EQUATIONS IN WORD PROBLEMS

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Presentation on theme: "INTERPRETING EQUATIONS IN WORD PROBLEMS"— Presentation transcript:

1 INTERPRETING EQUATIONS IN WORD PROBLEMS
EOC TEST STRATEGY- DAY 12 INTERPRETING EQUATIONS IN WORD PROBLEMS

2 TEST STRATEGIES FOR LINEAR RATE OF CHANGE
1). Recognize key terms & formats for linear rate of change. 2). Identify whether a problem will use linear rate of change. 3). Read the table to find the linear rate of change. 4). Read the table to find the linear rate of change w/out incremental change in x. 5). Read the table to find the linear rate of change between specified pts. 6). Interpret rate of change in context 7). Extend table to find equation 8). Use calculator to find equation. 9). Interpret the constant & coefficient. 10). Finding specific range value for a given domain value & vise versa.

3 TEST STRATEGY: IDENTIFY THE VARIABLES
Identify what the x & y intercepts represent. Time (days, weeks, minutes, etc.) is often “x” We will use this information to solve problems several different ways.

4 EXAMPLE #12.1 part one of three
A vintage car’s value is $45,000 when Bubba first purchased it in 1980 and the value appreciated at a constant rate of $200 every year.  What was the value in 1990? USE EQUATION from DAY 10: y = 200x WHAT IS X? ______________ WHAT IS Y? ______________ What should be substituted in the equation?_____

5 EXAMPLE #12.1 part two of three
A vintage car’s value is $45,000 when Bubba first purchased it in 1980 and the value appreciated at a constant rate of $200 every year.  What was the value in 1990? MAKE A TABLE: WHAT IS X? WHAT IS Y? x = ________ y = ________ What is the rate of change?_____ X (years) Y (value) 1980 45000 1981 45200 1982 45400 1983 …keep adding 1 …keep adding 200 1990

6 EXAMPLE #12.1 part three of three
A vintage car’s value is $45,000 when Bubba first purchased it in 1980 and the value appreciated at a constant rate of $200 every year.  What was the value in 1990? Math reasoning = 10 years 10 years times $200 = $2000 $45, ,000 = $47,000

7 EXAMPLE#12.2 part one of three
Big Betty wants to get a tattoo like her boyfriend Bubba. At “INK THIS” tattoo shop, the artist charges $45 to draw the design and $12 per 30 minutes to tattoo it. What is the maximum amount of time Big Betty can afford if she only has $110 to spend? EQUATION METHOD: X = minutes y = total cost y = x 110 = x 65 = x minutes = x HOW MANY HOURS?

8 EXAMPLE #12.2 part two of three
Big Betty wants to get a tattoo like her boyfriend Bubba. At “INK THIS” tattoo shop, the artist charges $45 to draw the design and $12 per 30 minutes to tattoo it. What is the maximum amount of time Big Betty can afford if she only has $110 to spend? TABLE METHOD: X = hours y = total cost Hours Total cost 45 0.5 57 1 69 1.5 81 … keep adding ½ hour …keep adding $12 until we go over $110

9 CHECK IT #12.2 part three of three
Big Betty wants to get a tattoo like her boyfriend Bubba. At “INK THIS” tattoo shop, the artist charges $45 to draw the design and $12 per 30 minutes to tattoo it. What is the maximum amount of time Big Betty can afford if she only has $110 to spend? Math reasoning $110 – 45 = $65 65 ÷ 12 = 5.4 sessions 5 times 12 = $60 6 times 12 = $72 So she can afford 5 sessions and each one is 30 minutes = 2.5 hours.

10 You Try: (wkst #12) *Identify the variables. *Choose equation, table or other method to solve.

11 Check it # A computer consulting company charges an hourly rate of $15 plus a $50 service fee. How long can a customer afford to allow the consultant to work if they only have a budget of $100?  

12 CHECK IT #12.4 The price for a gallon of milk was $0.78 in and has increased linearly by an average of $.20 yearly. What is the predicted price of milk in 2015?

13 CHECK IT #12.5 Big Betty wants to have her house painted. The painters charge an initial fee of $35 to come out and give an estimate and then $20 each hour they paint. How much should she expect to pay if the job will take 24 hours?


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