problem solving: form words to symblos
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PROBLEM SOLVING: FORM WORDS TO SYMBLOS. LESSON1.1. Keywords that indicate operations used in writing algebraic expressions. Keywords that indicate operations used in writing algebraic expressions. How to represent a word expression by algebraic symbols. - PowerPoint PPT PresentationTRANSCRIPT
PROBLEM SOLVING:
FORM WORDS TO SYMBLOS
LESSON1.1
Keywords that indicate operations used in writing algebraic
expressions.
Operation keywordsADDITION Sum, total, plus,
increased by, more than
SUBTRACTION
Difference, exceeds, less than, less, minus, decreased by, diminished by, subtracted from
Keywords that indicate operations used in writing algebraic
expressions.
Operation keywords
MULTIPLICATION
product, multiplied by, times, twice, doubled, tripled
DIVISION Quotient, divided by
How to represent a word expression by algebraic symbols.
First choose a variable to represent the unknowns.
Identify the key word(s) that indicate the operation.
Use algebraic symbols to represent the word expression.
Example 1. Write an algebraic expression for each word expression.
P29.50 less than the cost of a coat
Solution: Let x = the cost of the coat. P29.50 less than the cost : x – 29.50
Example 2. Write an algebraic expression for each word expression.
Five more than three times the number of pages
Solution: Let p = the number of pages. Five more than three times the
number :
3p +5
Many word problems involve more than one unknown .
In these problems, one condition (CONDITION1) tells how the unknowns related.
You can use this condition to represent the unknowns.
A second condition (CONDITION 2) tells what quantities are equal.
You can use this condition to write an equation for the problem.
Example 3.
In a recent year, the cost of electricity used for lighting a house was P9.00 less than 6 times the cost of operating a color tv set (CONDITION1). The total cost for these two items was P834.50 (CONDITION 2). Find the cost for lighting.
Solution
Use CONDITION1 to represent the unknowns.
Let c = the cost of operating a color tv set .
Then 6c - 9= the cost for lighting.
Solution Use CONDITION 2 to write an equation. c + 6c - 9= 834.50 total cost Solve the equation: c + 6c - 9= 834.50 7c -9 = 834.50 7c = 843.50 c = 120.50 6c – 9 = 614.00
Check: CONDITION 1.
Is the cost for lighting P9.00 less than 6 times the cost for the color tv?
614.00 =6(P120 – P9.00?
Yes.
Solution Use CONDITION 2 to write an equation. c + 6c - 9= 834.50 total cost Solve the equation: c + 6c - 9= 834.50 7c -9 = 834.50 7c = 843.50 c = 120.50 6c – 9 = 614.00
Check:CONDITION 2. Is the total cost P834.50?
P614.00 + P120 = P834.50?
Yes.
STEPS IN SOLVING WORD PROBLEMS.
Identify CONDITION1.Use this condition to represent the unknowns.
Identify CONDITION2. Use this condition to write an equation for the problem.
Solve the equation. Check in the original conditions of
the problem. Answer the question.